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<!DOCTYPE html>
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<title>A Gradient-Based Approach to Solving Mixed Binomial Distributions</title>
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<h1>A Gradient-Based Approach to Solving Mixed Binomial Distributions</h1>
<p class="date">Posted on September 4, 2024 • 5-minute read</p>
<figure class="cover">
<img src="blog-posts/binomial-cover.jpg" alt="A chalkboard with math">
</figure>
<section class="toc">
<h2>Contents</h2>
<div>
<ul>
<li><a href="#Problem">Problem Statement</a></li>
<li><a href="#Solution">Solution: Gradient Descent</a></li>
<li><a href="#Algorithm">Algorithm</a></li>
</ul>
</div>
</section>
<section id="Problem">
<h2>Problem Statement</h2>
<p>Let the probability mass function (PMF) of a binomial distribution \(B(n, p)\) be given by</p>
<div class="equation">
\begin{equation}
f(k;n,p)=\left( \begin{array}{c}
n \\
k
\end{array} \right)p^k (1-p)^{n-k}
\end{equation}
</div>
<p>Given a number of independent Bernoulli trials, \(n\), a number of distributions, \(m\), and an observed PMF</p>
<div class="equation">
\begin{equation}
t(k)=\frac{1}{m}\sum_{i=1}^m f(k;n,p_i)
\end{equation}
</div>
<p>compute \(p_i\ \forall i\in (1,...,m)\).</p>
</section>
<section id="Solution">
<h2>Solution: Gradient Descent</h2>
<p>Draw \(m\) values, \(p_i'\), from a uniform distribution between 0 and 1.</p>
<p>Let \(t'(k)\) be a PMF given by</p>
<div class="equation">
\begin{equation}
t'(k)=\frac{1}{m}\sum_{i=1}^m f(k;n,p_i')
\end{equation}
</div>
<p>Let \(J\) represent a cost function on the sum of the squared errors between \(t(k)\) and \(t'(k)\):</p>
<div class="equation">
\begin{equation}
J(p_1,...,p_m)=\sum_{j=0}^n \left( t(j)-t'(j) \right)^2
\end{equation}
</div>
<p>We can now restate the problem as a minimization of \(J\):</p>
<div class="equation">
\begin{equation}
\argmin_{p_1,...,p_m} J(p_1,...,p_m)
\end{equation}
<!-- (5) \label{eq:problem} -->
</div>
<p>We can solve this problem using gradient descent by computing the gradient of \(J\). For each \(p_l\):</p>
<div class="equation">
\begin{equation}
\begin{array}{rl}
\frac{\partial J}{\partial p_l} & =\frac{\partial}{\partial p_l} \sum_{j=0}^n \left( t(j)-\frac{1}{m}\sum_{i=1}^m f(k;n,p_i') \right)^2 \\
& =\sum_{j=0}^n \frac{\partial}{\partial p_l} \left( t(j)-\frac{1}{m}\sum_{i=1}^m f(k;n,p_i') \right)^2 \\
& =2\left( t(j)-\frac{1}{m}\sum_{i=1}^m f(k;n,p_i') \right)\left( -\frac{1}{m}\sum_{i=1}^m \frac{\partial}{\partial p_l}f(j;n,p_i) \right)
\end{array}
\end{equation}
<!-- (6) \label{eq:grad-J} -->
</div>
<p>Note that \(f(j;n,p_i)\) is a function of \(p_l\) if and only if \(i=l\). Therefore,</p>
<div class="equation">
\begin{equation}
\frac{\partial}{\partial p_l}f(j;n,p_i)=\begin{cases}
\begin{array}{ll}
\left. \frac{\partial}{\partial p} f(j;n,p) \right|_{p=p_l}& \mathrm{if}\ i=l \\
0 & \mathrm{otherwise}
\end{array}
\end{cases}
\end{equation}
<!-- (7) \label{eq:bin-derv-cases} -->
</div>
<p>Computing the derivative of the binomial distribution:</p>
<div class="equation">
\begin{equation}
\begin{array}{rl}
\frac{\partial}{\partial p} f(j;n,p) & =\frac{\partial}{\partial p} \left( \begin{array}{c} n \\ j \end{array} \right)p^j (1-p)^{n-j} \\
