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Sambit
01-26-2010, 10:48 AM
Let, X is a random variable with MGF = exp{ 3t + 8t^2}.
then P ( -4.84 < X < 9.6) = ?
(A) equal to 0.700
(B) equal to 0.925
(C) equal to 0.975
(D) greater than 0.999

Please solve the problem showing the steps....

AtlasFrysmith
01-26-2010, 11:38 AM
Please see the sticky note regarding homework help...

Riverdale27
01-26-2010, 02:29 PM
I would like to know the answer too...

In other words: how do you get from the MGF to the CDF or PDF?

fed1
01-26-2010, 02:44 PM
There is a one to one relation between distributions and there CDFs.

If you could recognize the MGF as belonging to a certain CDF you could solve this.

Guessing the right CDF is an option, but requires cleverness.

Analytical solution is possible using 'inverse laplace transform' but this is more math than i know.

Good luck!

AtlasFrysmith
01-26-2010, 03:11 PM
Yeah, if you have a table of distributions and their MGFs you can see if it has the same structure as one of those, and then if so you can determine the parameters.
This example looks familiar... Once you recognize the distribution it's an easy, or at least a familiar, problem.

More advanced note: There are some subtleties as to where the MGFs need to be equal for the distributions to be the same--it is necessary that the MGFs be equal for all t in some open interval around 0.

BGM
01-27-2010, 07:56 AM
If X ~ N(μ, σ^2), then M(t) = e^(μt + σ^2t^2/2)
Then inversely you can match the distribution and parameters
In your case μ = 3, σ^2 = 8*2 = 16

Sambit
01-27-2010, 10:53 AM
If X ~ N(μ, σ^2), then M(t) = e^(μt + σ^2t^2/2)
Then inversely you can match the distribution and parameters
In your case μ = 3, σ^2 = 8*2 = 16


i can understand this part.
but then, HOW TO FIND OUT the PROBABILITY using this?

do u suggest me to integrate the normal pdf from -4.84 to 9.6 ?

fed1
01-27-2010, 07:51 PM
i can understand this part.
but then, HOW TO FIND OUT the PROBABILITY using this?

do u suggest me to integrate the normal pdf from -4.84 to 9.6 ?

Yes. You must use direct integration of the pdf. It would be cheating to use

Normcdf( 9.6 ) - Normcdf( -4.84 ).