To obtain the MGF of a Neg Binomial, you just look at it as a sum of r independent geometrics r.v.s
with parameter p. By independence that just produces the MGF of your Neg Bi as the MGF
of a geo to the power r.
I am trying to find the moment generating function for the negative binomianl distribution. I know that it's baiscly finding <e^ty>. I therefore put e^ty into the sum from y=0 to infinity. I tried a few ways of solving it but I am stuck for quite sime time now. Here is what I tried:
- Expanding e^ty and use it's summation form. But this only led to showing how the moment generating function works.
- Use different substitutions to try and get the original form or one of it's varience back with a factor outside the summation. But this is not working too.
Am I going in the right direction?
Thanks for the solution. It is right?
But I am more interested in how these things work. I don't have a strong math background, so please forgive me if I'm asking the obvious.
Generally, I'm more comfortable with the continuous case:
.
I can evaluate this integral with different tools and techniques.
But the discrete case seems to be more problematic.
I can't use
to get anything useful. I think that this is due to a lack of math skills.
Are there any tricks that I should know in general?
Since I now know the answer, I tried working backwards:
Therefore, I need to show that:
Now I let:
because that's just another Neg Bi distribution for p'.
Is this correct?
Hello,
You have this sum :
Since has no y in it, you can factor it out :
Now, have a look here : Binomial theorem - Wikipedia, the free encyclopedia (at the end of the paragraph, when they say : "r=-s"). This gives in our case :
The red equality is obtained by the following identity :
Proof :