Question

Suppose that Y~DU(n), so that Y has pdf p(y)=1/n for y=1,...,n. Prove that E(e^(5Y))=(e^5(1-e^(5n)))/(n(1-e^5)). Hint: a geometric series is needed.

Answer 1

\[E(e^{kY})=\sum_{y=1}^n e^{ky} \frac{1}{n}=\frac{1}{n}\sum_{y=1}^n (e^k)^y\]
Sum of geometric series: \[\sum_{i=1}^nr^i=\frac{1-r^{n+1}}{1-r}\]

Answer 2

Oops, the summation should be
\[\sum_{i=1}^nr^i=\frac{r(1-r^{n+1})}{1-r}\]
The one I wrote previously would be right if the summation started at 0.

Answer 3

i can't remember the process for reaching the sum of a geometric series. Can you refresh me briefly?

Answer 4

You mean like deriving it from scratch?

Answer 5

yeah, just a brief explanation, not a full one. if you can? (me asking hopefully?)

Answer 6

A telescoping sum:
\[S=1+r+r^2+...+r^n\\
rS=r+r^2+r^3+...+r^{n+1}\\
S-rS=1-r^{n+1}\\
S=\frac{1-r^{n+1}}{1-r}\]

Answer 7

ohhhhs. I remember now. Thank you!!