Per wikipedia the Laplace transform of the gamma distribution is $$L_X(s) = (1+\theta s)^{-k} = \frac{\beta^\alpha}{(s+\beta)^\alpha}$$ As an exercise I would like to show this.The definition I have of the Laplace transform is: $$L_X(t) = \mathbb{E}[e^{-tX}]=\int_0^\infty e^{-Xt}f(t)\mathrm{dt}$$ where in this case the function $f(t)$ is the pdf of the Gamma distribution. There are two forms, which I guess correspond to the two versions of the Laplace transform on Wikipedia. These are

$$\frac{1}{\Gamma(k)\theta^k}x^{k-1}e^-\frac{x}{\theta} \:\:\:\text{ and }\:\:\: \frac{\beta^\alpha}{\Gamma(\alpha)}x^{\alpha-1}e^{-\beta x}$$ where $\Gamma(m) = \int_0^\infty x^{m-1}e^{-x}\mathrm{dx}$ is the gamma function. I think the second version might be have simpler notation so I combine the above as carefully as I can:

$$L_X(t) = \int_0^\infty e^{-xt} \frac{\beta^\alpha}{\int_0^\infty y^{\alpha-1}e^{-y}\mathrm{dy} }x^{\alpha-1}e^{-\beta x}\mathrm{dt} = \beta^\alpha\int_0^\infty \frac{ x^{\alpha-1} e^{-x(t+\beta)}}{\int_0^\infty y^{\alpha-1}e^{-y}\mathrm{dy} }\mathrm{dt} $$ I am not sure if I should have, or need, different variables for both each integral. I have no idea how to proceed. Do we have to assume alpha is an integer to get a factorial underneath? Or can we just ignore it? I guess if I am right to use a different variable, then I can just leave it as $\Gamma(\alpha)$ and hope it cancels later?

Trying this: (and noticing something that looks like a Gamma function to hopefully factor...)$$ \frac{\beta^\alpha}{\Gamma(\alpha)} \int_0^\infty x^{\alpha-1} e^{-x(t+\beta)}\mathrm{dt} = \frac{\beta^\alpha}{\Gamma(\alpha)} \int_0^\infty x^{\alpha-1}e^{-x} e^{(t+\beta)}\mathrm{dt} = \beta^\alpha \int_0^\infty e^{t+\beta} \mathrm{dt} $$

I am a bit dubious that I can take the gamma function out like that, the integral of a product isn't necessarily the product of the integrals $\int x^2\mathrm{dx}\neq \int x\mathrm{dx} \cdot \int x \mathrm{dx}$.

I am also skeptical that the remaining integral should be wrt x. My calculus is rusty and I might have lost track of what variable I am using.

So I think I spotted an error in the integration variable and have edited to be wrt t instead of x. I will try to continue...

$$\beta^\alpha\big[ e^{t+\beta} \big]_0^\infty$$

So I am definitely stuck now and can not see my mistake, maybe from spending too long looking at it, but I would appreciate any help.

  • $\begingroup$ Umm, your random variable is $X$, right? Why are you integrating over $t$? Further, if you are integrating over $t$, the equalities just above the struck out bit are incorrect. $\endgroup$ Dec 5, 2015 at 15:43

2 Answers 2


$$\mathbb{E}\left[e^{-sX} \right] = \int_{\mathbb{R}} e^{-sx} \frac{\beta^\alpha}{\Gamma(\alpha)}x^{\alpha-1}e^{-\beta x} \mathrm{d}x = \frac{\beta^\alpha}{\Gamma(\alpha)(\beta + s)^\alpha}\underbrace{\int_{\mathbb{R}} \left((\beta+s)x\right)^{\alpha-1}e^{-(\beta+s) x} \mathrm{d}\left((\beta+s)x\right)}_{=\Gamma(\alpha), \text{ use the substitution $u = (\beta + s) x $}} = \frac{\beta^\alpha}{(\beta + s)^\alpha}$$

  • $\begingroup$ With the substitution u = (b+s)x is it OK to immediately just say du = d((b+s)x). Is it really as simple as just applying the 'd' operator to both sides? $\endgroup$ Dec 5, 2015 at 16:01
  • $\begingroup$ Well, $\beta + s$ is a constant over the domain. Things are a little bit complicated in that the domain of integration is a line not parallel to the real line in the complex plane, but things work out just the same. You might want to read more about the Gamma function if you'd like to know why, the wiki page is a decent start. $\endgroup$ Dec 5, 2015 at 16:33

If the pdf of the distribution is $$f_{\alpha, \beta}(x) = \frac{\beta^\alpha}{\Gamma(\alpha)} x^{\alpha-1} e^{-\beta x},$$ then by definition, the distribution $\Gamma_{\alpha, \beta}$ defined by $$\Gamma_{\alpha, \beta}(A) = \int_A f_{\alpha, \beta}(x) \: d\lambda(x) = \int_A \frac{\beta^\alpha}{\Gamma(\alpha)} x^{\alpha-1} e^{-\beta x} \: d\lambda(x)$$ is a probability measure on the measurable space $([0,\infty), \mathcal B)$, where $\mathcal B$ is the Borel $\sigma$-algebra on $[0,\infty)$ and $\lambda$ is the one-dimensional Lebesgue measure. In particular, what this implies is $$\int_0^\infty \frac{\beta^\alpha}{\Gamma(\alpha)} x^{r-1} e^{-\beta x} \: dt = \Gamma_{\alpha, \beta}([0,\infty)) = 1$$ for any $\alpha, \beta > 0$. Then, if $X$ is a Gamma-distributed random variable with parameters $\alpha$ and $\beta$, \begin{align} L_X(s) &= \int_0^\infty e^{-sx} f_{\alpha, \beta}(x) \: dx = \int_0^\infty \frac{\beta^\alpha}{\Gamma(\alpha)} x^{\alpha-1} e^{-(s+\beta)x} \: dx \\ &= \frac{\beta^\alpha}{(s+\beta)^\alpha} \int_0^\infty \frac{(\beta+s)^\alpha}{\Gamma(\alpha)} x^{\alpha-1} e^{-(s+\beta)x} \: dx = \frac{\beta^\alpha}{(s+\beta)^\alpha}, \end{align} where we have used the fact that this integral equals $1$ (using $\beta+s$ instead of $\beta$, for $s \geq 0$, the domain of the Laplace transform).


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