# Thread: [SOLVED] Expectation - geometric series?

1. ## [SOLVED] Expectation - geometric series?

hi everyone, can anyone help me on what looks like a simple problem?
i dont need you guys to do the whole problem, only hint or tell me how to begin : )

im working with E(e^t*N) = (e^-lambda)*[(sum of n = 0:infinity)*((lambda*e^t)^n))/n!]

this is equal to exp(lambda(e^t - 1))

how do you equal this to exp(lambda(e^t - 1)) ? geometric progression?
i am trying geometric progression formula = a/1-r
but getting nowhere : (

thank you all. sorry i dont have word or latex atm so i couldnt type the whole formula in maths form

2. Use the series for e and let $x={\lambda}e^{t}$

$e=\sum_{n=0}^{\infty}\frac{x^{n}}{n!}$

Therefore, $e^{{\lambda}e^{t}}\cdot e^{-{\lambda}}=e^{{\lambda}(e^{t}-1)}$

3. Originally Posted by galactus
Use the series for e and let $x={\lambda}e^{t}$

$e=\sum_{n=0}^{\infty}\frac{x^{n}}{n!}$

Therefore, $e^{{\lambda}e^{t}}\cdot e^{-{\lambda}}=e^{{\lambda}(e^{t}-1)}$

i follow you on this, almost what i dont understand is..
where does the 'n!' go in ((lambda*e^t)^n)/n!? and about about the 'n' in (lambda*e^t)^n ? does it vanish?

thank you : )

sorry it's hard to follow and messy because i cant write up formulas here

i thought you can use the forumla a/1-r for geometric progression having infinity

Nvm. i got it
!