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<!DOCTYPE html>
<html>
<head>
<title>Some Common Distributions</title>
<meta charset="utf-8">
<meta name="description" content="Some Common Distributions">
<meta name="author" content="Brian Caffo, Jeff Leek, Roger Peng">
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<body style="opacity: 0">
<slides class="layout-widescreen">
<!-- LOGO SLIDE -->
<slide class="title-slide segue nobackground">
<aside class="gdbar">
<img src="../../assets/img/bloomberg_shield.png">
</aside>
<hgroup class="auto-fadein">
<h1>Some Common Distributions</h1>
<h2>Statistical Inference</h2>
<p>Brian Caffo, Jeff Leek, Roger Peng<br/>Johns Hopkins Bloomberg School of Public Health</p>
</hgroup>
<article></article>
</slide>
<!-- SLIDES -->
<slide class="" id="slide-1" style="background:;">
<hgroup>
<h2>The Bernoulli distribution</h2>
</hgroup>
<article data-timings="">
<ul>
<li>The <strong>Bernoulli distribution</strong> arises as the result of a binary outcome</li>
<li>Bernoulli random variables take (only) the values 1 and 0 with probabilities of (say) \(p\) and \(1-p\) respectively</li>
<li>The PMF for a Bernoulli random variable \(X\) is \[P(X = x) = p^x (1 - p)^{1 - x}\]</li>
<li>The mean of a Bernoulli random variable is \(p\) and the variance is \(p(1 - p)\)</li>
<li>If we let \(X\) be a Bernoulli random variable, it is typical to call \(X=1\) as a "success" and \(X=0\) as a "failure"</li>
</ul>
</article>
<!-- Presenter Notes -->
</slide>
<slide class="" id="slide-2" style="background:;">
<hgroup>
<h2>iid Bernoulli trials</h2>
</hgroup>
<article data-timings="">
<ul>
<li>If several iid Bernoulli observations, say \(x_1,\ldots, x_n\), are observed the
likelihood is
\[
\prod_{i=1}^n p^{x_i} (1 - p)^{1 - x_i} = p^{\sum x_i} (1 - p)^{n - \sum x_i}
\]</li>
<li>Notice that the likelihood depends only on the sum of the \(x_i\)</li>
<li>Because \(n\) is fixed and assumed known, this implies that the sample proportion \(\sum_i x_i / n\) contains all of the relevant information about \(p\)</li>
<li>We can maximize the Bernoulli likelihood over \(p\) to obtain that \(\hat p = \sum_i x_i / n\) is the maximum likelihood estimator for \(p\)</li>
</ul>
</article>
<!-- Presenter Notes -->
</slide>
<slide class="" id="slide-3" style="background:;">
<hgroup>
<h2>Plotting all possible likelihoods for a small n</h2>
</hgroup>
<article data-timings="">
<pre><code>n <- 5
pvals <- seq(0, 1, length = 1000)
plot(c(0, 1), c(0, 1.2), type = "n", frame = FALSE, xlab = "p", ylab = "likelihood")
text((0 : n) /n, 1.1, as.character(0 : n))
sapply(0 : n, function(x) {
phat <- x / n
if (x == 0) lines(pvals, ( (1 - pvals) / (1 - phat) )^(n-x), lwd = 3)
else if (x == n) lines(pvals, (pvals / phat) ^ x, lwd = 3)
else lines(pvals, (pvals / phat ) ^ x * ( (1 - pvals) / (1 - phat) ) ^ (n-x), lwd = 3)
}
)
title(paste("Likelihoods for n = ", n))
</code></pre>
</article>
<!-- Presenter Notes -->
</slide>
<slide class="" id="slide-4" style="background:;">
<article data-timings="">
<p><img src="assets/fig/unnamed-chunk-1.png" alt="plot of chunk unnamed-chunk-1"> </p>
</article>
<!-- Presenter Notes -->
</slide>
<slide class="" id="slide-5" style="background:;">
<hgroup>
<h2>Binomial trials</h2>
</hgroup>
<article data-timings="">
<ul>
<li>The <em>binomial random variables</em> are obtained as the sum of iid Bernoulli trials</li>
<li>In specific, let \(X_1,\ldots,X_n\) be iid Bernoulli\((p)\); then \(X = \sum_{i=1}^n X_i\) is a binomial random variable</li>
<li>The binomial mass function is
\[
P(X = x) =
\left(
\begin{array}{c}
n \\ x
\end{array}
\right)
p^x(1 - p)^{n-x}
\]
for \(x=0,\ldots,n\)</li>
</ul>
</article>
<!-- Presenter Notes -->
</slide>
<slide class="" id="slide-6" style="background:;">
<hgroup>
<h2>Choose</h2>
</hgroup>
