Finding the Probability distribution function given Moment Generating Function

  • Thread starter Mona1990
  • Start date
  • #1
13
0
Hi everyone,

So I am taking a statistics course and finding this concept kinda challenging. wondering if someone can help me with the following problem!

Suppose X is a discrete random variable with moment generating function
M(t) = 2/10 + 1/10e^t + 2/10e^(2t) + 3/10e^(3t) + 2/10e^(4t)
where t is a real number.

we want to find the probability function of X.

I know that M(t) = E(e^(tx)) = e^(tx)* f(x)
but not sure what to do from there.

Thanks for the help ^^
 

Answers and Replies

  • #2
statdad
Homework Helper
1,495
36
You know the definition of the mgf of a discrete random variable is

[tex]
m_X(t) = \sum_{k=0}^n {P(X=k) e^{kt}}
[/tex]

(I'm assuming the values of X are 0, 1, 2, ..., k for some integer k).

Match the terms of your mgf with this general form.
 
  • #3
13
0
hi,
so from matching i get P (X = 0 ) = 2/10, P (X=1) = 1/10....P(X=4) = 2/10
but i dont get how to find the probability function knowing these values.
 
  • #4
statdad
Homework Helper
1,495
36
hi,
so from matching i get P (X = 0 ) = 2/10, P (X=1) = 1/10....P(X=4) = 2/10
but i dont get how to find the probability function knowing these values.


Look again at the numbers you have in your first line.
 
  • #5
13
0
sorry I dont get it , what line?
 
  • #6
statdad
Homework Helper
1,495
36
P (x = 0 ) = 2/10, p (x=1) = 1/10....p(x=4) = 2/10
 
  • #7
13
0
is it:
f(x) = 2/10 if x is even , and x/10 if x is odd?
thanks for all your help!
 
  • #8
statdad
Homework Helper
1,495
36
You can give the distribution of a discrete r.v. as a table - one for the values, the other for the probabilities - you don't have to specify a "formula" for them.
 

Related Threads on Finding the Probability distribution function given Moment Generating Function

Replies
2
Views
891
Replies
1
Views
9K
  • Last Post
Replies
7
Views
2K
Replies
6
Views
2K
Replies
8
Views
3K
  • Last Post
Replies
4
Views
568
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
1
Views
1K
Top