& = \left( \begin{array}{c} n \\ j \end{array} \right) \frac{\partial}{\partial p} \left( p^j (1-p)^{n-j} \right)\\
& = \left( \begin{array}{c} n \\ j \end{array} \right) \left( p^j\frac{\partial}{\partial p} \left((1-p)^{n-j}\right) + (1-p)^{n-j}\frac{\partial}{\partial p}(p^j) \right) \\
& = \left( \begin{array}{c} n \\ j \end{array} \right) \left(-p^j(n-j)(1-p)^{n-j-1}+(1-p)^{n-j}kp^{j-1} \right) \\
& = \left( \begin{array}{c} n \\ j \end{array} \right)p^{j-1}(1-p)^{n-j-1}(j-pn)
\end{array}
\end{equation}
<!-- (8) \label{eq:bin-derv} -->
</div>
<p>Plugging our solution from Eq. (8) into Eq. (7) yields:</p>
<div class="equation">
\begin{equation}
\frac{\partial}{\partial p_l}f(j;n,p_i)=\begin{cases}
\begin{array}{ll}
\left( \begin{array}{c} n \\ j \end{array} \right)p_l^{j-1}(1-p_l)^{n-j-1}(j-p_ln) & \mathrm{if}\ i=l \\
0 & \mathrm{otherwise}
\end{array}
\end{cases}
\end{equation}
<!-- (9) \label{eq:cases-solved} -->
</div>
<p>Plugging our solution from Eq. (9) into Eq. (6) yields:</p>
<div class="equation">
\begin{equation}
\frac{\partial J}{\partial p_l}=\frac{2}{m}\sum_{j=0}^n\left( t(j)-\frac{1}{m}\sum_{i=1}^m f(j;n,p_i) \right)\left( \begin{array}{c} n \\ j \end{array} \right)p_l^{j-1}(1-p_l)^{n-j-1}(p_ln-j)
\end{equation}
</div>
</section>
<section id="Algorithm">
<h2>Algorithm</h2>
<p>Now that we have computed the gradient for \(J\), we can apply any gradient-based optimization algorithm to solve Eq. (5). For example, applying gradient descent with step size \(\alpha\) and stop condition \(\epsilon\):</p>
<div class="algorithm">
<div class="state">Given inputs \(t(k)\), \(m\), \(n\), \(\alpha\), and \(\epsilon\):</div>
<div class="state">\(i \gets 1\)</div>
<div class="state"><b>while</b> \(i \leq m\) <b>do</b>:</div>
<div class="indented">
<div class="state">\(p_i \sim U_{[0,1]}\)</div>
<div class="state">\(i\gets i+1\)</div>
</div>
<div class="state">\(D\gets \infty\)</div>
<div class="state"><b>while</b> \(D\geq \epsilon\) <b>do</b>:</div>
<div class="indented">
<div class="state">\(t'(k)\gets \sum_{i=1}^m f(k;n,p_i')\)</div>
<div class="state">\(l\gets 1\)</div>
<div class="state"><b>while</b> \(l\leq m\) <b>do</b>:</div>
<div class="indented">
<div class="state">\(D_l\gets \alpha\sum_{j=0}^n\left( t(j)-\frac{1}{m}\sum_{i=1}^m f(j;n,p_i) \right)\left( \begin{array}{c} n \\ j \end{array} \right)p_l^{j-1}(1-p_l)^{n-j-1}(p_ln-j)\)</div>
<div class="state">\(l\gets l+1\)</div>
</div>
<div class="state">\(l\gets 1\)</div>
<div class="state"><b>while</b> \(l\leq m\) <b>do</b>:</div>
<div class="indented">
<div class="state">\(p_l\gets p_l-D_l\)</div>
</div>
<div class="state">\(D\gets \sum_{i=1}^m D_i^2\)</div>
</div>
</div>
<p>The output of the algorithm is \(p_i\ \forall i\in (1,...,m)\), a list of probabilities defining the \(m\) binomial distributions whose average most closely matches \(t(k)\).</p>
<p>For unknown \(m\), plot \(J\) against \(m\) and select an appropriate cutoff point after which the difference between \(t(k)\) and \(t(j)\) is considered small enough. Alternatively, apply a penalty to \(m\), e.g. \(\lambda m\), and find the minimum of \(J+\lambda m\) iteratively.</p>
</section>
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