Finding the mean of the gamma distribution

1. Jan 22, 2012

thomas49th

Finding the mean of the gamma distribution ("reparameterise")

1. The problem statement, all variables and given/known data
Find the mean of the gamma distribution.
$$f_{X}(x) = \frac{e^{-kx}x^{r-1}k^{r}}{(r-1)!} x>0, r ε N*, k>0$$

2. Relevant equations
Interestingly this is the alternative form of the gamma distribution

3. The attempt at a solution

I didn't really know where to begin, so peeked at the answers. Apparently I have to reparameterise r→r+1. Can someone explain to me what this means? How does it differ from a substitution? Is there a way to solve it without reparameterising and how do I spot when to reparameterise?

Thanks
Thomas

Last edited: Jan 22, 2012
2. Jan 23, 2012

thomas49th

anybody?

3. Jan 23, 2012

LCKurtz

Re: Finding the mean of the gamma distribution ("reparameterise")

So$$E(X) = \int_0^\infty xf_X(x)\,dx=\int_0^\infty \frac{e^{-kx}x^{r}k^{r}}{(r-1)!}\, dx$$Try a change of variable $u=kx$ and remember the definition of $\Gamma(r)$ and that $\Gamma(r)=(r-1)!$.

4. Jan 23, 2012

awkward

Re: Finding the mean of the gamma distribution ("reparameterise")

Here's a hint for an easy no-calculus solution. By definition, the mean is the integral of

$$x f_{X}(x) = \frac{e^{-kx}x^{r}k^{r}}{(r-1)!} = \frac{e^{-kx}x^{r}k^{r+1}}{r!} \cdot \frac{r}{k}$$

Hmm... isn't that the pdf of a gamma distribution with parameters k and r+1, times r/k?

And doesn't a pdf integrate to 1?