Let the stochastic variable X be evenly distributed over the interval $[0,2π]$ calculate:
1) $E[\cos(X)]$
2) $E[\sin(X)^2]$
How does the integral look?
Does it look like this:
$\int_{0}^{2π}\frac{1}{2π}\cos(\frac{1}{2π})dx$
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asked Sep 20, 2015 at 12:15
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$\begingroup$ Did you check the formula for E(g(X)) when X has PDF f? Or is your problem to find f in the present case? $\endgroup$– DidSep 20, 2015 at 12:17
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$\begingroup$ You mean $\int_0^{2\pi}\frac{1}{2\pi}\cos(x)dx$? $\endgroup$– AugustinSep 20, 2015 at 12:17
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1$\begingroup$ If you are confused, a more satisfying way to evade your confusion would be to write down the formula for E(g(X)) when X has density f. Can you do that? $\endgroup$– DidSep 20, 2015 at 12:28
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1$\begingroup$ @Mathguy007 Almost. What is the $A$ standing for? $\endgroup$– drhabSep 20, 2015 at 12:34
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1$\begingroup$ If $f$ is defined as in the second comment of @Augustin then you can write the expectation as $\int g(x)f(x)dx$. That is the way you should think about it. A second step: since $f(x)=0$ if $x\notin[0,2\pi]$ and $f(x)=\frac1{2\pi}$ this integral equals $\int_{[0,2\pi]} g(x)\frac1{2\pi}dx$. $\endgroup$– drhabSep 20, 2015 at 12:42
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