Statistics |

Dose-Response Curves and Biomarker Diagnostic Power

Tue, 27 Aug 2019 11:12:32 -0700

The other day, one of my friends and colleagues (I’ll refer to him as “Dr. A”) asked me if I knew anything about assessing biomarker diagnostic power. He went on to describe his clinical problem, which I’ll try to recount here (but will likely mess up some of the relevant detail – his research pertains to generating induced pluripotent cardiac stem cells, which I have little to no experience with):

“So chemotherapy is meant to combat cancer. But some clinicians, and their patients, have found that some forms of chemotherapy and anti-cancer drugs later result in problems with the heart and vasculature – these problems are collectively referred to as ‘cardiotoxicity’.

We’re interested in developing biomarkers that will help us identify which patients might be susceptible to cardiotoxicity, and in assessing the predictive power of these biomarkers. Can you help me?”

What follows will be my exploration into what Dr. A called dose-response curves, and my approach on how to use these curves to assess biomarker diagnostic power. I’ll do a walkthrough of some Python code that I’ve written up, where I’ll examine dose-response curves and their diagnostic power using the receiver operating characteristic ( ROC) and the closely related area under the curve ( AUC) metrics.

If you want to see the actual dose-response curve analysis, skip to the Analyzing Synthetic Dose-Response Curves section down below.

Brief Introduction to the ROC and AUC

I first wanted to describe to Dr. A how to use the ROC for biomarkers, so I made a brief Python tutorial for him to look at. Below begins a more in-depth description of what I sent him.

We start by importing some necessary libraries:

%matplotlib inline
import matplotlib.pyplot as plt
from matplotlib.lines import Line2D

import numpy as np
from sklearn.metrics import auc

Next, we define a function to compute the ROC curve. While scikit-learn has a function to compute the ROC, computing it yourself makes it easier to understand what the ROC curve actually represents.

The ROC curve is essentially a curve that plots the true positive and false positive classification rates against one another, for various user-defined thresholds of the data. We pick a biomarker level threshold – let’s say $T$ – assign each sample a value of 0 or 1 depending on whether its measured biomarker level is greater than or less than $T$, and then compare these assignments to the true Case/Control classifications to compute the true and false positive rates. We plot these rates for many $T$ to generate the ROC curve.

The ROC lets you look at how sensitive your model is to various parameterizations – are you able to accurately identify Cases from Controls? How acceptable are misclassified results? In this situation, we don’t want to classify someone as healthy when in fact they might develop cardiotoxicity, so we want to allow some flexibility in terms of the number of false positives generated by our model – it’s a classic case of “better safe than sorry”, since the “sorry” outcome might be an accidental patient death.

def roc(case, control, npoints, gte=True):

 """
 Compute ROC curve for given set of Case/Control samples.

 Parameters:
 - - - - -
 case: float, array
 Samples from case patients
 control: float, array
 Samples from control patients
 npoints: int
 Number of TP/FP pairs to generate
 gte: boolean
 Whether control mean expected to be greater
 than case mean.

 Returns:
 - - - -
 specificity: float, array
 false positive rates
 sensitivity: float, array
 true positive rates
 """

 # make sure generating more
 # than 1 TP/FP pair
 # so we can plot an actual ROC curve
 assert npoints > 1


 # made sure case and control samples
 # are numpy arrays
 case = np.asarray(case)
 control = np.asarray(control)

 # check for NaN values
 # keep only indices without NaN
 case_nans = np.isnan(case)
 cont_nans = np.isnan(control)
 nans = (case_nans + cont_nans)

 specificity = []
 sensitivity = []

 # we'll define the min and max thresholds 
 # based on the min and max of our data
 conc = np.concatenate([case[~nans],control[~nans]])

 # function comparison map
 # use ```gte``` parameter
 # if we expect controls to be less than
 # cases, gte = False
 # otherwise gte = True
 comp_map = {'False': np.less,
 'True': np.greater}

 # generate npoints equally spaced threshold values
 # compute the false positive / true positive rates 
 # at each threshold
 for thresh in np.linspace(conc.min(), conc.max(), npoints):

 fp = (comp_map[gte](case[~nans], thresh)).mean()
 tn = 1-fp

 tp = (comp_map[gte](control[~nans], thresh)).mean()
 fn = 1-tp

 specificity.append(tn)
 sensitivity.append(tp)

 return [np.asarray(specificity), np.asarray(sensitivity)]

Next up, I generate 5 different datasets. Each dataset corresponds to fake samples from a “Control” distribution, and a “Case” distribution – each is distributed according to a univariate Normal distribution. The Control distribution remains the same in each scenario: $N(\mu = 10, \sigma = 1)$, but I change the $\mu_{Case}$ parameter of the Case distribution in each instance, such that $\mu_{Case} \in [5,6,7,8,9,10]$.

We plot the example datasets as follows, and then compute the ROC curves for each dataset.

# define unique means
m1 = np.arange(5,11)
samples = {}.fromkeys(m1)

# define control distribution (this stays the same across 
n2 = np.random.normal(loc=10, scale=1, size=2000)

fig, (ax) = plt.subplots(2, 3, figsize=(15, 6))

for i, ax in enumerate(fig.axes):

 n1 = np.random.normal(loc=m1[i], scale=1, size=2000)
 samples[m1[i]] = n1

 ax.hist(n1, 25, alpha=0.5, label='Case', density=True,)
 ax.hist(n2, 25, alpha=0.5, label='Control', density=True,)
 ax.set_title('Mean: {:}, Sigma: 1'.format(m1[i]), fontsize=15)
 ax.set_xlabel('Biomarker Measures Variable', fontsize=15)
 ax.set_ylabel('Density', fontsize=15)
 ax.legend()

plt.tight_layout()

Example Case/Control datasets.

fig, (ax1) = plt.subplots(1, 1)
for mean, case_data in samples.items():

 [spec, sens] = roc(case_data, n2, 100, gte=True)
 A = auc(spec, sens)

 L = 'Mean: %i, AUC: %.3f' % (mean, A)
 plt.plot(1-np.asarray(spec), np.asarray(sens), label=L)
 plt.legend(bbox_to_anchor=(1.04,1), fontsize=15)
 plt.xlabel('1-Specificity', fontsize=15)
 plt.ylabel('Sensitivity', fontsize=15)

 plt.title('ROC for Varying Case Distributions', fontsize=15)

ROC curves for example datasets.

We see that, as the distributions become less separated, the ability to distinguish points from either distribution is diminished. This is shown by 1) a flattening of the ROC curve towards the diagonal, along with 2) the integration of the ROC curve, which generates the AUC metric. When the distributions are far apart (as when the $\mu_{Case} = 5$), it is quite easy for a simple model to distinguish points sampled from either distribution, meaning this hypothetical model has good diagnostic power.

Analyzing Synthetic Dose-Response Curves

We now analyze some synthetic data, generated to look like dose-response curves. As a refresher, dose-response curves measure the behavior of some tissue cells in response to increasing levels of (generally) drugs. The $x$-axis is the drug dose, measured in some concentration or volume, and the $y$-axis is generally some measure of cell death or survival, generally ranging from 0% to 100%. The curves, however, look sigmoidal. So let’s first generate a function to create sigmoid curves.

def sigmoid(beta, intercept, x):

 """
 Fake sigmoid function, takes in coefficient, shift, and dose values.

 Parameters:
 - - - - -
 beta: float
 slope
 intercept: float
 negative exponential intercept
 x: float, array
 data samples

 Returns:
 - - - -
 dose_response: float, array
 single-subject dose-response vector
 Between 0 and 1.
 """

 dose_resonse = 1 / (1 + np.exp(-beta*x + intercept))

 return dose_resonse

Next, let’s actually generate some synthetic data for a dataset of fake subjects. I want to incorporate some variability into the Cases and Controls, so I’ll sample the subject parameters from distributions. In this case, for each subject, I’ll sample the logistic curve slope coefficient from $Beta$ distributions, and the intercept from $Normal$ distributions. We’ll sample 1000 Cases, and 1000 Controls.

