Let X be uniform on the interval [a, b].
@@ -587,6 +588,57 @@
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+ Let R be the value of a fair 6-sided die roll.
+ We've seen in that \E(R) = \frac{7}{2}.
+ To find \E\left(R^2\right):
+
+ \E\left(R^2\right) \amp = 1^2 \left(\frac{1}{6}\right) + 2^2 \left(\frac{1}{6}\right) + 3^2 \left(\frac{1}{6}\right) + 4^2 \left(\frac{1}{6}\right) + 5^2 \left(\frac{1}{6}\right) + 6^2 \left(\frac{1}{6}\right)
+ \amp = 1 \left(\frac{1}{6}\right) + 4 \left(\frac{1}{6}\right) + 9 \left(\frac{1}{6}\right) + 16 \left(\frac{1}{6}\right) + 25 \left(\frac{1}{6}\right) + 36 \left(\frac{1}{6}\right)
+ \amp = \frac{91}{6}.
+
+ Then
+
+ \Var(R) = \E\left(R^2\right) - \left(\E(R)\right)^2 = \frac{91}{6} - \frac{49}{4} = \frac{35}{12}.
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+ Notice that in the \E\left(R^2\right) calculation, we're just taking the basic structure of the \E(R) calculation and squaring the R values.
+ The analogous calculation for a continuous random variable looks similar:
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+ Let X be uniform on the interval [a, b].
+ So X has the pdf f(x) = \frac{1}{b - a}, and we saw in that \E(X) = \frac{a + b}{2}.
+ Now:
+
+ \E\left(X^2\right) \amp = \int_a^b x^2 f(x)\ dx
+ \amp = \int_a^b \frac{x^2}{b - a} \ dx
+ \amp = \frac{1}{b-a} \left(\frac{x^3}{3}\right)\bigg|_a^b
+ \amp = \frac{1}{b-a} \left( \frac{b^3}{3} - \frac{a^3}{3} \right)
+ \amp = \frac{1}{b - a}\left( \frac{(b - a)(a^2 + ab + b^2)}{3}\right)
+ \amp = \frac{a^2 + ab + b^2}{3}.
+
+ Then:
+
+ \Var(X) \amp = \E\left(X^2\right) - \left(\E(X)\right)^2
+ \amp = \frac{a^2 + ab + b^2}{3} - \frac{(a+b)^2}{4}
+ \amp = \frac{a^2 + ab + b^2}{3} - \frac{a^2 + 2ab + b^2}{4}
+ \amp = \frac{4a^2 + 4ab + 4b^2}{12} - \frac{3a^2 + 6ab + 3b^2}{12}
+ \amp = \frac{a^2 - 2ab + b^2}{12}
+ \amp = \frac{(a - b)^2}{12}.
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