Erlang Distribution - 109 Challenge Project

bookOutline.hjson

@@ -48,6 +48,7 @@
 			"continuous":"Continuous Distribution",
 			"uniform":"Uniform Distribution",
 			"exponential":"Exponential Distribution",
+			"erlang":"Erlang Distribution",
 			"normal":"Normal Distribution",
 			"binomial_approx":"Binomial Approximation"
 		},

chapters/part2/erlang/index.html

@@ -0,0 +1,43 @@
+
+% rebase('templates/chapter.html', title="Erlang Distribution")
+
+<center><h1>Erlang Distribution</h1></center>
+<hr/>
+
+<p>
+    If events are occurring sequentially with the same mean rate of occurrence after each event,
+    an Erlang random variable measures the amount of time until the $k^{th}$ event occurs.
+    The random variable is the summation of $k$ independent and identically distributed (IID) Exponential random variables.
+</p>
+
+<%
+  include('templates/rvCards/erlang.html')
+%>
+<div style="height: 20px"></div>
+
+<p>
+If you set $k$ equal to 50 up above, that is the equivalent of summing together 50 Exponential random variables with a mean rate of $\lambda$. Notice
+how the resulting PDF resembles that of a Gaussian. We will explore why that is when we cover the <a href="{{pathToLang}}part4/clt/">Central Limit Theorem</a>.
+</p>
+
+<div class="purpleBox">
+<p><b><i>Example</i></b>:
+    Agner Krarup Erlang wants to reward the 10th customer who walks into his gag foam-phone store, "Phoney Foam Phones".
+    Agner needs 15 minutes to prepare the prize. Since a YouTube video went viral showcasing how foam-phones can be used to clean dishes,
+    the store has become surprisingly popular. People are walking in at a mean rate
+    of one person per minute. What is the probability that Erlang will have the time to prepare his prize?
+</p>
+    Let $X$ be the number of minutes until the 10th customer walks in. Since a customer walking in is an event that is
+    occurring at a mean rate of once per minute, $X \sim {\rm Erlang}(k = 10, \lambda = 1)$.
+    The question is asking us to calculate P(X > 15):
+
+    \begin{align*}
+    P(X > 15) &= 1 - P(X < 15) \\
+    &= 1 - F_{X}(15) \\
+    &= 1 - (1 - \sum_{n=0}^{10-1}{\frac{1}{n!} e^{-1 \cdot 15} (1 \cdot 15)^{n}}) \\
+    &= 1 - (1 - \sum_{n=0}^{9}{\frac{1}{n!} e^{-15} (15)^{n}}) \\
+    &= 0.070
+    \end{align*}
+
+    Sorry Agner!
+</div>

templates/chapterList.html

@@ -47,6 +47,7 @@
 <a id=sidebar-continuous href="{{pathToLang}}part2/continuous">Continuous Distribution</a>
 <a id=sidebar-uniform href="{{pathToLang}}part2/uniform">Uniform Distribution</a>
 <a id=sidebar-exponential href="{{pathToLang}}part2/exponential">Exponential Distribution</a>
+<a id=sidebar-erlang href="{{pathToLang}}part2/erlang">Erlang Distribution</a>
 <a id=sidebar-normal href="{{pathToLang}}part2/normal">Normal Distribution</a>
 <a id=sidebar-binomial_approx href="{{pathToLang}}part2/binomial_approx">Binomial Approximation</a>
 

templates/rvCards/erlang.html

@@ -0,0 +1,104 @@
+<!-- Template that renders a single random variable card -->
+
+<div class="bordered">
+<p><b>Erlang Random Variable</b></p>
+
+
+<table>
+<tbody class="rvCardBody">
+	<tr>
+		<th style="width:150px">Notation:</td>
+		<td>$X \sim {\rm Erlang}(k, \lambda)$</td>
+	</tr>
+	<tr>
+		<th>Description:</td>
+		<td>Time until $k^{th}$ event occurs if (a) the events occur with a constant mean rate and (b) they occur independently of time since last event.</td>
+	</tr>
+	<tr>
+		<th>Parameters:</td>
+		<td>$k \in \{1, 2, \dots\}$, occurrence of event <br/>
+			$\lambda \in \{0, 1, \dots\}$, the constant average rate</td>
+	</tr>
+
+
+	<tr>
+		<th>Support:</td>
+		<td>$x \in \mathbb{R}^+$</td>
+	</tr>
+	<tr>
+		<th>PDF equation:</th>
+		<td class="mathLeft">$$f(x) = \frac{\lambda^{k}x^{k-1}e^{- \lambda x}}{(k-1)!}$$</td>
+	</tr>
+	<tr>
+		<th>CDF equation:</th>
+		<td class="mathLeft">$$F(x) = 1 - \sum_{n=0}^{k-1}{\frac{1}{n!} e^{-\lambda x} (\lambda x )^{n}}$$</td>
+	</tr>
+	<tr>
+		<th>Expectation:</th>
+		<td>$\E[X] = k/\lambda$</td>
+	</tr>
+	<tr>
+		<th>Variance:</th>
+		<td>$\var(X) = k/\lambda^2$</td>
+	</tr>
+	<tr>
+		<th>PDF graph:</th>
+	</tr>
+</tbody>
+</table>
+<div class="d-flex">
+	<div class="mr-2">
+Parameter $k$: <input onchange="redrawErlangPdf()" type="number" class="form-control example-inline-input mb-3" id="erlangPdfK" min="1" max="130" value="5" step = "1">
+</div>
+	<div class="mr-2">
+Parameter $\lambda$: <input onchange="redrawErlangPdf()" type="number" class="form-control example-inline-input mb-3" id="erlangPdfLambda" min="0" max="130" value="5" step = "1">
+</div>
+</div>
+<canvas id="exponentialPdf" style="max-height: 400px"></canvas>
+</div>
+
+<script>
+
+$(document).ready(function(){
+	redrawErlangPdf()
+})
+
+function redrawErlangPdf() {
+	let parentDivId = 'exponentialPdf'
+	let k = parseFloat($("#erlangPdfK").val())
+	let lambda = parseFloat($("#erlangPdfLambda").val())
+	drawErlangPdf(parentDivId, k, lambda)
+}
+
+function drawErlangPdf(parentDivId, k, lambda) {
+	if(window.myExpLine) {
+		window.myExpLine.destroy()
+	}
+
+	var expectation = k / lambda
+	var xValues = []
+	var yValues = []
+	var i = 0
+	if (k < 1 || lambda <= 0) {
+		xValues.push(0)
+		yValues.push(0)
+	}
+	else {
+		while (true) {
+			let pr_x = jStat.gamma.pdf(i, k, 1 / lambda).toFixed(5)
+			xValues.push(i.toFixed(3))
+			yValues.push(pr_x)
+			i += expectation / 100
+			if (pr_x < 0.001 && expectation < i) {
+				break;
+			}
+		}
+	}
+	let xLabel = 'Values that X can take on'
+	let yLabel = 'Probability'
+
+	var config = standardPDFConfig(xValues, yValues, xLabel, yLabel)
+	var ctx = document.getElementById(parentDivId).getContext('2d');
+	window.myExpLine = new Chart(ctx, config);
+}
+</script>