For the slopes, we have

$$ \begin{align} \beta_{Control} &\sim Beta(a=5, b=3) \ \beta_{Case} &\sim Beta(a=10, b=2) \end{align} $$

and for the intercepts, we have

$$ \begin{align} I_{Control} &\sim Normal(\mu=0, \sigma=1) \ I_{Case} &\sim Normal(\mu=4, \sigma=1) \end{align} $$

As such, the dose-response curve for individual, $k$, is generated as follows:

$$ dr_{k} = \frac{1}{1 + e^{-(\beta_{k}X + I_{k})}} $$

where $\beta_{k}$ and $I_{k}$ are the slope and intercept values for the given subject. These distributional parameterizations are arbitrary – I just wanted to be able to incorporate variability across subjects and groups.

Let’s generate some random Case/Control dose-response data and plot the coefficient histograms:

# n cases and controls
S = 1000

# dictionary of slopes and intercept values for each subject
controls = {k: {'beta': None,
 'intercept': None} for k in np.arange(S)}
cases = {k: {'beta': None,
 'intercept': None} for k in np.arange(S)}

# get lists of betas and intercepts
beta_control = []
beta_case = []

intercept_control = []
intercept_case = []

for i in np.arange(S):

 controls[i]['beta'] = np.random.beta(a=5, b=3)
 controls[i]['intercept'] = np.random.normal(loc=0, scale=1)

 intercept_control.append(controls[i]['intercept'])
 beta_control.append(controls[i]['beta'])

 cases[i]['beta'] = np.random.beta(a=10, b=2)
 cases[i]['intercept'] = np.random.normal(loc=4, scale=1)

 intercept_case.append(cases[i]['intercept'])
 beta_case.append(cases[i]['beta'])

Intercept histograms look like two different $$Normal$$ distributions:

fig = plt.figure(figsize=(12, 8))
plt.hist(intercept_control, 50, color='r', alpha=0.5, density=True, label='Control');
plt.hist(intercept_case, 50, color='b', alpha=0.5, density=True, label='Case');
plt.ylabel('Density', fontsize=15);
plt.xlabel('Intercepts', fontsize=15);
plt.title('Intercepts Coefficients By Group', fontsize=15);
plt.legend(fontsize=15);
plt.tight_layout()

Intercept distributions for synthetic dose-response curves.

Slope histograms look like two different $Beta$ distributions:

fig = plt.figure(figsize=(12, 8))
plt.hist(beta_control, 50, color='r', alpha=0.5, label='Control', density=True);
plt.hist(beta_case, 50, color='b', alpha=0.5, label='Case', density=True);
plt.ylabel('Density', fontsize=15);
plt.xlabel('Betas', fontsize=15);
plt.title('Slope Coefficients By Group', fontsize=15);
plt.legend(fontsize=15);
plt.tight_layout()

Slope distributions for synthetic dose-response curves.

Now we’ll generate some fake dose-response curves for each of the 1000 Controls, and 1000 Cases. We’ll plot a subset of these curves to visualize our cross-group curve variability.

# define synthetic dose range
doses = np.linspace(-15,15,500)
dose_min = doses.min()
shifted_dose = doses + np.abs(dose_min)

fig, (ax1) = plt.subplots(figsize=(12, 8))

ec50_control = []
ec50_case = []

for c in np.arange(S):

 control_sample = sigmoid(controls[c]['beta'], controls[c]['intercept'], doses)
 case_sample = sigmoid(cases[c]['beta'], cases[c]['intercept'], doses)

 ec50_control.append(shifted_dose[control_sample < 0.5].max())
 ec50_case.append(shifted_dose[case_sample < 0.5].max())

 if (c % 15) == 0:

 ax1.plot(shifted_dose, control_sample, c='r', linewidth=3, alpha=0.3)
 ax1.plot(shifted_dose, case_sample, c='b', linewidth=3, alpha=0.3)

plt.legend({})
plt.title('Dose Response Curve', fontsize=20);
plt.xlabel('Dose', fontsize=20);
plt.xticks(fontsize=15)
plt.ylabel('Response', fontsize=20);
plt.yticks(fontsize=15)

custom_lines = [Line2D([0], [0], color='r', lw=4),
 Line2D([0], [0], color='b', lw=4)]
plt.legend(custom_lines, ['Control', 'Case'], fontsize=20);

Case/Control dose-response curves.

For our preliminary biomarker of interest, let’s look at the ec50, which is the dose at which 50% of the cells show some response (i.e. where our $y$-axis = 0.5), for each sample in our dataset. We’ll plot these doses as a function of Cases and Controls.

fig = plt.figure(figsize=(12, 8))
plt.hist(ec50_control, 50, color='r', alpha=0.5, label='Control', density=True);
plt.hist(ec50_case, 50, color='b', alpha=0.5, label='Case', density=True);
plt.legend(fontsize=20);
plt.xlabel('ec50', fontsize=20);
plt.xticks(fontsize=15);
plt.ylabel('Density', fontsize=20);
plt.yticks(fontsize=15);

Biomarker distributions: ec50 for Cases and Controls.

If we select a different threshold – i.e. instead of 0.5, we can iterate over the range of 0.1 - 0.9, for example, in increments of 0.1 – we generate different biomarkers (ec10, ec20 … ec90). We can treat each biomarker as a different classification model, and assess how powerful that model is at assessing whether someone will develop cardiotoxicity or not. To do so, we’ll create distributions for each biomarker (not shown), and then generate ROC curves and AUC values for each curve.

ROC curves and AUC for each ec-X biomarker level.

This is where my limited domain knowledge comes at a cost – I’m not sure if the biomarkers I’ve chosen (i.e. incremental ec values) are actually biologically relevant. The point, however, is that each biomarker yields a different AUC, which theoretically shows that the Cases and Controls can be differentially distinguished, depending on which biomarker we choose to examine. In this case, ec10 has the most discriminative diagnostic power.

Something I did wonder about while exploring this data was how dependent the ROC curves and AUC statistics are on sample size. Previously, I’d looked at rates of convergence of various estimators – the AUC should also theoretically show some convergence to a “true” value as $n$ increases – but I’m not sure if it follows any sort of relevant distribution. I imagine the AUC is domain-dependent, in that it depends on the distribution of the biomarker of interest? Might be a good idea for another post…

Cheers.

The Delta Method

Tue, 19 Mar 2019 12:43:32 -0700

Here, we’ll look at various applications of the Delta Method, especially in the context of variance stabilizing transformations, along with looking at the confidence intervals of estimates.

The Delta Method is used as a way to approximate the Standard Error of transformations of random variables, and is based on a Taylor Series approximation.

In the univariate case, if we have a random variable, $X_{n}$, that converges in distribution to a $N(0, \sigma^{2})$ distribution, we can apply a function to this random variable as:

$$\begin{align} \sqrt{n}(X_{n} - \theta) \xrightarrow{d} N(0,\sigma^{2}) \\ \sqrt{n}(g(X_{n}) - g(\theta)) \xrightarrow{d} ; ? \end{align}$$

However, we don’t know the asymptotic variance of this transformed variable just yet. In this case, we can approximate our function $g(x)$ using a Taylor Series approximation, evaluated at $\theta$:

$$\begin{align} g(x) = g(\theta) + g\prime(\theta)(x-\theta) + O() \end{align}$$

where $O()$ is the remainder of higher-order Taylor Series terms that converges to 0.

By Slutsky’s Theorem and the Continuous Mapping Theorem, we know that since $\bar{\theta} \xrightarrow{p} \theta$, we know that $g\prime(\bar{\theta}) \xrightarrow{p} g\prime(\theta)$

Plugging this back in to our original equation and applying Slutsky’s Perturbation Theorem, we have:

$$\begin{align} &= \sqrt{n}(\Big[g(\theta) + g\prime(\theta)(x-\theta)\Big] - g(\theta)) \\ &= \sqrt{n}(g\prime(\theta)(x-\theta)) \\ &= g\prime(\theta)\sqrt{n}(X_{n} - \theta) \end{align}$$

and since we know that $\sqrt{n}(\bar{X_{n}} - \theta) \xrightarrow{d} N(0,\sigma^{2})$, we now know that $g\prime(\theta) \sqrt{n}(\bar{X_{n}} - \theta) \xrightarrow{d} N(0,g\prime(\theta)^{2} \sigma^{2})$. As such, we have that:

$$\begin{align} \sqrt{n}(g(X_{n}) - g(\theta)) \xrightarrow{d} N(0, g\prime(\theta)^{2}\sigma^{2}) \end{align}$$

The Delta Method can be generalized to the multivariate case, where, instead of the derivative, we use the gradient vector of our function:

$$\begin{align} \sqrt{n}(g(\bar{X_{n}} - g(\theta)) \xrightarrow{d} N(0, \nabla(g)^{T} \Sigma \nabla(g)) \end{align}$$

Below, I’m going to look at a few examples applying the Delta Method to simple functions of random variables. Then I’ll go into more involved examples applying the Delta Method via Variance Stabilizing Transformations. Oftentimes, the variance of an estimate depends on its mean, which can vary with the sample size. In this case, we’d like to find a function $g(\theta)$, such that, when applied via the Delta Method, the variance is constant as a function of the sample size.

