Calculating a Probability Density Function from a Moment Generating FunctionDate: 04/04/2009 at 11:25:13 From: Gabriel Subject: Moment Generating function The moment generating function of a random variable Y is M_Y(t) = (1/(1 - t))^2 Compute P(-2 < Y < 3), or explain why you cannot compute it. If I'm not mistaken, the random variable (X) is gamma with parameters (2,1). But I do not know how to carry on from there.... or is it un- computable? If so, how do I explain that? Can anyone help? Thanks. Date: 04/05/2009 at 00:44:31 From: Doctor Jordan Subject: Re: Moment Generating function Hi Gabriel, Let's try to compute P(-2 < Y < 3). We may get a probability, or we may run into a problem. (In a few places, I will use exp(x) instead of e^x, but they mean the same thing.) P(-2 < Y < 3) is equal to the integral of f(x) from -2 to 3, where f (x) is the probability density function of the random variable Y. Since M_Y(t) is defined in (-1,1), which is an open neighborhood around the origin, the distribution Y is uniquely determined (general fact). What this means is that there cannot be two different distributions with this moment generating function; and hence if I can find one distribution with this moment generating function, I know that this is the distribution of Y. Now looking up this moment generating function in a table, I see that it is, as you suggested, the moment generating function for the gamma distribution: http://planetmath.org/encyclopedia/GammaRandomVariable.html Thus the probability density function of Y is f(x) = x*exp(-x) for x >=0 and f(x) = 0 for x < 0. Generally, knowing the moment generating function, you could also compute the probability density function as f(x) = 1/(2pi) * integral from -infinity to infinity of M_Y(it) * exp(-itx) dt, but here we were able to just use a table. Since we have found the probability density function for Y, we can now find P(-2 < Y < 3): P(-2 < Y < 3) = integral of f(x) from -2 to 3 = integral of f(x) from 0 to 3 (since f(x) = 0 for x < 0) Using integration by parts, = -x*exp(-x) - exp(-x) evaluated from 0 to 3 = -4*exp(-3) + 1 = 0.80085... Thus, P(-2 < Y < 3) = 0.80085... I recommend carefully checking all of this yourself. I have double checked it, but there were lots of places to make mistakes. Also, it is essential to get your hands dirty with calculations in order to understand this material. In particular, make sure you see that M_Y (t) is indeed the moment generating function for the gamma distribution, and that I got the right probability density function. If you have any questions about this solution, please write back. - Doctor Jordan, The Math Forum http://mathforum.org/dr.math/ Date: 04/05/2009 at 04:27:48 From: Gabriel Subject: Moment Generating function It was very useful!! But what I don't understand is from this part: Since M_Y(t) is defined in (-1,1), which is an open neighborhood around the origin, the distribution Y is uniquely determined (general fact). What this means is that there cannot be two different distributions with this moment generating function; and hence if I can find one distribution with this moment generating function, I know that this is the distribution of Y. How do you know M_Y(t) is defined in (-1,1)? and what is an open neighbourhood around the origin? And another thing: I tried working out the product ... f(x) = 1/(2pi) * integral from -infinity to infinity of M_Y(it) * exp(-itx) dt, ... and I couldn't get x*exp(-x). I got stuck at 1/(2pi) exp(-ix) [{1 - ti)/i)] from -infinity to infinity.. On the PlanetMath website, there was a formula for f_x(x); but it requires Gamma(alpha, lamda) = x, and I do not know that. Thanks for all your help! Date: 04/05/2009 at 10:08:05 From: Doctor Jordan Subject: Re: Moment Generating function Hi Gabriel, For M_Y(t) to be defined in (-1,1) means that for any t between -1 and 1, M_Y(t) exists. M_Y(t) = (1/(1 - t))^2, so it is not defined for t = 1. To be defined in an open neighborhood of the origin is equivalent to being defined in an open interval around the origin, i.e., being defined on some interval of the form (-a,a). Here a = 1, but we could have chosen (-0.5,0.5) or (- 0.01,0.01), etc. All that matters is there is some interval. I stated the integral just as a general remark. We can solve this question without having to calculate this integral. Concerning Gamma(alpha,lambda), look carefully at note 4 of PlanetMath's page: The moment generating function of a gamma random variable is given by M_x(t) = [lambda/(lambda - t)]^alpha Spend at least half an hour agonizing over this before writing back! The way to remember and learn things is to really struggle over them and then get the answer. Even if you get nowhere, having struggled will make you remember it better. - Doctor Jordan, The Math Forum http://mathforum.org/dr.math/ Date: 04/06/2009 at 00:51:58 From: Gabriel Subject: Moment Generating function Thanks, sir!! Very sorry to keep bothering you, but I have to ask another question. The PlanetMath website says that, for a gamma random variable, the probability density function is f_x(x) = [something]/[gamma function]. After subbing in the values for gamma(2,1), I got f_x(x) = x*exp(-x)/[gamma function] Then I tried to compute the gamma function using the formula integral from zero to infinity of x^(t -1) * e^ x But try as I could, I could not get this result of yours: Thus the probability density function of Y is f(x) = x*exp(-x) for x >=0 and f(x) = 0 for x < 0. Date: 04/06/2009 at 08:32:38 From: Doctor Jordan Subject: Re: Moment Generating function Hi Gabriel, What do you find that Gamma(2) is equal to? - Doctor Jordan, The Math Forum http://mathforum.org/dr.math/ Date: 04/06/2009 at 10:18:37 From: Gabriel Subject: Thank you (Moment Generating function) Oh, I get it now!! It's Gamma(2) = (2 - 1)! = 1 Thank you, Doctor Jordan! I am really really grateful for all your patience and detailed explanations, especially with a slow learner like me... Thanks again! :D |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/