# Thread: Problem with an exercise (moment generating function)

1. ## Problem with an exercise (moment generating function)

Hi everybody,

I attached a pdf file with an exercise (moment generating function). I dont understand two steps which are described in this file.

I would be very glad if someone could help.

Many thanks,
Fab

2. ## Re: Problem with an exercise (moment generating function)

• $\int_0^\infty t^{z-1}e^{-t} dt = \Gamma(z)$
• $\Gamma(r+1) = r!$ (if r is an integer)

$\int \frac{a^2 x_2^{r+1}}{r+1} e^{-ax_2} dx_2$

move 1 power of "a" outside the fraction
= $\int a\frac{a x_2^{r+1}}{r+1} e^{-ax_2} dx_2$

multiply top and bottom by $a^r$
= $\int a\frac{a^{r+1} x_2^{r+1}}{a^{r}(r+1)} e^{-ax_2} dx_2$
= $\int a\frac{(ax_2)^{r+1}}{a^{r}(r+1)} e^{-ax_2} dx_2$

substitute v=ax, noting that $dx = dv/a$

= $\int \frac{a}{a} \frac{v^{r+1}}{a^{r}(r+1)} e^{-v} dv$
= $\int 1 \frac{v^{r+1}}{a^{r}(r+1)} e^{-v} dv$

move constants outside the integral
= $\frac{1}{a^r(r+1)} \int \frac{v^{r+1}}{1} e^{-v} dv$

The integral is now in the same form as the definition of the gamma function. i haven't written the limits in but they are still 0 and infinity, as required.

= $\frac{1}{a^r(r+1)} \Gamma(r+2)$

now (since r is in integer) we can use this property of the gamma function: $\Gamma(r+2) = (r+1)!$

=> $\frac{1}{a^r(r+1)} (r+1)!$

note that $\frac{(r+1)!}{r+1}=r!$
=> $\frac{1}{a^r} r!$

= $\frac{r!}{a^r}$

3. ## Re: Problem with an exercise (moment generating function)

SpringFan25,

This helped me a lot. Thank you very much for the comprehensive answer!!!

Kind regards