06_StatisticalInference/01_03_Expectations/index.html

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    <h1>Expected values</h1>
    <h2>Statistical Inference</h2>
    <p>Brian Caffo, Jeff Leek, Roger Peng<br/>Johns Hopkins Bloomberg School of Public Health</p>
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  <hgroup>
    <h2>Expected values</h2>
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  <article data-timings="">
    <ul>
<li>The <strong>expected value</strong> or <strong>mean</strong> of a random variable is the center of its distribution</li>
<li>For discrete random variable \(X\) with PMF \(p(x)\), it is defined as follows
\[
E[X] = \sum_x xp(x).
\]
where the sum is taken over the possible values of \(x\)</li>
<li>\(E[X]\) represents the center of mass of a collection of locations and weights, \(\{x, p(x)\}\)</li>
</ul>

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<slide class="" id="slide-2" style="background:;">
  <hgroup>
    <h2>Example</h2>
  </hgroup>
  <article data-timings="">
    <h3>Find the center of mass of the bars</h3>

<p><img src="assets/fig/unnamed-chunk-1.png" alt="plot of chunk unnamed-chunk-1"> </p>

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<slide class="" id="slide-3" style="background:;">
  <hgroup>
    <h2>Using manipulate</h2>
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  <article data-timings="">
    <pre><code>library(manipulate)
myHist &lt;- function(mu){
  hist(galton$child,col=&quot;blue&quot;,breaks=100)
  lines(c(mu, mu), c(0, 150),col=&quot;red&quot;,lwd=5)
  mse &lt;- mean((galton$child - mu)^2)
  text(63, 150, paste(&quot;mu = &quot;, mu))
  text(63, 140, paste(&quot;Imbalance = &quot;, round(mse, 2)))
}
manipulate(myHist(mu), mu = slider(62, 74, step = 0.5))
</code></pre>

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<slide class="" id="slide-4" style="background:;">
  <hgroup>
    <h2>The center of mass is the empirical mean</h2>
  </hgroup>
  <article data-timings="">
    <pre><code class="r">hist(galton$child, col = &quot;blue&quot;, breaks = 100)
meanChild &lt;- mean(galton$child)
lines(rep(meanChild, 100), seq(0, 150, length = 100), col = &quot;red&quot;, lwd = 5)
</code></pre>

<p><img src="assets/fig/lsm.png" alt="plot of chunk lsm"> </p>

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<slide class="" id="slide-5" style="background:;">
  <hgroup>
    <h2>Example</h2>
  </hgroup>
  <article data-timings="">
    <ul>
<li>Suppose a coin is flipped and \(X\) is declared \(0\) or \(1\) corresponding to a head or a tail, respectively</li>
<li>What is the expected value of \(X\)? 
\[
E[X] = .5 \times 0 + .5 \times 1 = .5
\]</li>
<li>Note, if thought about geometrically, this answer is obvious; if two equal weights are spaced at 0 and 1, the center of mass will be \(.5\)</li>
</ul>

<p><img src="assets/fig/unnamed-chunk-2.png" alt="plot of chunk unnamed-chunk-2"> </p>

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<slide class="" id="slide-6" style="background:;">
  <hgroup>
    <h2>Example</h2>
  </hgroup>
  <article data-timings="">
    <ul>
<li>Suppose that a die is rolled and \(X\) is the number face up</li>
<li>What is the expected value of \(X\)?
\[
E[X] = 1 \times \frac{1}{6} + 2 \times \frac{1}{6} +
3 \times \frac{1}{6} + 4 \times \frac{1}{6} +
5 \times \frac{1}{6} + 6 \times \frac{1}{6} = 3.5
\]</li>
<li>Again, the geometric argument makes this answer obvious without calculation.</li>
</ul>

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<slide class="" id="slide-7" style="background:;">
  <hgroup>
    <h2>Continuous random variables</h2>
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  <article data-timings="">
    <ul>
<li>For a continuous random variable, \(X\), with density, \(f\), the expected
value is defined as follows
\[
E[X] = \mbox{the area under the function}~~~ t f(t)
\]</li>
<li>This definition borrows from the definition of center of mass for a continuous body</li>
</ul>

