- Thread starter
- #1
If a r.v. X has m.g.f. $\displaystyle m_{X} (t) = E \{e^{t\ X}\}$, then...Suppose I have the MGF moment generating function mx(t) = (e^t -1)/t
How can I find EX?
Thank you for your help. For how I did it was I used the series expansion for e^t which is the summation of (t)^k / k!If a r.v. X has m.g.f. $\displaystyle m_{X} (t) = E \{e^{t\ X}\}$, then...
$\displaystyle E\{X^{n}\} = \lim_{t \rightarrow 0} \frac{d^{n} m_{X} (t)}{d t^{n}}\ (1)$
In Your case is...
$\displaystyle E\{X\} = \lim_{t \rightarrow 0} \frac{t\ e^{t} - e^{t} + 1}{t^{2}} = \frac{1}{2}\ (2)$
Kind regards
$\chi$ $\sigma$
The result You have obtained...Thank you for your help. For how I did it was I used the series expansion for e^t which is the summation of (t)^k / k!
so
\(\displaystyle [(t^k / t!) - 1 ] / t\) = [(1 + t/1 + t^2 / 2! + t^3 / 3! + ....) - 1] / t
then the 1s cancel out and I can factor out the t so it becomes
= 1 + t/2! + t^2 / 3! + ...
so when i use the moment generating function
mx^(k) (0) = EX
do i sub in t=0 and then the answer will be 1?
how do I go on from there? or is the answer e?
Confused!
How do you solve it using the series expansion?
okay so i forgot to mention i have to take the derivative after i factor/cancel out the t right. so then it becomesThe result You have obtained...
$\displaystyle \frac{e^{t}-1}{t} = 1 + \frac{t}{2!} + \frac{t^{2}}{3!} + ...\ (1)$
... is absolutely correct... what is the matter?...
Kind regards
$\chi$ $\sigma$