We’ll start by importing the necessary libraries and defining two functions:

%matplotlib inline
import matplotlib.pyplot as plt

from matplotlib import rc
rc('text', usetex=True)

from scipy.stats import norm, poisson, expon
import numpy as np

Here, we define two simple functions – one to compute the difference between our estimate and its population parameter, and the other to compute the function of our random variable as described by the Central Limit Theorem.

def conv_prob(n, est, pop):

 """
 Method to compute the estimate for convergence in probability.
 """

 return (est-pop)

def clt(n, est, pop):

 """
 Method to examine the Central Limit Theorem.
 """

 return np.sqrt(n)*(est-pop)

Let’s have a look at an easy example with the Normal Distribution. We’ll set $\mu = 0$ and $\sigma^{2} = 5$. Remember that when using the Scipy Normal distribution, the norm class accepts the standard deviation, not the variance. We’ll show via the Central Limit Theorem that the function $\sqrt{n}(\bar{X_{n}} - \mu) \xrightarrow{d} N(0,\sigma^{2})$.

# set sample sample sizes, and number of sampling iterations
N = [5,10,50,100,500,1000]
iters = 500

mu = 0; sigma = np.sqrt(5)

# store estimates
norm_clt = {n: [] for n in N}

samples = norm(mu,sigma).rvs(size=(iters,1000))

for n in N:
 for i in np.arange(iters):

 est_norm = np.mean(samples[i,0:n])
 norm_clt[n].append(clt(n, est_norm, mu))

Now let’s plot the results.

# Plot results using violin plots
fig = plt.subplots(figsize=(8,5))

for i,n in enumerate(N):
 temp = norm_clt[n]
 m = np.mean(temp)
 v = np.var(temp)
 print('Sample Size: %i has empirical variance: %.2f' % (n, v.mean()))

 plt.violinplot(norm_clt[n], positions=[i],)

Central Limit Theorem applied to Normal Distribution.

As expected, we see that the Normal distribution mean and variance estimates are independent of the sample size. In this case, we don’t need to apply a variance stabilizing transformation. We also see that the variance fluctuates around $5$. Now, let’s apply a simple function $g(\theta) = \theta^{2}$ to our data. So $g\prime(\theta) = 2\theta$, and the variance of our function becomes $g\prime(\mu)^{2}\sigma^{2} = (2\mu)^{2} \sigma^{2} = 4\mu^{2}\sigma^{2}$. Let’s look at a few plots, as a function of changing $\mu$.

# set sample sample sizes, and number of sampling iterations
mus = [1,2,3,4]

N = [5,10,50,100,500,1000]
iters = 2000
sigma = np.sqrt(5)


fig, ([ax1,ax2], [ax3,ax4]) = plt.subplots(2,2, figsize=(14,9))
for j ,m in enumerate(mus):

 # store estimates
 norm_clt = {n: [] for n in N}
 samples = norm(m, sigma).rvs(size=(iters, 1000))


 plt.subplot(2,2,j+1)
 for k, n in enumerate(N):
 np.random.shuffle(samples)
 for i in np.arange(iters):

 est_norm = np.mean(samples[i, 0:n])
 norm_clt[n].append(clt(n, est_norm**2, m**2))

 plt.violinplot(norm_clt[n], positions=[k],)

Central Limit Theorem applied to function of Normal Distribution.

We see that the variance increases as the mean increases, and that, as the sample sizes increase, the distributions converge to the $$N(0, 4\mu^{2}\sigma^{2})$$ asymptotic distribution.

Variance Stabilization for the Poisson Distribution

Now let’s look at an example where the variance depends on the sample size. We’ll use the Poisson distribution in this case. We know that for the Poisson distribution, the variance is dependent on the mean, so let’s define a random variable, $X_{\lambda}$, where $\lambda = n*\theta$. $n$ is the sample size, and $\theta$ is a fixed constant.

We’ll define $ X_{\lambda } = \sum_{i=1}^{n} X_{\theta}$, the sum of $n$ independent Poisson random variables, so that the expected value and variance of $X_{\lambda } = n\theta$

If we wanted to apply the Central Limit Theorem to $X_{\lambda }$, our convergence would be as follows:

$$\begin{align} \sqrt{n}(X_{\lambda} - \lambda) \xrightarrow{d} N(0,\sigma^{2}(\lambda)) \end{align}$$

where the variance $\sigma^{2}(\lambda)$ depends on the mean, $\lambda$. In order to stabilize the variance of this variable, we can apply the Delta Method, in order to generate a variable that converges to a standard Normal distribution asymptotically.

$$\begin{align} \sqrt{n}(g(X_{\lambda}) - g(\lambda)) \xrightarrow{d} N(0,g\prime(\theta)^{2}\sigma^{2}) \ \end{align}$$

where

$$\begin{align} &g\prime(\theta)^{2} \theta = 1 \\ &g\prime(\theta)^{2} = \frac{1}{\theta} \\ &g\prime(\theta) = \frac{1}{\sqrt{\theta}} \\ &g(\theta) = \int \frac{\partial{\theta}}{\sqrt{\theta}} \\ &g(\theta) = 2\sqrt{\theta} \end{align}$$

is our variance stabilizing function.

def p_lambda(n, theta=0.5):

 """
 Function to compute lambda parameter for Poisson distribution.
 Theta is constant.
 """
 return n*theta

theta = 0.5

N = [5,10,50,100,250,500,750,1000]
iters = 500

clt_pois = {n: [] for n in N}
pois_novar= {n: [] for n in N}
pois_var = {n: [] for n in N}

for n in N:
 for i in np.arange(iters):
 est_mu = np.mean(poisson(mu=(n*theta)).rvs(n))

 pois_novar[n].append(clt(n, est_mu, p_lambda(n)))
 pois_var[n].append(clt(n, 2*np.sqrt(est_mu), 2*np.sqrt(p_lambda(n))))

 clt_pois[n].append(conv_prob(n, est_mu, n*theta))

fig,([ax1, ax2]) = plt.subplots(2,1, figsize=(15,6))

plt.subplot(1,2,1)
for i,n in enumerate(N):
 plt.violinplot(pois_novar[n], positions=[i])

plt.subplot(1,2,2)
for i,n in enumerate(N):
 plt.violinplot(pois_var[n], positions=[i])

Variance stabilization of Poisson distribution.

Variance Stabilization for the Exponential Distribution

Applying the same method to the Exponential distribution, we’ll find that the variance stabilizing transformation is $g(\theta) = log(\theta)$. We’ll apply that here:

theta = 0.5

N = [5,10,50,100,250,500,750,1000]
iters = 500

clt_exp = {n: [] for n in N}
exp_novar= {n: [] for n in N}
exp_var = {n: [] for n in N}

for n in N:
 for i in np.arange(iters):
 samps = expon(scale=n*theta).rvs(n)

 est_mu = np.mean(samps)
 est_var = np.var(samps)

 exp_novar[n].append(clt(n, est_mu, (n*theta)))
 exp_var[n].append(clt(n, np.log(est_mu), np.log(n*theta)))

 clt_exp[n].append(conv_prob(n, est_mu, n*theta))

fig,([ax1, ax2]) = plt.subplots(2,1, figsize=(15,6))

plt.subplot(1,2,1)
for i,n in enumerate(N):
 plt.violinplot(exp_novar[n], positions=[i])

plt.subplot(1,2,2)
for i,n in enumerate(N):
 plt.violinplot(exp_var[n], positions=[i])

Variance stabilization of Exponential distribution.

Example of Standard Error Computation Using Delta Method for Polynomial Regression

As an example of applying the Delta Method to a real-world dataset, I’ve downloaded the banknote dataset from the UCI Machine Learning Repository. In this exercise, I’ll apply the logistic function via logistic regression to assess whether or not a banknote is real or fake, using a set of features. I’ll compute confidence intervals of our prediction probabilities using the Delta Method. There are four unique predictors in this case: the variance, skew, kurtosis, and entropy of the Wavelet-transformed banknote image. I’ll treat each of these predictors independently, using polynomial basis functions of degree 3.

