Probability- moment generating functions.

[C.E]

1. A random variable Y is called gamma($$\theta$$, n) for $$\theta$$>0 and natural n if it takes positive values and takes the following PDF:

f(y)=$$\frac{1}{\theta (n-1)!}$$($$\frac{y}{\theta}$$)$$^{n-1}$$ exp$$\frac{-y}{\theta}$$

Show how to find the moment generating function, expectation and variance of Y.

3. I have not got very far I am stil stuck on finding the moment generating function.

I know it is given by the following:

$$\int_{0}^{\infty} \exp(ty)f(y) dy$$

but I have no idea how to evaluate it and get the correct answer (can somebody please show me?)

The answer you should get is: G$$_{y}$$(t)=1/(1-t$$\theta$$)$$^{n}$$


I think I will know how to find the expectation and variance once I have the moment generating function.

 

[Billy Bob]

Substitute $$u=\frac{(1-\theta t) y}{\theta}$$ .

(The restriction on t is that $$1-\theta t>0$$ .)

 

[C.E]

I do not see how that helps, could someone please elaborate?

 

[Billy Bob]

Write down the integral you are trying to evaluate.

Don't just write "f(y)." Write down what f(y) is.

You now have two factors with e to a power. Combine them into one factor of e to a power, using properties of exponents.

Now make a u-substitution, the one I suggested, as in calc 1 or calc 2.

Write this out and show us how far you can get.

 

[C.E]

Ok this is as far as I keep getting:

$$\int_{0}^{\infty} e^{\frac{-y}{\theta}}e^{tx} {\frac{y^{n-1}}{(t\theta-1)^{n-1}}{\frac{1}{\theta(n-1)!} dy$$

=$$\int_{0}^{\infty} e^{\frac{y(t\theta-1)}{\theta}}{\frac{y^{n-1}}{\theta^{n-1}}{\frac{1}{\theta(n-1)!}dy$$

let u=$${\frac{y(t\theta-1)}{\theta}$$

Then du=$${\frac{(t\theta-1)}{\theta}$$dy

=$$\int_{0}^{\infty} {\frac{1}{(t\theta-1)(n-1)!}}{\frac{u^{n-1}}{(t\theta-1)^{n-1}}e^u du$$

 

[C.E]

= $$\frac{1}{(t\theta-1)^n}$$ $$\int_{0}^{\infty} e^u \frac{u^{n-1}}{(n-1)!}$$

Is this right so far? Is there some reason why $$\int_{0}^{\infty} e^u \frac{u^{n-1}}{(n-1)!}$$ should be one?

 

[Billy Bob]

C.E, that's mostly pretty good!

Here is an important technical detail. The m.g.f. is defined for t near 0, so $$\theta t -1$$ is negative. This means that for the substitution you used, y=infinity would become u=-infinity. Thus, your integral with du should become from 0 to -infinity.

Now you can find $$\int u^{n-1} e^u\,du$$ using integration by parts. Or use a table of integrals. The first time you use IBP you obtain a formula involving $$\int u^{n-2} e^u\,du$$. Then do IBP again and you get $$\int u^{n-3} e^u\,du$$. Pretty soon you'll see what happens with the factorial.

Do the IBP with definite integrals, $$\int_a^b u\,dv=uv\bigr|_a^b-\int_a^b v\,du$$ so that you can take care of those "uv" terms at each step.

 

[C.E]

How do you know that the MGF is for t near to 0 (it does not say in the question)? I assume from the limits you told me it is for 0<t<1?

 

[Billy Bob]

MGF is always for t near 0, because that's where you need it to find the mean and moments. M'(0), M''(0), etc.

$$\theta t -1<0$$ i.e. $$1-\theta t>0$$ will give you $$t<1/ \theta$$, and this last inequality includes t=0.

You need to use that condition to get your integral du from 0 to -infinity, and the -infinity will be used to get convergence of the improper integral.