I am stuck with the question below.
If X is a negative binomial random variable, then $$ Y=r+x $$ is the total number of trails necessary to obtain r S's. Obtain the moment generating function of Y and then it's mean value and variance. Are the mean and variance intuitively constant with the expressions for E(X) and V(X)? Explain.
I know that the moment generating function of a negative binomial is: $$ M_X(t)=\frac{p^r}{[1-e^t(1-p)]^r}. $$ How do I use relate this to the moment generating function I need when I add a constant? I can find the mean by deriving this moment generating function and plugging in 0 for t. I also know how to use the shortcut to find variance. Does the last part mean I just say that the mean and variance are as expected because I derive away the constant?
I think you are looking for a connection to the geometric random variable? If $X\sim Geo(p)$ counts the failures and the first final success, we get $$M_X(t)=\frac{pe^t}{1−e^t(1−p)}$$ But also $$M_{X-1}(t)=\mathbf E[exp(Xt-t)]=\frac{pe^t}{1−e^t(1−p)}e^{-t}$$
For the negative binomial $Y\sim NB(r,p)$ you got $$M_Y(t)=\left(\frac{p}{1−e^t(1−p)}\right)^r=\left(\frac{pe^t}{1−e^t(1−p)}\cdot e^{-t}\right)^r$$
So $$Y\sim \sum_{i=1}^r(X_i-1)=\left(\sum_{i=1}^rX_i\right)-r$$ and $X_i\sim Geo(p)$ independent. This $Y$ does not count the r successes, only the failures.