Finding probability using moment-generating functions

[brogrammer]

I'm working through Schaum's Outline of Probability, Random Variables, and Random Processes, and am stuck on a question about moment-generating functions. If anyone has the 2nd edition, it is question 4.60, part (b).

The question gives the following initial information: E[X^k]=0.8 for k = 1, 2, ... and the moment generating function is: 0.2+0.8\sum_{k=0}^{\infty}\frac{t^k}{k!}=0.2+0.8e^t.

The question is asking to find P(X=0) and P(X=1). I'm trying to do the first part and solve P(X=0). By the definition of a moment-generating function for discrete random variables, I know I can use the following equation:

\sum_{i}e^{tx_i}p_X(x_i)=0.2+0.8e^t

For P(X=0), the above equation becomes: e^{t(0)}p_X(0)=0.2+0.8e^t. The LHS simplifies to p_X(0) which means P(X=0)=0.2+0.8e^t. But I know that is not the right answer. The right answer is P(X=0)=0.2.

Can someone please show me where I'm going wrong? Thanks in advance for your help.

 

[chiro]

Hey brogrammer and welcome to the forums.

If you expand the sum and plug in the values for i you will get e^(0t)P(X=0) + e^(1t)P(X=1) = 0.2 + 0.8e^t = P(X=0) + e^(t)*P(X=1)

Now can you equate like terms with co-effecients?

 

[brogrammer]

Chiro -

Thanks for the reply. That makes sense. My thick brain didn't realize that the question wants me to see that the sample space for the r.v. X is just 0 and 1, i.e. a Bernoulli trial. Now it makes sense.

Thanks man.