If a Gamma distribution is parameterized with $\alpha$ and $\beta$, then:

$$ E(\Gamma(\alpha, \beta)) = \frac{\alpha}{\beta} $$

I would like to calculate the expectation of a squared Gamma, that is:

$$ E(\Gamma(\alpha, \beta)^2) = ? $$

I think it is:

$$ E(\Gamma(\alpha, \beta)^2) = \left(\frac{\alpha}{\beta}\right)^2 + \frac{\alpha}{\beta^2} $$

Does anyone know if this latter expression is correct?

  • 1
    $\begingroup$ This was related to a simulation study I am working on where I am drawing standard deviations from a Gamma, and then wanted the mean of the variances (i.e., squared Gammas). $\endgroup$
    – Joshua
    Commented Aug 6, 2014 at 13:54

2 Answers 2


The expectation of the square of any random variable is its variance plus its expectation squared, as

$\mathbb{D}^2(X)=\mathbb{E}([X-\mathbb{E}(X)]^2)=\mathbb{E}(X^2)-[\mathbb{E}(X)]^2 \Rightarrow \mathbb{E}(X^2) = \mathbb{D}^2(X)+[\mathbb{E}(X)]^2$.

The expectation of the $\Gamma$-distribution parametrized as above is $\alpha/\beta$ (like you mentioned), the variance is $\alpha/\beta^2$, hence, the expectation of its square is


That is: you are right.

  • $\begingroup$ I appreciate the response, although I am not sure I follow your equation---if you follow it through D2(X) ends up equaling D2(X) + E(X)^2 $\endgroup$
    – Joshua
    Commented Aug 6, 2014 at 13:57
  • 3
    $\begingroup$ That line is not a single equation! Note the arrow in the middle. The first part (on the left side of the arrow) is one equation which implies the second equation (on the right side of the arrow). (By adding $[\mathbb{E}(X)]^2$ to both sides.) $\endgroup$ Commented Aug 6, 2014 at 14:00

For the sake of completeness, I will directly compute the raw moments from the density. First, under a shape/rate parametrization, the gamma distribution has density $$f_X(x) = \frac{\beta^\alpha x^{\alpha-1} e^{-\beta x}}{\Gamma(\alpha)}, \quad x > 0.$$ We will take for granted that for any choice of parameters $\alpha, \beta > 0$, we have $$\int_{x=0}^\infty f_X(x) \, dx = 1,$$ although this result is easily derived from the identity $$\int_{z=0}^\infty x^{z-1} e^{-z} \, dz = \Gamma(z).$$ Then it follows that for a positive integer $k$, $$\begin{align*} \mathrm{E}[X^k] &= \int_{x=0}^\infty x^k f_X(x) \, dx \\ &= \frac{1}{\Gamma(\alpha)} \int_{x=0}^\infty \beta^\alpha x^{\alpha+k-1} e^{-\beta x} \, dx \\ &= \frac{\Gamma(\alpha+k)}{\beta^k \Gamma(\alpha)} \int_{x=0}^\infty \frac{\beta^{\alpha+k} x^{\alpha+k-1} e^{-\beta x}}{\Gamma(\alpha+k)} \, dx \\ &= \frac{\Gamma(\alpha+k)}{\beta^k \Gamma(\alpha)}, \end{align*}$$ where in the penultimate step we observe that the integral equals $1$ because it is the integral of a gamma density with parameters $\alpha+k$ and $\beta$. For $k = 2$, we immediately obtain $\mathrm{E}[X^2] = \frac{\Gamma(\alpha+2)}{\beta^2 \Gamma(\alpha)} = \frac{(\alpha+1)\alpha}{\beta^2}.$ Another approach is via the moment generating function: $$\begin{align*} M_X(t) = \mathrm{E}[e^{tX}] &= \int_{x=0}^\infty \frac{\beta^\alpha x^{\alpha-1} e^{-\beta x + tx}}{\Gamma(\alpha)} \, dx \\ &= \frac{\beta^\alpha}{(\beta-t)^\alpha} \int_{x=0}^\infty \frac{(\beta-t)^\alpha x^{\alpha-1} e^{-(\beta-t)x}}{\Gamma(\alpha)} \, dx \\ &= \biggl(\frac{\beta}{\beta-t}\biggr)^{\!\alpha}, \quad t < \beta, \end{align*}$$ where the condition on $t$ is required for the integral to converge. We may rewrite this as $$M_X(t) = (1 - t/\beta)^{-\alpha},$$ and it follows that $$\mathrm{E}[X^k] = \left[ \frac{d^k M_X(t)}{dt^k} \right]_{t=0} = \left[(1-t/\beta)^{-\alpha-k}\right]_{t=0} \prod_{j=0}^{k-1} \frac{\alpha+j}{\beta} = \frac{\Gamma(\alpha+k)}{\beta^k \Gamma(\alpha)}.$$

  • $\begingroup$ Very clear, and helpful derivation. $\endgroup$
    – Joshua
    Commented Aug 7, 2014 at 22:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.