# Moment Generating Function for Sum of Independent Random Variables

I'm taking a graduate course in probability and statistics using Larsen and Marx, 4th edition and looking specifically at estimation methods this week. I ran into a homework problem that is related to moment generating functions and I can't quite connect the dots on how they arrived at the solution.

If you have three independent random variables $$Y_{1}, Y_{2}, Y_{3}$$ and you would like to determine the moment-generating function of $$W = Y_{1} + Y_{2} + Y_{3}$$ knowing that each of the three independent random variables have the same pdf $$f_{y} = \lambda y e^{-\lambda y}, y \geq 0$$

The easy part of the this problem is applying the theorem that says for $$W = W_{1} + W_{2} + W_{3}$$ the moment generating function of the sum is: $$M_{W}(t) = M_{W_{1}}(t)* M_{W_{2}}(t)* M_{W_{3}}(t)$$

Where I run into trouble is getting the individual moment generating functions for the Y's. The problem directs you to apply yet another theorem where you would let, for example, another random variable V equal to $$aY_{1}+b$$ and it follows that $$M_{V}(t) = e^{bt}M_{W}(at)$$

The solution states that if you allow $$V = (1/\lambda)*W$$ then the pdf of V then becomes $$f_{V}(y) = ye^{-y}, y \geq 0$$ and subsequently, you can get the moment generating function using a simple integration by parts but I can't quite follow the application of the theorem used to get to the pdf of V.

Any insight? Likely a fundamental property I missed along the way...

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If $V = aY_1+b$, then $M_V(t) = e^{bt}M_{Y_1}(at)$ certainly holds, but for $M_V(t)$ to equal $e^{bt}M_{W}(at)$ where $W = Y_1+Y_2+Y_3$ seems rather far-fetched. –  Dilip Sarwate Jul 1 '12 at 19:00

You have not made use of the definition of a moment generating function. The moment generating function for any random variable $X$ is usually defined as $$M_X(t) = \mathbb{E} \left( e^{tX} \right)$$

EDIT

First, your $f_y$ is incorrect. It should be $f_Y(y) = \lambda e^{- \lambda y}$. We get $$M_Y(t) = \displaystyle \int_0^{\infty} e^{ty} \lambda e^{-\lambda y} dy = \displaystyle \int_0^{\infty} \lambda e^{(t-\lambda) y} dy = \dfrac{\lambda}{\lambda-t}$$ Hence, the moment generating function for $Y = Y_1 + Y_2 + Y_3$ is $M_Y(t) = \dfrac{\lambda_1}{\lambda_1-t} \dfrac{\lambda_2}{\lambda_2-t} \dfrac{\lambda_3}{\lambda_3-t}$

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Isn't $\mathbb{E} \left( e^{itX} \right)$ usually called the characteristic function of the random variable and distinguished from the moment-generating function? –  Dilip Sarwate Jul 1 '12 at 17:19
@DilipSarwate Ya. But I actually don't see the difference and I have seen people interchangeably using them. I have now removed that statement now. –  user17762 Jul 1 '12 at 17:22
Maybe I'm not seeing how it applies to arriving at the simpler PDF. I'm only using the definition to develop the integral for getting the MGF once I have the simpler PDF. –  PatternMatching Jul 1 '12 at 17:43
I corrected his pdf to $\lambda (\lambda y)e^{-\lambda y}$ instead of your simpler version because with the simpler version, integration by parts is not really needed. So I figured that Larson and Marx did really intend the Gamma pdf of order $2$ where integration by parts can be used, and not the Gamma pdf of order $1$ –  Dilip Sarwate Jul 1 '12 at 18:30
@DilipSarwate Ok. Thanks Dilip for letting me know. –  user17762 Jul 1 '12 at 18:32

You might want to check your pdf of $Y$ for typographical errors. Is it a valid pdf?

First, if $Y = aX+b$, then $$f_Y(y) = \frac{1}{|a|}f_X\left(\frac{y-b}{a}\right)$$ and so if $f_X(x) = xe^{-x}$ for $x \geq 0$ $b = 0$, $a = \lambda^{-1}$, then $f_Y(y) = \lambda (\lambda y) e^{-\lambda y}$ for $y \geq 0$. (This is the correct pdf, not what you have, but for an alternative, see the answer by Marvis). Thus, if you can figure out $M_X(t)$, the MGF of $X$, then you can deduce the MGF of $Y$ as $M_Y(t) = M_X(\lambda^{-1}t)$, from which it follows that $M_W(t) = [M_X(\lambda^{-1}t)]^3$.

So what is $M_X(t)$? As Marvis points out to you,

$$M_X(t) = E[e^{tX}] = \int_0^\infty e^{tx} x e^{-x}\,\mathrm dx = \int_0^\infty x e^{(t-1)x}\,\mathrm dx$$ which integral you should be able to compute via integration by parts.

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