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I have another problem in a statistics past paper that goes as follows:

Let $X$ be a continuous random variable, taking values in the range $[0,1]$ with pdf given up to proportionality by $f(x) \sim x^4$, $0 \le x \le 1$. What is the value of $E(X)$?

Normally $E(X)$ should be $$\int_0^1 xf(x) ~dx,$$ right? In which case $E(X)$ would be $1/6$. But for some reason that's not the answer in the marking scheme. Does the keyword "proportionality" have anything to do with it? If so, what?

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In case the $f$ is not known to be normalized $\mathbb{E}(X) = \frac{\int_0^1 x f(x) dx}{\int_0^1 f(x) dx}$. – Sasha Jul 24 '12 at 23:58
up vote 4 down vote accepted

Begin as follows. the density is of the form $x\mapsto cx^4$. It must integrate to 1, so $$1 = \int_0^1 cx^4\,dx = {c\over 5}.$$ We conclude that $c = 5,$ so the density is $$f(x) = 5x^4 \qquad 0\le x\le 1.$$ Now you can compute the expectation.

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Thanks. I forgot I could do the integral = 1 bit. – Sorin Cioban Jul 25 '12 at 6:16

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