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Finding expected value from the moment generating function

oyth94

Member
Jun 2, 2013
33
Suppose I have the MGF moment generating function mx(t) = (e^t -1)/t
How can I find EX?
 

chisigma

Well-known member
Feb 13, 2012
1,704
Suppose I have the MGF moment generating function mx(t) = (e^t -1)/t
How can I find EX?
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


$\chi$ $\sigma$
 

oyth94

Member
Jun 2, 2013
33
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


$\chi$ $\sigma$
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?
 

chisigma

Well-known member
Feb 13, 2012
1,704
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?
The 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

$\chi$ $\sigma$
 

oyth94

Member
Jun 2, 2013
33
The 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

$\chi$ $\sigma$
okay so i forgot to mention i have to take the derivative after i factor/cancel out the t right. so then it becomes
1 + t/2! + t^2 /3! + ...
derivative --> 1/2 + 2t/6 + ...
and then after i take the derivative i set t=0 because mx1(0) = EX
so from there i will get
EX = 1/2 as you said before.