<article data-timings="">
<ul>
<li>Recall that the notation
\[\left(
\begin{array}{c}
n \\ x
\end{array}
\right) = \frac{n!}{x!(n-x)!}
\] (read "\(n\) choose \(x\)") counts the number of ways of selecting \(x\) items out of \(n\)
without replacement disregarding the order of the items</li>
</ul>
<p>\[\left(
\begin{array}{c}
n \\ 0
\end{array}
\right) =
\left(
\begin{array}{c}
n \\ n
\end{array}
\right) = 1
\] </p>
</article>
<!-- Presenter Notes -->
</slide>
<slide class="" id="slide-7" style="background:;">
<hgroup>
<h2>Example justification of the binomial likelihood</h2>
</hgroup>
<article data-timings="">
<ul>
<li>Consider the probability of getting \(6\) heads out of \(10\) coin flips from a coin with success probability \(p\) </li>
<li>The probability of getting \(6\) heads and \(4\) tails in any specific order is
\[
p^6(1-p)^4
\]</li>
<li>There are
\[\left(
\begin{array}{c}
10 \\ 6
\end{array}
\right)
\]
possible orders of \(6\) heads and \(4\) tails</li>
</ul>
</article>
<!-- Presenter Notes -->
</slide>
<slide class="" id="slide-8" style="background:;">
<hgroup>
<h2>Example</h2>
</hgroup>
<article data-timings="">
<ul>
<li>Suppose a friend has \(8\) children (oh my!), \(7\) of which are girls and none are twins</li>
<li>If each gender has an independent \(50\)% probability for each birth, what's the probability of getting \(7\) or more girls out of \(8\) births?
\[\left(
\begin{array}{c}
8 \\ 7
\end{array}
\right) .5^{7}(1-.5)^{1}
+
\left(
\begin{array}{c}
8 \\ 8
\end{array}
\right) .5^{8}(1-.5)^{0} \approx 0.04
\]</li>
</ul>
<pre><code class="r">choose(8, 7) * 0.5^8 + choose(8, 8) * 0.5^8
</code></pre>
<pre><code>## [1] 0.03516
</code></pre>
<pre><code class="r">pbinom(6, size = 8, prob = 0.5, lower.tail = FALSE)
</code></pre>
<pre><code>## [1] 0.03516
</code></pre>
</article>
<!-- Presenter Notes -->
</slide>
<slide class="" id="slide-9" style="background:;">
<article data-timings="">
<pre><code class="r">plot(pvals, dbinom(7, 8, pvals)/dbinom(7, 8, 7/8), lwd = 3, frame = FALSE, type = "l",
xlab = "p", ylab = "likelihood")
</code></pre>
<p><img src="assets/fig/unnamed-chunk-3.png" alt="plot of chunk unnamed-chunk-3"> </p>
</article>
<!-- Presenter Notes -->
</slide>
<slide class="" id="slide-10" style="background:;">
<hgroup>
<h2>The normal distribution</h2>
</hgroup>
<article data-timings="">
<ul>
<li>A random variable is said to follow a <strong>normal</strong> or <strong>Gaussian</strong> distribution with mean \(\mu\) and variance \(\sigma^2\) if the associated density is
\[
(2\pi \sigma^2)^{-1/2}e^{-(x - \mu)^2/2\sigma^2}
\]
If \(X\) a RV with this density then \(E[X] = \mu\) and \(Var(X) = \sigma^2\)</li>
<li>We write \(X\sim \mbox{N}(\mu, \sigma^2)\)</li>
<li>When \(\mu = 0\) and \(\sigma = 1\) the resulting distribution is called <strong>the standard normal distribution</strong></li>
<li>The standard normal density function is labeled \(\phi\)</li>
<li>Standard normal RVs are often labeled \(Z\)</li>
</ul>
</article>
<!-- Presenter Notes -->
</slide>
<slide class="" id="slide-11" style="background:;">
<article data-timings="">
<pre><code class="r">zvals <- seq(-3, 3, length = 1000)
plot(zvals, dnorm(zvals), type = "l", lwd = 3, frame = FALSE, xlab = "z", ylab = "Density")
sapply(-3:3, function(k) abline(v = k))
</code></pre>
<p><img src="assets/fig/unnamed-chunk-4.png" title="plot of chunk unnamed-chunk-4" alt="plot of chunk unnamed-chunk-4" style="display: block; margin: auto;" /></p>
</article>
<!-- Presenter Notes -->
</slide>
<slide class="" id="slide-12" style="background:;">
<hgroup>
<h2>Facts about the normal density</h2>
</hgroup>
<article data-timings="">
<ul>
<li>If \(X \sim \mbox{N}(\mu,\sigma^2)\) the \(Z = \frac{X -\mu}{\sigma}\) is standard normal</li>
<li>If \(Z\) is standard normal \[X = \mu + \sigma Z \sim \mbox{N}(\mu, \sigma^2)\]</li>
