# Thread: Expected value of a complicated density function

1. ## Expected value of a complicated density function

Let X be a randon variable with density function:

$\displaystyle f(x) = \frac {cx^6}{1+7x^{20}} \ \ \ \ \ -1 \leq x \leq 1$

and $\displaystyle f(x) = 0 \ \ \ otherwise$

where c is such that this is a density.

Find $\displaystyle E[X]$

First I tried to take the integral of $\displaystyle \int _{-1}^{1} \frac {cx^6}{1+7x^{20}} = 1$ to find c, but I'm stuck here unless I get to use a calculator.

Are there anyway to find c? Or rather, do I need to find c in order to find the expected value, that is, $\displaystyle \int _{-1}^{1} \frac {cx^7}{1+7x^{20}}$

2. Use partial fractions to integrate?

3. Hello,

Why being complicated ?

$\displaystyle E(X)=\int_{-1}^1 \frac{cx^7}{1+7x^{20}} ~dx$

since we integrate an odd function over a symmetric interval over 0, the integral is 0 !

More generally, if the pdf of a rv is an even function, then all the odd moments of the rv are 0

4. Originally Posted by Moo
Hello,

Why being complicated ?

$\displaystyle E(X)=\int_{-1}^1 \frac{cx^7}{1+7x^{20}} ~dx$

since we integrate an odd function over a symmetric interval over 0, the integral is 0 !

More generally, if the pdf of a rv is an even function, then all the odd moments of the rv are 0
There's a restriction required on this statement (otherwise what are we to make of the Cauchy Distribution ....?)

5. Oh, indeed !
Then if the moment exists... which is okay here since there's a continuous function in a compact.

So it doesn't work for the Cauchy distribution because the x*pdf near infinity is equivalent to 1/x, which is not integrable.

(hey I hate justifying things like that ^^)