Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If we have a moment generating function, $M_X(t)=(1-\frac{t}{\alpha})^{-\beta}$ where $\alpha$ is any positive number $\alpha\in\mathbb{R}$ and $\beta$ is any positive integer, how do we find $E(X^k)$?

share|cite|improve this question
Would you kindly review answers extended to you in 25 out of 37 questions and accept some of them. – Sasha Apr 26 '12 at 20:14

Differentiate $(1-\frac{t}{\alpha})^{-\beta}$ $k$ times with respect to $t$, then set $t=0$.

share|cite|improve this answer

The moment generating function $M_X(t)$ is defined as $\mathbb{E}\left(\mathrm{e}^{t X}\right)$. Then, assuming the interchanging of expectation and differentiation is warranted $$ \frac{\mathrm{d}^k}{\mathrm{d} t^k} M_X(t) = \mathbb{E}\left( \frac{\mathrm{d}^k}{\mathrm{d} t^k} \mathrm{e}^{t X} \right) = \mathbb{E}\left( X^k \mathrm{e}^{t X} \right) $$ Evaluating the derivative at $t=0$ gives: $$ \left. \frac{\mathrm{d}^k}{\mathrm{d} t^k} M_X(t) \right|_{t=0} = \mathbb{E}\left( X^k \right) $$

share|cite|improve this answer

As has been mentioned, $$\mathbb{E}\left( X^k \right) = \left. \frac{d ^k}{d t^k} M_X(t) \right|_{t=0}.$$ This follows directly from the definition of the moment-generating function.

For your particular $M_X$ this can be computed in closed form. First, note that
$$\begin{eqnarray*} \left(1-\frac{t}{\alpha}\right)^{-\beta} &=& \sum_{j=0}^\infty {-\beta\choose j} \left(- \frac{t}{\alpha}\right)^j \\ &=& \sum_{j=0}^\infty \frac{(-1)^j(\beta)_j}{j!} \left(- \frac{t}{\alpha}\right)^j \\ &=& \sum_{j=0}^\infty \frac{(\beta)_j}{\alpha^j} \frac{t^j}{j!}, \end{eqnarray*}$$ where $(\beta)_j = \Gamma(\beta+j)/\Gamma(\beta) = \beta(\beta+1)\cdots(\beta+j-1)$ is Pochhammer's symbol. (This is the rising factorial, in notation commonly used for special functions.) Of course, we analytically continue ${n\choose m} = n!/\left(m!(n-m)!\right)$ to $$\frac{\Gamma(n+1)}{\Gamma(m+1)\Gamma(n-m+1)}.$$ Therefore, $$\begin{eqnarray*} \mathbb{E}\left( X^k \right) &=& \left. \frac{d ^k}{d t^k} \left(1-\frac{t}{\alpha}\right)^{-\beta} \right|_{t=0} \\ &=& \frac{(\beta)_k}{\alpha^k} \end{eqnarray*}$$ since $(d/dt)^k t^j |_{t=0} = j! \delta_{jk}$.

share|cite|improve this answer
You don't need to actually take the derivative: given the Maclaurin series $M_X(t) = \sum_{k=0}^\infty a_k t^k$, ${\mathbb E}(X^k) = a_k k!$, since $\mathbb{E}(e^{tX}) = \mathbb{E} \left( \sum_{k=0}^\infty t^k X^k/k! \right) = \sum_{k=0}^\infty \mathbb{E}(X^k) t^k/k!$ – Robert Israel Apr 26 '12 at 23:14
@RobertIsrael: Of course, one way or another we are just looking at the coefficient of $t^k/k!$. – user26872 Apr 26 '12 at 23:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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