• Support PF! Buy your school textbooks, materials and every day products Here!

Probability- moment generating functions.

  • Thread starter C.E
  • Start date
  • #1
C.E
102
0
1. A random variable Y is called gamma([tex]\theta[/tex], n) for [tex]\theta[/tex]>0 and natural n if it takes positive values and takes the following PDF:

f(y)=[tex]\frac{1}{\theta (n-1)!}[/tex]([tex]\frac{y}{\theta}[/tex])[tex]^{n-1}[/tex] exp[tex]\frac{-y}{\theta}[/tex]

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:

[tex]\int_{0}^{\infty} \exp(ty)f(y) dy[/tex]

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[tex]_{y}[/tex](t)=1/(1-t[tex]\theta[/tex])[tex]^{n}[/tex]


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

Answers and Replies

  • #2
392
0
Substitute [tex]u=\frac{(1-\theta t) y}{\theta} [/tex] .

(The restriction on t is that [tex]1-\theta t>0[/tex] .)
 
  • #3
C.E
102
0
I do not see how that helps, could someone please elaborate?
 
  • #4
392
0
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.
 
  • #5
C.E
102
0
Ok this is as far as I keep getting:

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

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

let u=[tex]{\frac{y(t\theta-1)}{\theta}[/tex]

Then du=[tex]{\frac{(t\theta-1)}{\theta}[/tex]dy

=[tex]\int_{0}^{\infty} {\frac{1}{(t\theta-1)(n-1)!}}{\frac{u^{n-1}}{(t\theta-1)^{n-1}}e^u du[/tex]
 
  • #6
C.E
102
0
= [tex]\frac{1}{(t\theta-1)^n}[/tex] [tex]\int_{0}^{\infty} e^u \frac{u^{n-1}}{(n-1)!}[/tex]

Is this right so far? Is there some reason why [tex]\int_{0}^{\infty} e^u \frac{u^{n-1}}{(n-1)!}[/tex] should be one?
 
  • #7
392
0
C.E, that's mostly pretty good!

Here is an important technical detail. The m.g.f. is defined for t near 0, so [tex]\theta t -1[/tex] 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 [tex]\int u^{n-1} e^u\,du[/tex] using integration by parts. Or use a table of integrals. The first time you use IBP you obtain a formula involving [tex]\int u^{n-2} e^u\,du[/tex]. Then do IBP again and you get [tex]\int u^{n-3} e^u\,du[/tex]. Pretty soon you'll see what happens with the factorial.

Do the IBP with definite integrals, [tex]\int_a^b u\,dv=uv\bigr|_a^b-\int_a^b v\,du[/tex] so that you can take care of those "uv" terms at each step.
 
  • #8
C.E
102
0
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?
 
  • #9
392
0
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.

[tex]\theta t -1<0[/tex] i.e. [tex]1-\theta t>0[/tex] will give you [tex]t<1/ \theta[/tex], 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.
 

Related Threads for: Probability- moment generating functions.

Replies
2
Views
772
Replies
6
Views
2K
  • Last Post
Replies
4
Views
689
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
4
Views
378
Replies
4
Views
2K
Top