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
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
You need to know these 2 facts to follow this answer:
- $\displaystyle \int_0^\infty t^{z-1}e^{-t} dt = \Gamma(z)$
- $\displaystyle \Gamma(r+1) = r!$ (if r is an integer)
Answer
$\displaystyle \int \frac{a^2 x_2^{r+1}}{r+1} e^{-ax_2} dx_2$
move 1 power of "a" outside the fraction
=$\displaystyle \int a\frac{a x_2^{r+1}}{r+1} e^{-ax_2} dx_2$
multiply top and bottom by $\displaystyle a^r$
=$\displaystyle \int a\frac{a^{r+1} x_2^{r+1}}{a^{r}(r+1)} e^{-ax_2} dx_2$
=$\displaystyle \int a\frac{(ax_2)^{r+1}}{a^{r}(r+1)} e^{-ax_2} dx_2$
substitute v=ax, noting that $\displaystyle dx = dv/a$
=$\displaystyle \int \frac{a}{a} \frac{v^{r+1}}{a^{r}(r+1)} e^{-v} dv$
=$\displaystyle \int 1 \frac{v^{r+1}}{a^{r}(r+1)} e^{-v} dv$
move constants outside the integral
=$\displaystyle \frac{1}{a^r(r+1)} \int \frac{v^{r+1}}{1} e^{-v} dv$
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.
=$\displaystyle \frac{1}{a^r(r+1)} \Gamma(r+2)$
now (since r is in integer) we can use this property of the gamma function: $\displaystyle \Gamma(r+2) = (r+1)!$
=>$\displaystyle \frac{1}{a^r(r+1)} (r+1)!$
note that $\displaystyle \frac{(r+1)!}{r+1}=r!$
=>$\displaystyle \frac{1}{a^r} r!$
=$\displaystyle \frac{r!}{a^r}$