marginal moment for random sums - question about proof

**Volga** · January 31st 2011, 03:29 AM

Suppose that { $X_j$ } is a sequence of intedependent identically distributed random variables and that N is a random variable taking non-negative integer values.

If $Y=\Sigma_{j=1}^N$ , then $M_Y(t)=M_N(lnM_X(t))$

Proof.

$M_Y(t)=E[M_{Y|N}(t|N)]=E[(M_X(t)^N]=E[e^{NXt}]=E[exp({N*ln(e^{xt})]=$
$=E(e^{N*lnM_X(t)})=M_N(lnM_X(t))$

The beginning and the end is from the textbook, the part in the middle with $E[e^{NXt}]=E[exp({N*ln(e^{xt})]$ is mine.

I am not quite sure about the last transition

$E[exp({N*ln(e^{xt})]=E(e^{N*lnM_X(t)})=M_N(lnM_X(t))$ .

I can see how $E[e^{N*something}]$ becomes $M_N(something)$ . I am not sure why $ln(e^{xt})=lnM_X(t)$ if there is no expected value E anywhere around (since the definition of MGF is $M_X(t)=E(e^{Xt}$ ).

**Moo** · January 31st 2011, 11:59 AM

Hello !

Originally Posted by Volga

The beginning and the end is from the textbook, the part in the middle with $E[e^{NXt}]=E[exp({N*ln(e^{xt})]$ is mine.

(...)

I can see how $E[e^{N*something}]$ becomes $M_N(something)$ . I am not sure why $ln(e^{xt})=lnM_X(t)$ if there is no expected value E anywhere around (since the definition of MGF is $M_X(t)=E(e^{Xt}$ ).

That's because it's $E[e^{NXt}]=E[\exp(N*\ln(e^{Xt}))]$ (capital X)

P.S. : HK

\o///

**Volga** · January 31st 2011, 03:43 PM

It's not so obvious to me.

Instead, I would write $E[exp(N*ln(e^{Xt})]=M_N(ln(e^{Xt}))=M_N(Xt)$

?? I suspect there is something here about the expected value operator that I don't know...

**Moo** · January 31st 2011, 10:09 PM

Oh right sorry, I misunderstood your question.

There is a big mistake for the third '=' sign :
$E\left[M_X(t)^N\right]=E\left[\left(E\left[e^{Xt}\right]\right)^N\right]$ , which is completely different from what you wrote.

I would like to know what you mean by "the beginning" and "the end". Do you mean that the book only provides $M_Y(t)=M_N(\ln M_X(t))$ ?

**Volga** · February 1st 2011, 12:12 AM

Originally Posted by Moo

Oh right sorry, I misunderstood your question.

There is a big mistake for the third '=' sign :
$E\left[M_X(t)^N\right]=E\left[\left(E\left[e^{Xt}\right]\right)^N\right]$ , which is completely different from what you wrote.

I would like to know what you mean by "the beginning" and "the end". Do you mean that the book only provides $M_Y(t)=M_N(\ln M_X(t))$ ?

This is from the book:

$M_Y(t)=E[M_{Y|N}(t|N)]=E[(M_X(t)^N]=E(e^{N*lnM_X(t)})=M_N(lnM_X(t))$

The mistake you pointed out is mine ))) I now understand completely, from the double expected value expression you wrote. Thank you!!

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