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<slide class="" id="slide-8" style="background:;">
  <hgroup>
    <h2>Example</h2>
  </hgroup>
  <article data-timings="">
    <ul>
<li>Consider a density where \(f(x) = 1\) for \(x\) between zero and one</li>
<li>(Is this a valid density?)</li>
<li>Suppose that \(X\) follows this density; what is its expected value?<br>
<img src="assets/fig/unnamed-chunk-3.png" alt="plot of chunk unnamed-chunk-3"> </li>
</ul>

  </article>
  <!-- Presenter Notes -->
</slide>

<slide class="" id="slide-9" style="background:;">
  <hgroup>
    <h2>Rules about expected values</h2>
  </hgroup>
  <article data-timings="">
    <ul>
<li>The expected value is a linear operator </li>
<li>If \(a\) and \(b\) are not random and \(X\) and \(Y\) are two random variables then

<ul>
<li>\(E[aX + b] = a E[X] + b\)</li>
<li>\(E[X + Y] = E[X] + E[Y]\)</li>
</ul></li>
</ul>

  </article>
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</slide>

<slide class="" id="slide-10" style="background:;">
  <hgroup>
    <h2>Example</h2>
  </hgroup>
  <article data-timings="">
    <ul>
<li>You flip a coin, \(X\) and simulate a uniform random number \(Y\), what is the expected value of their sum? 
\[
E[X + Y] = E[X] + E[Y] = .5 + .5 = 1
\] </li>
<li>Another example, you roll a die twice. What is the expected value of the average? </li>
<li>Let \(X_1\) and \(X_2\) be the results of the two rolls
\[
E[(X_1 + X_2) / 2] = \frac{1}{2}(E[X_1] + E[X_2])
= \frac{1}{2}(3.5 + 3.5) = 3.5
\]</li>
</ul>

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<slide class="" id="slide-11" style="background:;">
  <hgroup>
    <h2>Example</h2>
  </hgroup>
  <article data-timings="">
    <ol>
<li>Let \(X_i\) for \(i=1,\ldots,n\) be a collection of random variables, each from a distribution with mean \(\mu\)</li>
<li>Calculate the expected value of the sample average of the \(X_i\)
\[
\begin{eqnarray*}
E\left[ \frac{1}{n}\sum_{i=1}^n X_i\right]
& = & \frac{1}{n} E\left[\sum_{i=1}^n X_i\right] \\
& = & \frac{1}{n} \sum_{i=1}^n E\left[X_i\right] \\
& = & \frac{1}{n} \sum_{i=1}^n \mu =  \mu.
\end{eqnarray*}
\]</li>
</ol>

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<slide class="" id="slide-12" style="background:;">
  <hgroup>
    <h2>Remark</h2>
  </hgroup>
  <article data-timings="">
    <ul>
<li>Therefore, the expected value of the <strong>sample mean</strong> is the population mean that it&#39;s trying to estimate</li>
<li>When the expected value of an estimator is what its trying to estimate, we say that the estimator is <strong>unbiased</strong></li>
</ul>

  </article>
  <!-- Presenter Notes -->
</slide>

<slide class="" id="slide-13" style="background:;">
  <hgroup>
    <h2>The variance</h2>
  </hgroup>
  <article data-timings="">
    <ul>
<li>The variance of a random variable is a measure of <em>spread</em></li>
<li>If \(X\) is a random variable with mean \(\mu\), the variance of \(X\) is defined as</li>
</ul>

<p>\[
Var(X) = E[(X - \mu)^2]
\]</p>

<p>the expected (squared) distance from the mean</p>

<ul>
<li>Densities with a higher variance are more spread out than densities with a lower variance</li>
</ul>

  </article>
  <!-- Presenter Notes -->
</slide>

<slide class="" id="slide-14" style="background:;">
  <article data-timings="">
    <ul>
<li>Convenient computational form
\[
Var(X) = E[X^2] - E[X]^2
\]</li>
<li>If \(a\) is constant then \(Var(aX) = a^2 Var(X)\)</li>
<li>The square root of the variance is called the <strong>standard deviation</strong></li>
<li>The standard deviation has the same units as \(X\)</li>
</ul>