In this example, we’re interested in the standard error of our probability estimate. Our function is the Logistic Function, as follows:

$$\begin{align} g(\beta) &= \frac{1}{1+e^{-x^{T}\beta}} \\ &= \frac{e^{x^{T}\beta}}{1+e^{x^{T}\beta}} \end{align}$$

where the gradient of this multivariate function is:

$$\begin{align} \nabla g(\beta) &= \frac{\partial g}{\partial \beta} e^{x^{T}\beta}(1+e^{x^{T}\beta})^{-1} \\ &= x^{T}e^{x^{T}\beta}(1+e^{x^{T}\beta})^{-1} - x^{T}e^{x^{T}\beta}e^{x^{T}\beta} \\ &= x^{T}\Big(e^{x^{T}\beta}(1+e^{x^{T}\beta})^{-1} - e^{x^{T}\beta}e^{x^{T}\beta}\Big)(1+e^{x^{T}\beta})^{-2} \\ &= x^{T} \frac{e^{x^{T}\beta}}{(1+e^{x^{T}\beta})^{2}} \\ \nabla g(\beta) &= x^{T} g(\beta)(1-g(\beta)) \end{align}$$

so that the final estimate of our confidence interval becomes

$$\begin{align} & \sim N(0,x^{T} g(\beta)(1-g(\beta)) \Sigma g(\beta)(1-g(\beta))x) \\ & \sim N(0, \nabla g(\beta)^{T} \Sigma \nabla g(\beta)) \end{align}$$

import pandas as pd
import statsmodels.api as sm
from sklearn.preprocessing import PolynomialFeatures

import pandas as pd
import statsmodels.api as sm
from sklearn.preprocessing import PolynomialFeatures

bank = pd.read_csv('/Users/kristianeschenburg/Documents/Statistics/BankNote.txt',
 sep=',',header=None, names=['variance', 'skew', 'kurtosis', 'entropy','class'])
bank.head()

fig = plt.subplots(2,2, figsize=(12,8))
for j, measure in enumerate(['variance', 'kurtosis', 'skew', 'entropy']):

 predictor = np.asarray(bank[measure])
 response = np.asarray(bank['class'])

 idx = (response == 1)

 # plot test set
 plt.subplot(2,2,j+1)
 plt.violinplot(predictor[idx], positions=[1]);
 plt.violinplot(predictor[~idx], positions=[0])
 plt.title('{:} By Classification'.format(measure), fontsize=18)
 plt.ylabel('Measure: {:}'.format(measure),fontsize=15)
 plt.yticks(fontsize=13)
 plt.xticks([0,1],['Fake','Real'], fontsize=15)

plt.tight_layout()

Bank note feature distributions, based on note class.

Based on the above plot, we can see that variance, skew, and kurtosis seem to be the most informative, while the entropy distributions do not seem to be that different based on bank note class.

Next, we fit a logistic regression model of note classification on note feature, with polynomial order of degree 3. We then compute the standard errors of the transformed variance. It was transformed using the logistic function, so we’ll need to compute the gradient of this function.

fig = plt.subplots(2,2, figsize=(12,8))
for j, measure in enumerate(['variance', 'kurtosis', 'skew', 'entropy']):

 # Generate polynomial object to degree 
 # transform age to 4-degree basis function
 poly = PolynomialFeatures(degree=2)
 idx_order = np.argsort(bank[measure])

 predictor = bank[measure][idx_order]
 response = bank['class'][idx_order]

 features = poly.fit_transform(predictor.values.reshape(-1,1));

 # fit logit curve to curve
 logit = sm.Logit(response, features).fit();

 test_features = np.linspace(np.min(predictor), np.max(predictor), 100)
 test_features = poly.fit_transform(test_features.reshape(-1,1))
 # predict on test set
 class_prob = logit.predict(test_features)

 cov = logit.cov_params()
 yx = (class_prob*(1-class_prob))[:,None] * test_features
 se = np.sqrt(np.diag(np.dot(np.dot(yx, cov), yx.T)))

 # probability can't exceed 1, or be less than 0
 upper = np.maximum(0, np.minimum(1, class_prob+1.96*se))
 lower = np.maximum(0, np.minimum(1, class_prob-1.96*se))

 # plot test set
 plt.subplot(2,2,j+1)
 plt.plot(test_features[:, 1], class_prob);
 plt.plot(test_features[:, 1], upper, color='red', linestyle='--', alpha=0.5);
 plt.plot(test_features[:, 1], lower, color='red', linestyle='--', alpha=0.5);
 plt.title(r'P(isReal \Big| X)', fontsize=18)
 plt.xlabel('{:}'.format(measure),fontsize=15)
 plt.ylabel('Probability',fontsize=15)
 plt.grid(True)

plt.tight_layout()

Confidence intervals for each feature, computed using Delta Method.

Mahalanobis Distances of Brain Connectivity

Fri, 07 Dec 2018 05:12:32 -0700

For one of the projects I’m working on, I have an array of multivariate data relating to brain connectivity patterns. Briefly, each brain is represented as a surface mesh, which we represent as a graph $G = (V,E)$, where $V$ is a set of $n$ vertices, and $E$ is the set of edges between vertices.

Additionally, for each vertex $v \in V$, we also have an associated scalar label, which we’ll denote $l(v)$, that identifies what region of the cortex each vertex belongs to, the set of regions which we define as $L = {1, 2, … k}$. And finally, for each vertex $v \in V$, we also have a multivariate feature vector $r(v) \in \mathbb{R}^{1 \times k}$, that describes the strength of connectivity between it, and every region $l \in L$.

Example of cortical map, and array of connectivity features.

I’m interested in examining how “close” the connectivity samples of one region, $l_{j}$, are to another region, $l_{k}$. In the univariate case, one way to compare a scalar sample to a distribution is to use the $t$-statistic, which measures how many standard deviations away from the mean a given sample is:

$$\begin{align} t_{s} = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \end{align}$$

where $\mu$ is the population mean, and $s$ is the sample standard deviation. If we square this, we get:

$$\begin{align} t^{2} = \frac{(\bar{x} - \mu)^{2}}{\frac{s^{2}}{n}} = \frac{n (\bar{x} - \mu)^{2}}{S^{2}} \sim F(1,n) \end{align}$$

We know the last part is true, because the numerator and denominator are independent $\chi^{2}$ distributed random variables. However, I’m not working with univariate data – I have multivariate data. The multivariate generalization of the $t$-statistic is the Mahalanobis Distance:

$$\begin{align} d &= \sqrt{(\bar{x} - \mu)\Sigma^{-1}(\bar{x}-\mu)^{T}} \end{align}$$

where the squared Mahalanobis Distance is:

$$\begin{align} d^{2} &= (\bar{x} - \mu)\Sigma^{-1}(\bar{x}-\mu)^{T} \end{align}$$

where $\Sigma^{-1}$ is the inverse covariance matrix. If our $X$’s were initially distributed with a multivariate normal distribution, $N_{p}(\mu,\Sigma)$ (assuming $\Sigma$ is non-degenerate i.e. positive definite), the squared Mahalanobis distance, $d^{2}$ has a $\chi^{2}_{p}$ distribution. We show this below.

