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September 16th, 2017, 01:08 PM  #1 
Senior Member Joined: Oct 2015 From: Antarctica Posts: 128 Thanks: 0  Problems finding the Moment Generating Function
Let the random variable X equal the number of flips of a fair coin that are required to observe heads  tails on consecutive flips. Find the pmf of X and find the moment generating function of X. The PMF is straightforward since there are always two desired outcomes divided by 2^n (where n >= 2). So the PMF should simply be 1/(2^(n1)). The moment generating function is where I'm struggling. I should be able to use use properties of geometric series to obtain the moment generating function, but something keeps going wrong and I'm not sure what. I'm supposed to obtain: M(t) = e^(2t)/(e^t  2)^2 but I keep ending up with M(t) = e^(2t)/(2  e^t) I honestly don't know what's going wrong, as I've checked over my work several times. Do I have the wrong PMF, perhaps? If not, how did you calculate the moment generating function? 
September 16th, 2017, 01:24 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,038 Thanks: 1063 
I get... $P[m] = \left . p(1p)^{m1}+p^{m1}(1p) \right_{p=\frac 1 2} = 2^{1m},~m=2,3, \dots$ $M(t) = E[e^{t X}] = \displaystyle \sum_{m=2}^\infty~2^{1m}e^{m t} = \dfrac{e^{2t}}{2e^t}$ 
September 16th, 2017, 02:12 PM  #3 
Senior Member Joined: Oct 2015 From: Antarctica Posts: 128 Thanks: 0 
Yeah, that's exactly what I get! But I don't understand what's wrong. Here's the problem statement: Yet our answers are clearly different than the one they want us to find. 
September 16th, 2017, 03:29 PM  #4 
Senior Member Joined: Sep 2015 From: USA Posts: 2,038 Thanks: 1063 
Ok, I did it assuming you could either go headstails or tailsheads. This problem is different. Let me look at this a bit further.

September 16th, 2017, 04:12 PM  #5 
Senior Member Joined: Sep 2015 From: USA Posts: 2,038 Thanks: 1063 
Ok I have it. Legit sequences that take $m$ flips to obtain a $HT$ transition are of the following form. $k$ occurrences of $T$ followed by $mk1$ occurences of $H$, followed by $T$ $k = 0, 1, \dots m2$ These have probability $pr_k(m) = (1p)^{k+1}p^{mk1}$ $pr(m) = \displaystyle \sum_{k=0}^{m2} pr_k(m) = \dfrac{p^{m+1}p^m+p (1p)^m}{12 p}$ notice you can't just substitute $p=\dfrac 1 2$ addenda: You can however take the limit as $p \to \dfrac 1 2$ and proceed that way. You get $p(m) = 2^{m}(m1)$ You can then find $M(t)$ using that expression. This is probably much easier. : end addenda Then $M(t) = \displaystyle \sum_{m=2}^\infty ~e^{t m} pr(m) = \sum_{m=2}^\infty ~e^{t m} \dfrac{p^{m+1}p^m+p (1p)^m}{12 p} = \dfrac{(p1) p e^{2 t}}{\left(p e^t1\right) \left(p e^te^t+1\right)}$ Now, letting $p=\dfrac 1 2$, and probably churning through a ton of algebra we get. $M(t) = \dfrac{e^{2 t}}{\left(e^t2\right)^2}$ Last edited by romsek; September 16th, 2017 at 04:17 PM. 
September 16th, 2017, 06:18 PM  #6 
Senior Member Joined: Oct 2015 From: Antarctica Posts: 128 Thanks: 0 
Although I understand how to find the MGF from the PMF you said the limit was equal to (2^m * (m1)), I'm still not quite sure how to interpret the PMF in the context of the problem. To me, when I read the problem, I don't understand how to derive the PMF from it. Edit; I understand it much more intuitively now. Without going through the binomial stuff, the m1/2^m makes sense if you write out a few sets. Last edited by John Travolski; September 16th, 2017 at 06:31 PM. 

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