Finding expected value from the moment generating function

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Discussion Overview

The discussion revolves around finding the expected value from a moment generating function (MGF), specifically the MGF given by mx(t) = (e^t - 1)/t. Participants explore different methods to derive the expected value, including the use of derivatives and series expansions.

Discussion Character

  • Mathematical reasoning
  • Exploratory

Main Points Raised

  • Some participants propose using the definition of the MGF, stating that E{X^n} can be found using the n-th derivative of the MGF evaluated at t=0.
  • One participant calculates E{X} by taking the limit of the second derivative of the MGF and arrives at the value of 1/2.
  • Another participant expresses confusion about the process of using series expansion for e^t and how it relates to finding the expected value.
  • There is a discussion about the correctness of the series expansion and its implications for calculating the expected value.
  • One participant confirms the correctness of the series expansion result but seeks clarification on the next steps involving derivatives.
  • Another participant discusses the need to take derivatives after simplifying the series expansion and evaluates the expected value at t=0.

Areas of Agreement / Disagreement

Participants express some agreement on the method of using the MGF and series expansion, but there is ongoing confusion and uncertainty regarding the steps to take after the series expansion and the interpretation of the results.

Contextual Notes

Some assumptions about the convergence of the series and the validity of the derivatives taken at t=0 are not explicitly stated, leaving room for further clarification.

oyth94
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Suppose I have the MGF moment generating function mx(t) = (e^t -1)/t
How can I find EX?
 
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oyth94 said:
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 $\chi$ $\sigma$
 
chisigma said:
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
$$[(t^k / t!) - 1 ] / t$$ = [(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?
 
oyth94 said:
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
$$[(t^k / t!) - 1 ] / t$$ = [(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

$\chi$ $\sigma$
 
chisigma said:
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.
 

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