We know that $(X-\mu)$ is distributed $N_{p}(0,\Sigma)$. We also know that, since $\Sigma$ is symmetric and real, that we can compute the eigendecomposition of $\Sigma$ as:

$$\begin{align} \Sigma = U \Lambda U^{T} \end{align}$$

and consequently, because $U$ is an orthogonal matrix, and because $\Lambda$ is diagonal, we know that $\Sigma^{-1}$ is:

$$\begin{align} \Sigma^{-1} &= (U \Lambda U^{T})^{-1} \\ &= U \Lambda^{-1} U^{T} \\ &= (U \Lambda^{\frac{-1}{2}}) (U \Lambda^{\frac{-1}{2}})^{T} \\ &= R R^{T} \end{align}$$

Therefore, we know that $R^{T}(X-\mu) \sim N_{p}(0,I_{p})$:

$$\begin{align} X &\sim N_{p}(\mu,\Sigma) \\ (X-\mu) = Y &\sim N_{p}(0,\Sigma)\\ R^{T}Y = Z &\sim N_{p}(0, R^{T} \Sigma R) \\ &\sim N_{p}(0, \Lambda^{\frac{-1}{2}} U^{T} (U \Lambda U^{T}) U \Lambda^{\frac{-1}{2}}) \\ &\sim N_{p}(0, \Lambda^{\frac{-1}{2}} I_{p} \Lambda I_{p} \Lambda^{\frac{-1}{2}}) \\ &\sim N_{p}(0,I_{p}) \end{align}$$

so that we have

$$\begin{align} &= (X-\mu)\Sigma^{-1}(X-\mu)^{T} \\ &= (X-\mu)RR^{T}(X-\mu)^{T} \\ &= Z^{T}Z \end{align}$$

the sum of $p$ squared standard Normal random variables, which is the definition of a $\chi_{p}^{2}$ distribution with $p$ degrees of freedom. So, given that we start with a $MVN$ random variable, the squared Mahalanobis distance is $\chi^{2}_{p}$ distributed. Because the sample mean and sample covariance are consistent estimators of the population mean and population covariance parameters, we can use these estimates in our computation of the Mahalanobis distance.

Also, of particular importance is the fact that the Mahalanobis distance is not symmetric. That is to say, if we define the Mahalanobis distance as:

$$\begin{align} M(A, B) = \sqrt{(A - \mu(B))\Sigma(B)^{-1}(A-\mu(B))^{T}} \end{align}$$

then $M(A,B) \neq M(B,A)$, clearly. Because the parameter estimates are not guaranteed to be the same, it’s straightforward to see why this is the case.

Now, back to the task at hand. For a specified target region, $l_{T}$, with a set of vertices, $V_{T} = {v \; : \; l(v) \; = \; l_{T}, \; \forall \; v \in V}$, each with their own distinct connectivity fingerprints, I want to explore which areas of the cortex have connectivity fingerprints that are different from or similar to $l_{T}$’s features, in distribution. I can do this by using the Mahalanobis Distance. And based on the analysis I showed above, we know that the data-generating process of these distances is related to the $\chi_{p}^{2}$ distribution.

First, I’ll estimate the covariance matrix, $\Sigma_{T}$, of our target region, $l_{T}$, using the Ledoit-Wolf estimator (the shrunken covariance estimate has been shown to be a more reliable estimate of the population covariance), and mean connectivity fingerprint, $\mu_{T}$. Then, I’ll compute $d^{2} = M^{2}(A,A)$ for every $\{v: v \in V_{T}\}$. The empirical distribution of these distances should follow a $\chi_{p}^{2}$ distribution. If we wanted to do hypothesis testing, we would use this distribution as our null distribution.

Next, in order to assess whether this intra-regional similarity is actually informative, I’ll also compute the similarity of $l_{T}$ to every other region, $\{ l_{k} \; : \; \forall \; k \in L \setminus \{T\} \}$ – that is, I’ll compute $M^{2}(A, B) \; \forall \; B \in L \setminus T$. If the connectivity samples of our region of interest are as similar to one another as they are to other regions, then $d^{2}$ doesn’t really offer us any discriminating information – I don’t expect this to be the case, but we need to verify this.

Then, as a confirmation step to ensure that our empirical data actually follows the theoretical $\chi_{p}^{2}$ distribution, I’ll compute the location and scale Maximum Likelihood(MLE) parameter estimates of our $d^{2}$ distribution, keeping the d.o.f. (i.e. $p$) fixed.

See below for Python code and figures…

Step 1: Compute Parameter Estimates

%matplotlib inline
import matplotlib.pyplot as plt

from matplotlib import rc
rc('text', usetex=True)

import numpy as np
from scipy.spatial.distance import cdist
from scipy.stats import chi2, probplot

from sklearn import covariance

# lab_map is a dictionary, mapping label values to sample indices
# our region of interest has a label of 8
LT = 8

# get indices for region LT, and rest of brain
lt_indices = lab_map[LT]
rb_indices = np.concatenate([lab_map[k] for k in lab_map.keys() if k != LT])

data_lt = conn[lt_indices, :]
data_rb = conn[rb_indices, :]

# fit covariance and precision matrices
# Shrinkage factor = 0.2
cov_lt = covariance.ShrunkCovariance(assume_centered=False, shrinkage=0.2)
cov_lt.fit(data_lt)
P = cov_lt.precision_

Next, compute the Mahalanobis Distances:

# LT to LT Mahalanobis Distance
dist_lt = cdist(data_lt, data_lt.mean(0)[None,:], metric='mahalanobis', VI=P)
dist_lt2 = dist_lt**2

# fit covariance estimate for every region in cortical map
EVs = {l: covariance.ShrunkCovariance(assume_centered=False,
 shrinkage=0.2) for l in labels}

for l in lab_map.keys():
 EVs[l].fit(conn[lab_map[l],:])

# compute d^2 from LT to every cortical region
# save distances in dictionary
lt_to_brain = {}.fromkeys(labels)
for l in lab_map.keys():

 temp_data = conn[label_map[l], :]
 temp_mu = temp_data.mean(0)[None, :]

 temp_mh = cdist(data_lt, temp_mu, metric='mahalanobis', VI=EVs[l].precision_)
 temp_mh2 = temp_mh**2

 lt_to_brain[l] = temp_mh2

# plot distributions seperate (scales differ)
fig = plt.subplots(2,1, figsize=(12,12))
plt.subplot(2,1,1)
plt.hist(lt_to_brain[LT], 50, density=True, color='blue',
 label='Region-to-Self', alpha=0.7)

plt.subplot(2,1,2)
for l in labels:
 if l != LT:
 plt.hist(lt_to_brain[l], 50, density=True, linewidth=2,
 alpha=0.4, histtype='step')

Empirical distributions of within-region (top) and between-region (bottom) $d^{2}$ values. Each line is the distribution of the distance of samples in our ROI to a whole region.

As expected, the distribution of $d^{2}$, the distance of samples in our region of interest, $l_{T}$, to distributions computed from other regions, is (considerably) larger and much more variable, while the profile of points within $l_{T}$ looks to have much smaller variance – this is good! This means that we have high intra-regional similarity when compared to inter-regional similarities. This fits what’s known in neuroscience as the “cortical field hypothesis”.

Step 2: Distributional QC-Check

Because we know that our data should follow a $\chi^{2}_{p}$ distribution, we can fit the MLE estimate of our location and scale parameters, while keeping the $df$ parameter fixed.

p = data_lt.shape[1]
mle_chi2_theory = chi2.fit(dist_lt2, fdf=p)

xr = np.linspace(data_lt.min(), data_lt.max(), 1000)
pdf_chi2_theory(xr, *mle_chi2_theory)

fig = plt.subplot(1,2,2,figsize=(18, 6))

# plot theoretical vs empirical null distributon
plt.subplot(1,2,1)
plt.hist(data_lt, density=True, color='blue', alpha=0.6,
 label = 'Empirical')
plt.plot(xr, pdf_chi2_theory, color='red',
 label = '$\chi^{2}_{p}')

# plot QQ plot of empirical distribution
plt.subplot(1,2,2)
probplot(D2.squeeze(), sparams=mle_chi2_theory, dist=chi2, plot=plt);

Density and QQ plot of null distribution.

From looking at the QQ plot, we see that the empirical density fits the theoretical density pretty well, but there is some evidence that the empirical density has heavier tails. The heavier tail of the upper quantile could probability be explained by acknowledging that our starting cortical map is not perfect (in fact there is no “gold-standard” cortical map). Cortical regions do not have discrete cutoffs, although there are reasonably steep gradients in connectivity. If we were to include samples that were considerably far away from the the rest of the samples, this would result in inflated densities of higher $d^{2}$ values.

Likewise, we also made the distributional assumption that our connectivity vectors were multivariate normal – this might not be true – in which case our assumption that $d^{2}$ follows a $\chi^{2}_{p}$ would also not hold.

Finally, let’s have a look at some brains! Below, is the region we used as our target – the connectivity profiles from vertices in this region were used to compute our mean vector and covariance matrix – we compared the rest of the brain to this region.

Region of interest.

Estimated squared Mahalanobis distances, overlaid on cortical surface.

