how to calculate E(1/x) and E(1/X^2)when given E(X)=3

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- Dec 9th 2009, 12:28 AMrajrhow to calculate E(1/x)
how to calculate E(1/x) and E(1/X^2)when given E(X)=3

- Dec 9th 2009, 12:30 AMrajr
Am i right in saying e(1/x)=1/e(x)=1/3 thank you

- Dec 9th 2009, 02:20 AMMoo
No, that's false ;)

And without any further information, it's not possible to calculate E[1/X] - Dec 9th 2009, 02:23 AMrajr
Sorry I should have explained the question in detail

X~Gamma(a,3)

i.e. am i right in saying the inverse of the distribution is 1/X~Gamma(a,1/3).

Therefore E(1/X)=1/3a and Var(1/X)=1/3(a^2)

Thanks for your response - Dec 9th 2009, 02:31 AMMoo
No you are not...

For any bounded function g, we have $\displaystyle E[g(X)]=\int_{\mathbb{R}} f(x)g(x) ~dx$

where f is the pdf of X.

So just substitute here :

$\displaystyle E[1/X]=\int_{\mathbb{R}} \tfrac 1x \cdot f(x) ~dx$

I didn't explicitly write f, the pdf of a Gamma distribution, because there exist 2 versions of it. - Dec 9th 2009, 03:42 AMmr fantastic
Does it agree with what is here: Inverse-gamma distribution - Wikipedia, the free encyclopedia