<li>The non-standard normal density is \[\phi\{(x - \mu) / \sigma\}/\sigma\]</li>
</ul>
</article>
<!-- Presenter Notes -->
</slide>
<slide class="" id="slide-13" style="background:;">
<hgroup>
<h2>More facts about the normal density</h2>
</hgroup>
<article data-timings="">
<ol>
<li>Approximately \(68\%\), \(95\%\) and \(99\%\) of the normal density lies within \(1\), \(2\) and \(3\) standard deviations from the mean, respectively</li>
<li>\(-1.28\), \(-1.645\), \(-1.96\) and \(-2.33\) are the \(10^{th}\), \(5^{th}\), \(2.5^{th}\) and \(1^{st}\) percentiles of the standard normal distribution respectively</li>
<li>By symmetry, \(1.28\), \(1.645\), \(1.96\) and \(2.33\) are the \(90^{th}\), \(95^{th}\), \(97.5^{th}\) and \(99^{th}\) percentiles of the standard normal distribution respectively</li>
</ol>
</article>
<!-- Presenter Notes -->
</slide>
<slide class="" id="slide-14" style="background:;">
<hgroup>
<h2>Question</h2>
</hgroup>
<article data-timings="">
<ul>
<li>What is the \(95^{th}\) percentile of a \(N(\mu, \sigma^2)\) distribution?
<ul>
<li>Quick answer in R <code>qnorm(.95, mean = mu, sd = sd)</code></li>
</ul></li>
<li>We want the point \(x_0\) so that \(P(X \leq x_0) = .95\)
\[
\begin{eqnarray*}
P(X \leq x_0) & = & P\left(\frac{X - \mu}{\sigma} \leq \frac{x_0 - \mu}{\sigma}\right) \\ \\
& = & P\left(Z \leq \frac{x_0 - \mu}{\sigma}\right) = .95
\end{eqnarray*}
\]</li>
<li>Therefore
\[\frac{x_0 - \mu}{\sigma} = 1.645\]
or \(x_0 = \mu + \sigma 1.645\)</li>
<li>In general \(x_0 = \mu + \sigma z_0\) where \(z_0\) is the appropriate standard normal quantile</li>
</ul>
</article>
<!-- Presenter Notes -->
</slide>
<slide class="" id="slide-15" style="background:;">
<hgroup>
<h2>Question</h2>
</hgroup>
<article data-timings="">
<ul>
<li>What is the probability that a \(\mbox{N}(\mu,\sigma^2)\) RV is 2 standard deviations above the mean?</li>
<li>We want to know
\[
\begin{eqnarray*}
P(X > \mu + 2\sigma) & = &
P\left(\frac{X -\mu}{\sigma} > \frac{\mu + 2\sigma - \mu}{\sigma}\right) \\ \\
& = & P(Z \geq 2 ) \\ \\
& \approx & 2.5\%
\end{eqnarray*}
\]</li>
</ul>
</article>
<!-- Presenter Notes -->
</slide>
<slide class="" id="slide-16" style="background:;">
<hgroup>
<h2>Other properties</h2>
</hgroup>
<article data-timings="">
<ul>
<li>The normal distribution is symmetric and peaked about its mean (therefore the mean, median and mode are all equal)</li>
<li>A constant times a normally distributed random variable is also normally distributed (what is the mean and variance?)</li>
<li>Sums of normally distributed random variables are again normally distributed even if the variables are dependent (what is the mean and variance?)</li>
<li>Sample means of normally distributed random variables are again normally distributed (with what mean and variance?)</li>
<li>The square of a <em>standard normal</em> random variable follows what is called <strong>chi-squared</strong> distribution </li>
<li>The exponent of a normally distributed random variables follows what is called the <strong>log-normal</strong> distribution </li>
<li>As we will see later, many random variables, properly normalized, <em>limit</em> to a normal distribution</li>
</ul>
</article>
<!-- Presenter Notes -->
</slide>
<slide class="" id="slide-17" style="background:;">
<hgroup>
<h2>Final thoughts on normal likelihoods</h2>
</hgroup>
<article data-timings="">
<ul>
<li>The MLE for \(\mu\) is \(\bar X\).</li>
<li>The MLE for \(\sigma^2\) is
\[
\frac{\sum_{i=1}^n (X_i - \bar X)^2}{n}
\]
(Which is the biased version of the sample variance.)</li>
<li>The MLE of \(\sigma\) is simply the square root of this
estimate</li>
</ul>
</article>
<!-- Presenter Notes -->
</slide>
<slide class="" id="slide-18" style="background:;">
<hgroup>