Here, larger $d^{2}$ values are in red, and smaller $d^{2}$ are in black. Interestingly, we do see pretty large variance of $d^{2}$ spread across the cortex – however the values are smoothly varying, but there do exists sharp boundaries. We kind of expected this – some regions, though geodesically far away, should have similar connectivity profiles if they’re connected to the same regions of the cortex. However, the regions with connectivity profiles most different than our target region are not only contiguous (they’re not noisy), but follow known anatomical boundaries, as shown by the overlaid boundary map.

This is interesting stuff – I’d originally intended on just learning more about the Mahalanobis Distance as a measure, and exploring its distributional properties – but now that I see these results, I think it’s definitely worth exploring further!

Convergence In Probability

Wed, 28 Nov 2018 13:12:32 -0700

I’m going over Chapter 5 in Casella and Berger’s (CB) “Statistical Inference”, specifically Section 5.5: Convergence Concepts, and wanted to document the topic of convergence in probability with some plots demonstrating the concept.

From CB, we have the definition of convergence in probability: a sequence of random variables $X_{1}, X_{2}, … X_{n}$ converges in probability to a random variable $X$, if for every $\epsilon > 0$,

$$\begin{align} \lim_{n \to \infty} P(| X_{n} - X | \geq \epsilon) = 0 \\ \end{align}$$

Intuitively, this means that, if we have some random variable $X_{k}$ and another random variable $X$, the absolute difference between $X_{k}$ and $X$ gets smaller and smaller as $k$ increases. The probability that this difference exceeds some value, $\epsilon$, shrinks to zero as $k$ tends towards infinity. Using convergence in probability, we can derive the Weak Law of Large Numbers (WLLN):

$$\begin{align} \lim_{n \to \infty} P(|\bar{X}_{n} - \mu | \geq \epsilon) = 0 \end{align}$$

which we can take to mean that the sample mean converges in probability to the population mean as the sample size goes to infinity. If we have finite variance (that is $Var(X) < \infty$), we can prove this using Chebyshev’s Inequality

$$\begin{align} &= P(|\bar{X}_{n} - \mu | \geq \epsilon) \\ &= P((\bar{X}_{n} - \mu)^{2} \geq \epsilon^{2}) \leq \frac{E\Big[(\bar{X}_{n} - \mu)^{2}\Big]}{\epsilon^{2}} \\ &= P((\bar{X}_{n} - \mu)^{2} \geq \epsilon^{2}) \leq \frac{Var(\bar{X_{n}})}{\epsilon^{2}} \\ &= P((\bar{X}_{n} - \mu)^{2} \geq \epsilon^{2}) \leq \frac{\sigma^{2}}{n^{2}\epsilon^{2}} \end{align}$$

where $\frac{\sigma^{2}}{n^{2} \epsilon^{2}} \rightarrow 0$ as $n \rightarrow \infty$. Intuitively, this means, that the sample mean converges to the population mean – and the probability that their difference is larger than some value is bounded by the variance of the estimator. Because we showed that the variance of the estimator (right hand side) shrinks to zero, we can show that the difference between the sample mean and population mean converges to zero.

We can also show a similar WLLN result for the sample variance using Chebyshev’s Inequality, as:

$$\begin{align} S_{n}^{2} = \frac{1}{n-1} \sum_{i=1}^{n} (X_{i} - \bar{X}_{n})^{2} \end{align}$$

using the unbiased estimator, $S_{n}^{2}$, of $\sigma^{2}$ as follows:

$$\begin{align} P(|S_{n}^{2} - \sigma^{2}| \geq \epsilon) \leq \frac{E\Big[(S_{n}^{2} - \sigma^{2})^{2}\Big]}{\epsilon^{2}} = \frac{Var(S_{n}^{2})}{\epsilon^{2}} \end{align}$$

so all we need to do is show that $Var(S_{n}^{2}) \rightarrow 0$ as $n \rightarrow \infty$.

Let’s have a look at some (simple) real-world examples. We’ll start by sampling from a $N(0,1)$ distribution, and compute the sample mean and variance using their unbiased estimators.

# Import numpy and scipy libraries
import numpy as np
from scipy.stats import norm

%matplotlib inline
import matplotlib.pyplot as plt

plt.rc('text', usetex=True)

# Generate set of samples sizes
samples = np.concatenate([np.arange(0, 105, 5),
 10*np.arange(10, 110, 10),
 100*np.arange(10, 210, 10)])

# number of repeated samplings for each sample size
iterations = 500

# store sample mean and variance
means = np.zeros((iterations, len(samples)))
vsrs = np.zeros((iterations, len(samples)))

for i in np.arange(iterations):
 for j, s in enumerate(samples):

 # generate samples from N(0,1) distribution
 N = norm.rvs(loc=0, scale=1, size=s)
 mu = np.mean(N)

 # unbiased estimate of variance
 vr = ((N - mu)**2).sum()/(s-1)

 means[i, j] = mu
 vsrs[i, j] = vr

Let’s have a look at the sample means and variances as a function of the sample size. Empirically, we see that both the sample mean and variance estimates converge to their population parameters, 0 and 1.

Sample mean estimates as a function of sample size.

Sample variance estimates as a function of sample size.

Below is a simple method to compute the empirical probability that an estimate exceeds the epsilon threshold.

def ecdf(data, pparam, epsilon):

 """
 Compute empirical probability P( |estimate - pop-param| < epsilon).

 Parameters:
 - - - - -
 data: array, float
 array of samples
 pparam: float
 true population parameter
 epsilon: float
 threshold value
 """

 compare = (np.abs(data - pparam) < epsilon)
 prob = compare.mean(0)

 return prob

# test multiple epsilon thresholds
e = [0.9, 0.75, 0.5, 0.25, 0.1, 0.05, 0.01]

mean_probs = []
vrs_probs = []
# compute empirical probabilities at each threshold
for E in e:
 mean_probs.append(1 - ecdf(means, pparam=0, epsilon=E))
 vrs_probs.append(1-ecdf(vsrs, pparam=1, epsilon=E))

Empirical probability that mean estimate exceeds population mean by epsilon.

Empirical probability that variance estimate exceeds population variance by epsilon.

The above plots show that, as sample size increases, the mean estimator and variance estimator both converge to their true population parameters. Likewise, examining the empirical probability plots, we can see that the probability that either estimate exceeds the epsilon thresholds shrinks to zero as the sample size increases.

If we wish to consider a stronger degree of convergence, we can consider convergence almost surely, which says the following:

$$\begin{align} P(\lim_{n \to \infty} |X_{n} - X| \geq \epsilon) = 0 \ \end{align}$$

which considers the entire joint distribution of estimates $( X_{1}, X_{2}…X_{n}, X)$, rather than all pairwise estimates $(X_{1},X), (X_{2},X)… (X_{n},X)$ – the entire set of estimates must converge to $X$ as the sample size approaches infinity.

Overview of Poisson-Multinomial Relationship

Thu, 08 Nov 2018 01:12:32 -0700

In this post, I’m going to briefly cover the relationship between the Poisson distribution and the Multinomial distribution.

Let’s say that we have a set of independent, Poisson-distributed random variables $Y_{1}, Y_{2}… Y_{k}$ with rate parameters $\lambda_{1}, \lambda_{2}, …\lambda_{k}$. We can model the sum of these random variables as a new random variable $N = \sum_{i=1}^{k} Y_{i}$.