  </article>
  <!-- Presenter Notes -->
</slide>

<slide class="" id="slide-15" style="background:;">
  <hgroup>
    <h2>Example</h2>
  </hgroup>
  <article data-timings="">
    <ul>
<li><p>What&#39;s the sample variance from the result of a toss of a die? </p>

<ul>
<li>\(E[X] = 3.5\) </li>
<li>\(E[X^2] = 1 ^ 2 \times \frac{1}{6} + 2 ^ 2 \times \frac{1}{6} + 3 ^ 2 \times \frac{1}{6} + 4 ^ 2 \times \frac{1}{6} + 5 ^ 2 \times \frac{1}{6} + 6 ^ 2 \times \frac{1}{6} = 15.17\) </li>
</ul></li>
<li><p>\(Var(X) = E[X^2] - E[X]^2 \approx 2.92\)</p></li>
</ul>

  </article>
  <!-- Presenter Notes -->
</slide>

<slide class="" id="slide-16" style="background:;">
  <hgroup>
    <h2>Example</h2>
  </hgroup>
  <article data-timings="">
    <ul>
<li><p>What&#39;s the sample variance from the result of the toss of a coin with probability of heads (1) of \(p\)? </p>

<ul>
<li>\(E[X] = 0 \times (1 - p) + 1 \times p = p\)</li>
<li>\(E[X^2] = E[X] = p\) </li>
</ul></li>
<li><p>\(Var(X) = E[X^2] - E[X]^2 = p - p^2 = p(1 - p)\)</p></li>
</ul>

  </article>
  <!-- Presenter Notes -->
</slide>

<slide class="" id="slide-17" style="background:;">
  <hgroup>
    <h2>Interpreting variances</h2>
  </hgroup>
  <article data-timings="">
    <ul>
<li>Chebyshev&#39;s inequality is useful for interpreting variances</li>
<li>This inequality states that
\[
P(|X - \mu| \geq k\sigma) \leq \frac{1}{k^2}
\]</li>
<li>For example, the probability that a random variable lies beyond \(k\) standard deviations from its mean is less than \(1/k^2\)
\[
\begin{eqnarray*}
2\sigma & \rightarrow & 25\% \\
3\sigma & \rightarrow & 11\% \\
4\sigma & \rightarrow &  6\% 
\end{eqnarray*}
\]</li>
<li>Note this is only a bound; the actual probability might be quite a bit smaller</li>
</ul>

  </article>
  <!-- Presenter Notes -->
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<slide class="" id="slide-18" style="background:;">
  <hgroup>
    <h2>Example</h2>
  </hgroup>
  <article data-timings="">
    <ul>
<li>IQs are often said to be distributed with a mean of \(100\) and a sd of \(15\)</li>
<li>What is the probability of a randomly drawn person having an IQ higher than \(160\) or below \(40\)?</li>
<li>Thus we want to know the probability of a person being more than \(4\) standard deviations from the mean</li>
<li>Thus Chebyshev&#39;s inequality suggests that this will be no larger than 6\%</li>
<li>IQs distributions are often cited as being bell shaped, in which case this bound is very conservative</li>
<li>The probability of a random draw from a bell curve being \(4\) standard deviations from the mean is on the order of \(10^{-5}\) (one thousandth of one percent)</li>
</ul>

  </article>
  <!-- Presenter Notes -->
</slide>

<slide class="" id="slide-19" style="background:;">
  <hgroup>
    <h2>Example</h2>
  </hgroup>
  <article data-timings="">
    <ul>
<li>A former buzz phrase in industrial quality control is Motorola&#39;s &quot;Six Sigma&quot; whereby businesses are suggested to control extreme events or rare defective parts</li>
<li>Chebyshev&#39;s inequality states that the probability of a &quot;Six Sigma&quot; event is less than \(1/6^2 \approx 3\%\)</li>
<li>If a bell curve is assumed, the probability of a &quot;six sigma&quot; event is on the order of \(10^{-9}\) (one ten millionth of a percent)</li>
</ul>

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