<h2>The Poisson distribution</h2>
</hgroup>
<article data-timings="">
<ul>
<li>Used to model counts</li>
<li>The Poisson mass function is
\[
P(X = x; \lambda) = \frac{\lambda^x e^{-\lambda}}{x!}
\]
for \(x=0,1,\ldots\)</li>
<li>The mean of this distribution is \(\lambda\)</li>
<li>The variance of this distribution is \(\lambda\)</li>
<li>Notice that \(x\) ranges from \(0\) to \(\infty\)</li>
</ul>
</article>
<!-- Presenter Notes -->
</slide>
<slide class="" id="slide-19" style="background:;">
<hgroup>
<h2>Some uses for the Poisson distribution</h2>
</hgroup>
<article data-timings="">
<ul>
<li>Modeling event/time data</li>
<li>Modeling radioactive decay</li>
<li>Modeling survival data</li>
<li>Modeling unbounded count data </li>
<li>Modeling contingency tables</li>
<li>Approximating binomials when \(n\) is large and \(p\) is small</li>
</ul>
</article>
<!-- Presenter Notes -->
</slide>
<slide class="" id="slide-20" style="background:;">
<hgroup>
<h2>Poisson derivation</h2>
</hgroup>
<article data-timings="">
<ul>
<li>\(\lambda\) is the mean number of events per unit time</li>
<li>Let \(h\) be very small </li>
<li>Suppose we assume that
<ul>
<li>Prob. of an event in an interval of length \(h\) is \(\lambda h\)
while the prob. of more than one event is negligible</li>
<li>Whether or not an event occurs in one small interval
does not impact whether or not an event occurs in another
small interval
then, the number of events per unit time is Poisson with mean \(\lambda\) </li>
</ul></li>
</ul>
</article>
<!-- Presenter Notes -->
</slide>
<slide class="" id="slide-21" style="background:;">
<hgroup>
<h2>Rates and Poisson random variables</h2>
</hgroup>
<article data-timings="">
<ul>
<li>Poisson random variables are used to model rates</li>
<li>\(X \sim Poisson(\lambda t)\) where
<ul>
<li>\(\lambda = E[X / t]\) is the expected count per unit of time</li>
<li>\(t\) is the total monitoring time</li>
</ul></li>
</ul>
</article>
<!-- Presenter Notes -->
</slide>
<slide class="" id="slide-22" style="background:;">
<hgroup>
<h2>Poisson approximation to the binomial</h2>
</hgroup>
<article data-timings="">
<ul>
<li>When \(n\) is large and \(p\) is small the Poisson distribution
is an accurate approximation to the binomial distribution</li>
<li>Notation
<ul>
<li>\(\lambda = n p\)</li>
<li>\(X \sim \mbox{Binomial}(n, p)\), \(\lambda = n p\) and</li>
<li>\(n\) gets large </li>
<li>\(p\) gets small</li>
<li>\(\lambda\) stays constant</li>
</ul></li>
</ul>
</article>
<!-- Presenter Notes -->
</slide>
<slide class="" id="slide-23" style="background:;">
<hgroup>
<h2>Example</h2>
</hgroup>
<article data-timings="">
<p>The number of people that show up at a bus stop is Poisson with
a mean of \(2.5\) per hour.</p>
<p>If watching the bus stop for 4 hours, what is the probability that \(3\)
or fewer people show up for the whole time?</p>
<pre><code class="r">ppois(3, lambda = 2.5 * 4)
</code></pre>
<pre><code>## [1] 0.01034
</code></pre>
</article>
<!-- Presenter Notes -->
</slide>
<slide class="" id="slide-24" style="background:;">
<hgroup>
<h2>Example, Poisson approximation to the binomial</h2>
</hgroup>
<article data-timings="">
<p>We flip a coin with success probablity \(0.01\) five hundred times. </p>
<p>What's the probability of 2 or fewer successes?</p>
<pre><code class="r">pbinom(2, size = 500, prob = 0.01)
</code></pre>
<pre><code>## [1] 0.1234
</code></pre>
<pre><code class="r">ppois(2, lambda = 500 * 0.01)
</code></pre>
<pre><code>## [1] 0.1247
</code></pre>
</article>
<!-- Presenter Notes -->
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data-slide=1 title='The Bernoulli distribution'>
1
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2
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3
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