Let start with $k=2$. We can define the distrbution of $F_{N}(n)$ as follows:

$$\begin{align} &= P(N \leq n) \\ &= P(Y_{1} + Y_{2} \leq n) \\ &= P(Y_{1} = y_{1}, Y_{2} = n - y_{1}) \\ &= P(Y_{1} = y_{1}) \cdot P(Y_{2} = n-y_{1}) \\ &= \sum_{y_{1}=0}^{n} \frac{e^{-\lambda_{1}}\lambda_{1}^{y_{1}}}{y_{1}!} \cdot \frac{e^{-\lambda_{2}}\lambda_{2}^{n-y_{1}}}{(n-y_{1})!} \\ &= e^{-(\lambda_{1}+\lambda_{2})} \sum_{y_{1}=0}^{n} \frac{\lambda_{1}^{y_{1}}\lambda_{2}^{n-y_{1}}}{y_{1}!(n-y_{1})!} \\ &= e^{-(\lambda_{1}+\lambda_{2})} \sum_{y_{1}=0}^{n} \frac{n!}{n!}\frac{\lambda_{1}^{y_{1}}\lambda_{2}^{n-y_{1}}}{y_{1}!(n-y_{1})!} \\ &= \frac{e^{-(\lambda_{1}+\lambda_{2})}}{n!} \sum_{y_{1}=0}^{n} {n\choose y_{1}} \lambda_{1}^{y_{1}}\lambda_{2}^{n-y_{1}} \end{align}$$

Here, we can apply the Binomial Theorem to the summation to get the following (remember that the Binomial Theorem says, for two numbers $x$ and $y$, that $(x+y)^{n} = \sum_{i=0}^{n} {n \choose i}x^{i}y^{n-i}$):

$$\begin{align} \frac{e^{-(\lambda_{1}+\lambda_{2})}(\lambda_{1} + \lambda_{2})^{n}}{n!} \\ \end{align}$$

which we see is in fact just another Poisson distribution with rate parameter equal to $\lambda_{1} + \lambda_{2}$. This shows that the sum of independent Poisson distributed random variables is also a Poisson random variable, with rate parameter equal to the sum of the univariate rates. By induction, we see that for $k$ independent Poisson distributed random variables $Y_{1}…Y_{k}$, their sum $\sum_{i=1}^{k} Y_{i} \sim Poisson(\sum_{i=1}^{k} \lambda_{i})$.

Now let’s say we’re interested in modeling the conditional distribution of $(Y_{1}…Y_{k}) \mid \sum_{i=1}^{k} = n$. By definition of conditional probability, we have that

$$\begin{align} P(\bar{Y} \mid N=n) &= \frac{P(\bar{Y} ; \cap ; N=n)}{P(N=n)} \\ &= \frac{P(\bar{Y})}{P(N=n)} \end{align}$$

We have the following:

$$\begin{align} P(\bar{Y} \mid N=n) &= \frac{P(\bar{Y} \; \cap \; N=n)}{P(N=n)} \\ &= \Big( \prod_{i=1}^{k} \frac{e^{-\lambda_{i}} \cdot \lambda_{i}^{y_{i}}}{y_{i}!} \Big) \Big/ \frac{e^{-\sum_{i=1}^{k} \lambda_{i}}(\sum_{i}^{k} \lambda_{i})^{n}}{n!} \\ &= \Big( \frac{ e^{-\sum_{i=1}^{k}} \prod_{i=1}^{k} \lambda_{i}^{y_{i}}}{\prod_{i=1}^{k} y_{i}!} \Big) \Big/ \frac{e^{-\sum_{i=1}^{k} \lambda_{i}}(\sum_{i}^{k} \lambda_{i})^{n}}{n!} \\ &= { n \choose y_{1}, y_{2}, …y_{k}} \frac{\prod_{i=1}^{k} \lambda_{i}^{y_{i}}} { \sum_{i}^{k} \lambda_{i})^{n}} \\ &= { n \choose y_{1}, y_{2}, …y_{k}} \prod_{i=1}^{k} \Big( \frac{ \lambda_{i} }{\sum_{i}^{k} \lambda_{i}} \Big)^{y_{i}} \\ &\sim MultiNom(n; \frac{\lambda_{1}}{\sum_{i=1}^{k}}, \frac{\lambda_{2}}{\sum_{i=1}^{k}}, … \frac{\lambda_{k}}{\sum_{i=1}^{k}}) \end{align}$$

So finally, we see that, given the sum of independent Poisson random variables, that conditional distribution of each element of the Poisson vector is Multinomial distributed, with count probabilities scaled by the sum of the individual rates. Importantly, we can extend these ideas (specifically the sum of independent Poisson random variables) to other models, such as splitting and merging homogenous and non-homogenous Poisson Point Processes.

Multivariate Normal Distribution

Sat, 12 May 2018 03:14:14 -0700

In this post, I’ll be covering the basics of Multivariate Normal Distributions, with special emphasis on deriving the conditional and marginal distributions.

Given a random variable under the usual Gauss-Markov assumptions, with $y_{i} \sim N(\mu, \sigma^{2})$ with $e \sim N(0,\sigma^{2})$, and $N$ independent samples $y_{1}…y_{n}$, we can define vector $\mathbf{y} = [y_{1}, y_{2},…y_{n}] \sim N_{n}(\mathbf{\mu},\sigma^{2}I)$ with $\mathbf{e} \sim N_{n}(\mathbf{0},\sigma^{2}I)$. We can see from the covariance structure of the errors that all off-diagonal elements are 0, indicating that our samples are independent with equal variances.

Marginal Distributions

Now assume that $\mathbf{y} = [\mathbf{y_{1}}, \mathbf{y_{2}} ] \sim N(\mathbf{\mu},\Sigma)$, where $\mathbf{\mu} = \begin{bmatrix} \mu_{1} \ \mu_{2} \end{bmatrix}$, and $\Sigma$ is an arbitrary covariance matrix, where we cannot assume independence. If $\Sigma$ is non-singular, we can decompose $\Sigma$ as

$$ \Sigma = \begin{bmatrix} \Sigma_{11} & \Sigma_{21}^{T} \\ \Sigma_{21} & \Sigma_{22} \end{bmatrix}$$

and, using the inversion lemmas from Blockwise Matrix Inversion, define its inverse $\Sigma^{-1} = V$ as

$$ V = \begin{bmatrix} V_{11} & V_{21}^{T} \\ V_{21} & V_{22} \\ \end{bmatrix} \begin{bmatrix} (\Sigma_{11} - \Sigma_{12}\Sigma_{22}^{-1}\Sigma_{21})^{-1} & -\Sigma^{-1}\Sigma_{12}(\Sigma_{22} - \Sigma_{21}\Sigma_{11}^{-1}\Sigma_{12})^{-1} \\ -\Sigma_{22}^{-1}\Sigma_{21}(\Sigma_{11}-\Sigma_{12}\Sigma_{22}^{-1}\Sigma_{21})^{-1} & (\Sigma_{22} - \Sigma_{21}\Sigma_{11}^{-1}\Sigma_{12})^{-1} \end{bmatrix}$$

From the properties of transformations of Normal random variables, we can define the marginal of $$y_{1}$$ as

$$\begin{align} By \sim N(B\mu,B\Sigma B^{T}) \end{align}$$

where $B = \begin{bmatrix} \mathbf{I} & 0 \end{bmatrix}$ such that

$$ \begin{bmatrix} \mathbf{I} & 0 \end{bmatrix} \begin{bmatrix} \mathbf{\mu_{1}} \ \mathbf{\mu_{2}} \end{bmatrix} = \mathbf{\mu_{1}}$$ $$\begin{bmatrix} \mathbf{I} & 0 \end{bmatrix} \begin{bmatrix} \Sigma_{11} & \Sigma_{12} \\ \Sigma_{21} & \Sigma_{22} \end{bmatrix} \begin{bmatrix} \mathbf{I} \\ 0 \end{bmatrix} = \Sigma_{11}$$

so that $\mathbf{y_{1}} \sim N(\mathbf{\mu_{1}},\Sigma_{11})$.

Conditional Distributions

Showing the conditional distribution is a bit long-winded, so bear with me. We are interested in finding the distribution of $y_{2}\mid y_{1}$, which we can explicitly represent as

$$\begin{align} f_{y_{1}}(y_{2} \mid y_{1}) = \frac{f_{y_{1},y_{2}}(y_{1},y_{2})}{f_{y_{1}}(y_{1})} \end{align}$$

Writing out the joint density for $y$, we have the following

$$\begin{align} f(y) = \frac{1}{(2\pi)^{n/2}\mid \Sigma \mid ^{1/2}}\exp^{(-1/2)(y-\mu)^{T}\Sigma^{-1}(y-\mu)} \end{align}$$

Partitioning this expression up into the individual terms related to $y_{1}$ and $y_{2}$, the exponent becomes

$$ (y-\mu)^{T}V(y-\mu) = \begin{bmatrix} y_{1} - \mu_{1} \\ y_{2} - \mu_{2} \end{bmatrix}^{T} \begin{bmatrix} V_{11} & V_{12} \\ V_{21} & V_{22} \end{bmatrix} \begin{bmatrix} y_{1} - \mu_{1} \\ y_{2} - \mu_{2} \end{bmatrix}$$

