Finding $\mathbb{E}[X^s e^{tX}]$ where $X \sim \mathrm{Gamma} (\alpha, \beta)$

Question

Question

Find $\mathbb{E}[X^s e^{tX}]$, where $X \sim \mathrm{Gamma} (\alpha, \beta)$ and $\alpha$, $\beta$ are the shape and scale parameters respectively. Specify also the real values of $s$ and $t$ for which the required expectation exists.

My working

Let $Y \sim \mathrm{Gamma} (\alpha, 1)$, then $X = \beta Y$ and $f_Y(y) = \frac {y^{\alpha - 1} e^{-y}} {\Gamma(\alpha)}\ \forall\ y \in \mathbb{R}^+$.

$\begin{aligned}[t] \mathbb{E}[X^s e^{tX}] & = \mathbb{E}[(\beta Y)^s e^{t\beta Y}] \\[1 mm] & = \beta^s \int^{\infty}_{-\infty} y^s e^{t\beta y} f_Y(y)\ \mathrm{d}y \\[1 mm] & = \frac {\beta^s} {\Gamma(\alpha)} \int^{\infty}_{0} y^{\alpha - 1 + s} e^{\left(t\beta - 1\right)y}\ \mathrm{d}y \end{aligned}$

However, this is where I am stuck. I am having difficulty in evaluating the integral and finding the values of $s$ and $t$ for which there is convergence. I have tried relating the integral to the gamma function, but to no avail, in particular because of the exponential term in the integrand.

Answer 1

Are you allowed to just evaluate the integral?

$$\int_0^\infty y^{\alpha-1}e^{-\beta y}dy=\frac{\Gamma(\alpha)}{\beta^\alpha}$$ so

$$\int_0^\infty y^{\alpha+s-1}e^{(t\beta-1)y}dy=\frac{\Gamma(\alpha+s)}{(1-t\beta)^{\alpha+s}}$$

It must be that

$$\alpha+s>0\rightarrow s>-\alpha\\ 1-t\beta>0\rightarrow t<\frac 1 \beta$$

Answer 2

If $\beta$ is supposed to be the "scale parameter" the density is the following.

$$f_X(x)=\frac{1}{\Gamma{(\alpha)}\beta^{\alpha}}x^{\alpha-1}e^{-x/\beta}$$

thus your expectation is

$$\mathbb{E}[X^s e^{tX}]=\frac{\Gamma{(a+s)}}{\Gamma{(\alpha)}\beta^{\alpha}\left(\frac{1}{\beta}-t\right)^{\alpha+s}}\underbrace{\int_0^{\infty}\frac{\left(\frac{1}{\beta}-t\right)^{\alpha+s}}{\Gamma{(a+s)}}x^{[(\alpha+s)-1]} e^{-(1/\beta-t)x}dx}_{=1}$$

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further details on the solution

$$\mathbb{E}[X^s e^{tX}]=\frac{1}{\Gamma{(\alpha)}\beta^{\alpha}}\int_0^{\infty}x^{[(\alpha+s)-1]} e^{-(1/\beta-t)x}dx$$

Observe that the integrand is the kernel of a Gamma distribution with different parameter that you can easy recognize. Thus all you have to do is to fit the integrand with the missing parmeters' expressions and if you muliply for something, in the meantime divide out of the integral for the same quantity...very very easy

Using the definition of Gamma distribution you can immediately specify the parameters' range