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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 $E(X^k)$?

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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
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3 Answers 3

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

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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: $$ \left. \frac{\mathrm{d}^k}{\mathrm{d} t^k} M_X(t) \right|_{t=0} = \mathbb{E}\left( X^k \right) $$

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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 $(d/dt)^k t^j |_{t=0} = j! \delta_{jk}$.

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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
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