Expanding this quadratic form out, we see that we end up with

$$\begin{align} (y_{1} - \mu_{1})^{T} V_{11}^{-1}(y_{1}-\mu_{1}) + 2(y_{1}-\mu_{1})^{T}V_{12}(y_{2}-\mu_{2}) + (y_{2} - \mu_{2})^{T}V_{22}(y_{2}-\mu_{2}) \end{align}$$

Let us, for simplicity, set $z_{1} = (y_{1} - \mu_{1})$ and $z_{2} = (y_{2} - \mu_{2})$. Substituting back in our definitions of $V_{11}$,$V_{12}$,$V_{21}$, and $V_{22}$, and using the Sherman-Morrison-Woodbury definition for $V_{11}$, we have the following

$$\begin{align} &z_{1}^{T}(\Sigma_{11}^{-1} + \Sigma_{11}^{-1}\Sigma_{12}(\Sigma_{22} - \Sigma_{21}\Sigma_{11}^{-1}\Sigma_{12})^{-1}\Sigma_{21}\Sigma_{11})z_{1} \\ &- 2z_{1}^{T}(\Sigma_{11}^{-1}\Sigma_{12}(\Sigma_{22} - \Sigma_{21}\Sigma_{11}^{-1}\Sigma_{12}^{-1})^{-1})z_{2} \\ &+ z_{2}^{T}(\Sigma_{22} - \Sigma_{21}\Sigma_{11}^{-1}\Sigma_{12})^{-1})z_{2} \end{align}$$

which, by distribution of $z_{1}$ across the first term and splitting the second term into its two sums, we have

$$\begin{align} &z_{1}^{T}\Sigma_{11}^{-1}z_{11} + z_{1}^{T}\Sigma_{11}^{-1}\Sigma_{12}(\Sigma_{22} - \Sigma_{21}V_{11}^{-1}\Sigma_{12})^{-1}\Sigma_{21}\Sigma_{11}^{-1}z_{1} \\ &- z_{1}^{T}(\Sigma_{11}^{-1}\Sigma_{12}(\Sigma_{22} - \Sigma_{21}\Sigma_{11}^{-1}\Sigma_{12})^{-1})z_{2} - z_{1}^{T}(\Sigma_{11}^{-1}\Sigma_{12}(\Sigma_{22} - \Sigma_{21}\Sigma_{11}^{-1}\Sigma_{12})^{-1})z_{2} \\ &+ z_{2}^{T}(\Sigma_{22} - \Sigma_{21}\Sigma_{11}^{-1}\Sigma_{12})^{-1})z_{2} \end{align}$$

We can pull out forms $z_{1}^{T}\Sigma_{11}^{-1}\Sigma_{12}(\Sigma_{22}-\Sigma_{21}\Sigma_{11}^{-1}\Sigma_{12})^{-1}$ to the left and $(\Sigma_{22}-\Sigma_{21}\Sigma_{11}^{-1}\Sigma_{12})^{-1}z_{2}$ to the right and, after applying a transpose, have

$$\begin{align} =z_{1}^{T}\Sigma_{11}^{-1}z_{11} + (z_{2} -\Sigma_{21}\Sigma_{11}^{-1}z_{1})^{T}(\Sigma_{22} - \Sigma_{21}\Sigma_{11}^{-1}\Sigma_{12})^{-1}(z_{2} - \Sigma_{21}\Sigma_{11}^{-1}z_{1}) \end{align}$$

Plugging the above back into our exponential term in our original density function, we see that we have a product of two exponential terms

$$\begin{align} &\frac{1}{C_{1}} \exp(\frac{-1}{2}(z_{1}^{T}\Sigma_{11}^{-1}z_{11})) \\ \end{align}$$

and

$$\begin{align} &\frac{1}{C_{2}}\exp(\frac{-1}{2}(z_{2} - z_{1}\Sigma_{11}^{-1}\Sigma_{12})^{T}(\Sigma_{22} - \Sigma_{21}\Sigma_{11}^{-1}\Sigma_{12})^{-1}(z_{2} - \Sigma_{21}\Sigma_{11}^{-1}z_{1})) \end{align}$$

where

$$\begin{align} C_{1} &= (2\pi)^{p/2}\mid \Sigma_{11} \mid^{1/2} \\ C_{2} &= (2\pi)^{q/2}\mid \Sigma_{22} - \Sigma_{21}\Sigma_{11}^{-1}\Sigma_{12} \mid ^{1/2} \end{align}$$

The first term is the marginal density of $y_{1}$ and the second is the conditional density of $y_{2} \mid y_{1}$ with conditional mean $\mu_{2\mid 1} = \mu_{2} + \Sigma_{11}^{-1}\Sigma_{12}(y_{1} - \mu_{1})$ and conditional variance $\Sigma_{2\mid 1} = \Sigma_{22} - \Sigma_{21}\Sigma_{11}^{-1}\Sigma_{12}$.

While long and drawn out, the formulas show that the conditional distribution of any subset of Normal random variables, given another subset, is also a Normal distribution, with conditional mean and variance defined by functions of the means and covariances of the original random vector.

Blockwise Matrix Inversion

Tue, 08 May 2018 23:24:17 -0700

I’m taking a Statistics course on the theory of linear models, which covers Gauss-Markov models and various extensions of them. Sometimes, when dealing with partitioned matrices, and commonly Multivariate Normal Distributions, we’ll often need to invert matrices in a blockwise manner. This has happened often enough during this course (coincidentally was necessary knowledge for a midterm question), so I figured I should just document some of the inversion lemmas.

Let’s define our partitioned matrix as

$$ R = \begin{bmatrix} A & B \\ C & D \end{bmatrix}$$

We are specifically interested in finding

$$ R^{-1} = \begin{bmatrix} W & X \\ Y & Z \end{bmatrix}$$

such that

$$ R R^{-1} = R^{-1}R = \begin{bmatrix} I & 0 \\ 0 & I \end{bmatrix}$$

Part 1: $R R^{-1}$

For the right inverse ($R R^{-1}$), we can define

$$ \begin{aligned} AW + BY = I \\ AX + BZ = 0 \\ CW + DY = 0 \\ CX + DZ = I \\ \end{aligned} $$

and, assuming $A$ and $D$ are invertible,

$$\begin{aligned} X = -A^{-1}BZ \\ Y = -D^{-1}CW \\ \end{aligned}$$

We can plug these identities back into the first system of equations as

$$\begin{aligned} AW + B(-D^{-1}CW) &= (A - BD^{-1}C)W = I \\ C(-A^{-1}BZ) + DZ &= (D - CA^{-1}B)Z = I \\ \end{aligned}$$

so that

$$\begin{aligned} W = (A-BD^{-1}C)^{-1} \\ Z = (D-CA^{-1}B)^{-1} \\ \end{aligned}$$

and finally

$$ R^{-1} = \begin{bmatrix} W & X \\ Y & Z \end{bmatrix} = \begin{bmatrix} (A-BD^{-1}C)^{-1} & -A^{-1}B(D-CA^{-1}B)^{-1} \\ -D^{-1}C(A-BD^{-1}C)^{-1} & (D-CA^{-1}B)^{-1} \\ \end{bmatrix}$$

It is important to note that the above result only holds if $A$, $D$, $(D-CA^{-1}B)$, and $(A-BD^{-1}C)$ are invertible.

Part 2: $R^{-1} R$

Following the same logic as above, we have the following systems of equations for the left inverse ($R^{-1}R$)

$$\begin{aligned} WA + XC = I \\ WB + XD = 0 \\ YA + ZC = 0 \\ YB + ZD = I \\ \end{aligned}$$

so that

$$\begin{aligned} X = WBD^{-1} = A^{-1}BZ \\ Y = ZCA^{-1} = D^{-1}CW \\ \end{aligned}$$

which indicates that

$$\begin{aligned} W = (A-BD^{-1}C)^{-1} = C^{-1}D(D-CA^{-1}B)^{-1}CA^{-1} \\ X = (A-BD^{-1}C)^{-1}BD^{-1} = A^{-1}B(D-CA^{-1}B)^{-1} \\ \end{aligned}$$

Importantly, blockwise matrix inversion allows us to define the inverse of a larger matrix, with respect to its subcomponents. Likewise, from here, we can go on to derive the Sherman-Morrison formula and Woodbury theorem, which allow us to do all kinds of cool stuff, like rank-one matrix updates. In the next few posts, I’ll go over a few examples of where blockwise matrix inversions are useful, and common scenarios where rank-one updates of matrices are applicable.