Discrete Geometric Moment Generating Function

**angie2macau** · November 7th 2009, 07:21 PM

Hi everybody,

I have a question regarding how to proceed with the following problem. I am required to produce a moment generating function for the following problem. I would like to know if the operation I performed is valid for this question. I never understood mgf's that well.

Question:
A balanced coin is flipped repeatedly until a head appears. Let X denote the number of flips needed. Define Y = 2X-1.

Find the moment generating function for Y.

My reasoning:
This is a geometric distribution (discrete random variable).

I applied the definition of the m.g.f.

$m_{y}(t) = \Sigma e^t(2X-1) \times q^x-1 \times p$

Note: I'm new to latex. Please consider the (2X-1) as raised to the power of e, along with the t.

Is this the valid first step for this problem, and others like it?

Thanks for your time!

**matheagle** · November 7th 2009, 07:30 PM

I do not understand your MGF at all.

$M_Y(t) =E(e^{Yt}) =E(e^{(2X-1)t}) =e^{-t} E(e^{2Xt}) =e^{-t} E(e^{X(2t)}) =e^{-t} M_X(2t)$

**mr fantastic** · November 7th 2009, 07:46 PM

Originally Posted by matheagle

I do not understand your MGF at all.

$M_Y(t) =E(e^{Yt}) =E(e^{(2X-1)t}) =e^{-t} E(e^{2Xt}) =e^{-t} E(e^{X(2t)}) =e^{-t} M_X(2t)$

And the mgf for a geometric distribution has been done, among other places, here: http://www.mathhelpforum.com/math-he...tribution.html

**angie2macau** · November 8th 2009, 04:38 AM

Originally Posted by matheagle

I do not understand your MGF at all.

$M_Y(t) =E(e^{Yt}) =E(e^{(2X-1)t}) =e^{-t} E(e^{2Xt}) =e^{-t} E(e^{X(2t)}) =e^{-t} M_X(2t)$

Haha, boy was I off. Thanks matheagle. I had that in my notes, as I'm guessing that's what my prof did on the blackboard, but I couldn't see how it applied to this problem. This stuff is confusing.

Thanks

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