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mgf of continuous random variables

nacho

Active member
Sep 10, 2013
156
i have a simple enough question

Find the MGF of a continuous random variable with the PDF:

f(x) = 2x, 0<x<1

I understand MGF is calculated as:

$$M(S) = \int_{-\infty}^{+\infty} e^{Sx} f(x)dx$$

which would give me

$$\int_{-\infty}^{+\infty} e^{Sx} 2xdx$$
but how would i compute this integral?


edit: scratch that. I think i am on the right track here, someone check?

if Y = aX + b, then
$$M_{y}(S) = E[e^{2S}] = e^{S}E[E^{2Sx}]$$

$$M_{y}(S) = e^S \int_{0}^{1}e^{2Sx}dx
= ... $$

$$= e^{S}(\frac{e^{2S}}{2s} - \frac{1}{2S}) $$
Is this correct?/ Am I on the correct track?

On another note, let's celebrate me getting the hang of latex! Yay (Clapping)
 
Last edited:

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,493
I understand MGF is calculated as:

$$M(S) = \int_{-\infty}^{+\infty} e^{Sx} f(x)dx$$

which would give me

$$\int_{-\infty}^{+\infty} e^{Sx} 2xdx$$
but how would i compute this integral?
Integrating by parts gives \[\int_{0}^{1} e^{Sx}2x\,dx =2\frac{e^s(s-1)+1}{s^2}\]You can check WolframAlpha.

if Y = aX + b, then
$$M_{y}(S) = E[e^{2S}] = e^{S}E[E^{2Sx}]$$
What? Where did $a$ and $b$ go? Also, $M_Y(S)=E[e^{YS}]$, not $E[e^{2S}]$.
 

springfan25

New member
Mar 3, 2012
14
Edit: previous poster already replied