For $X$ a gamma random variable with parameters $\alpha >2$ and $\beta> 0$

a) Prove that the mean of $\dfrac1X$ is $\dfrac{\beta}{\alpha-1}$.

b) Prove that the variance of $\dfrac1X$ is $\dfrac{\beta^2}{(\alpha-1)^2 (\alpha - 2)}$.

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What did you try? Can you write a density? Which integral do have to compute? – Davide Giraudo Sep 13 '12 at 12:57

As a digression, if $X$ follows gamma distribution, its reciprocal $X^{-1}$ follows inverse gamma distribution.
By the definition of the expectation, since $X$ is non-negative: $$\mathbb{E}\left(\frac{1}{X}\right) = \int_0^\infty \frac{1}{x} f_X(x) \mathrm{d}x$$ Recall that $f_X(x) = C_{\alpha, \beta} x^{\alpha-1} \mathrm{e}^{-x \beta}$, for some normalization constant $C_{\alpha, \beta}$. Use it in the integral above: $$\mathbb{E}\left(\frac{1}{X}\right) = C_{\alpha, \beta} \int_0^\infty \frac{1}{x} \cdot x^{\alpha-1} \mathrm{e}^{-x \beta} \mathrm{d}x = C_{\alpha, \beta} \int_0^\infty x^{\alpha-2} \mathrm{e}^{-x \beta} \mathrm{d}x$$ The latter integral is the Euler integral of the second kind. With simplification you should arrive at the requested expression for item a).
For the item b), note that $$\mathbb{Var}\left(\frac{1}{X}\right) = \mathbb{E}\left(\left(\frac{1}{X}\right)^2\right) - \mathbb{E}\left(\frac{1}{X}\right)^2$$ and repeat calculations above: $$\mathbb{E}\left(\left(\frac{1}{X}\right)^2\right) = C_{\alpha, \beta} \int_0^\infty x^{\alpha-3} \exp(-x \beta) \mathrm{d}x$$