0
$\begingroup$

So my text book on probability theory gives an example of a distribution with random parameters, which after a few paragraphs ends up with this integral:

$$ \begin{split} \int_0^{\infty}\frac{x^k}{k!}e^{-2x}dx &= \frac{1}{2^{k+1}} \int_0^{\infty}\frac{1}{\Gamma(k+1)}2^{k+1}x^{k+1-1}e^{-2x}dx\\ &= \frac{1}{2^{k+1}} \cdot 1 \end{split} $$

where $\Gamma$ is the Gamma function.

My question is this: How in the world did the author calculate this integral?


For those of you who are curious, that integral is the result of calculating $P(X=k|M=x)$ with $M \in \text{Exp}(1)$ and $X|M=m \in \text{Po}(m)$. The example can be found in An intermediate course in Probability, 2nd Ed. (Gut, 2009, pp 39).

$\endgroup$

1 Answer 1

4
$\begingroup$

Just doing the substitution $u=2x$ and using $$\Gamma (t)=\int_0^\infty x^{t-1}e^{-x}\,\mathrm d x.$$

$\endgroup$
4
  • $\begingroup$ I don't see how you can use that equivalence to solve the integral. Can you elaborate? $\endgroup$
    – Mossmyr
    Sep 18, 2019 at 12:59
  • $\begingroup$ $$\int_0^\infty x^ke^{-2x}\,\mathrm d x\underset{u=2x}{=}\frac{1}{2^{k+1}}\int_0^\infty u^ke^{-u}\,\mathrm d u=\frac{1}{2^{k+1}}\Gamma (k+1).$$@Mossmyr $\endgroup$
    – Surb
    Sep 18, 2019 at 13:02
  • $\begingroup$ Nooow I see. Then you can cancel the $\frac{1}{\Gamma(k+1)}$ found in the equation in my post. But I think you meant to write $\frac{1}{2^k}$ in your comment. $\endgroup$
    – Mossmyr
    Sep 18, 2019 at 13:11
  • 1
    $\begingroup$ No, it's $\frac{1}{2^{k+1}}$ (that come from the fact that $dx=\frac{1}{2}du$) $\endgroup$
    – Surb
    Sep 18, 2019 at 13:14

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .