Solve f(x) = ce^-x, Find E(x) and Probability Generating Function

[JoanneTan]

Here is the question,

f(x) = ce^-x , x = 1, 2, 3...

Find the value of c.
Find the moment generating function of X.
Use the result obtained, find E(x).
Find the probability generating function of X.
Verify that E(x) obtained using probability generating function is same as the first E(x) founded.

I check the answer of the book, but it's wrong. Can someone help me? I'm looking for the working.

 

[Ray Vickson]

Here is the question,

f(x) = ce^-x , x = 1, 2, 3...

Find the value of c.
Find the moment generating function of X.
Use the result obtained, find E(x).
Find the probability generating function of X.
Verify that E(x) obtained using probability generating function is same as the first E(x) founded.

I check the answer of the book, but it's wrong. Can someone help me? I'm looking for the working.

PF rules require that you show your work.

 

[JoanneTan]

Ok.. For the first question which is to find value of c.

But the answer given is e - 1.
It's not e^-1.. I'm confusing how to get e - 1.

 

[HallsofIvy]

Yes, you must have $ce^{-1}+ ce^{-2}+ ce^{-3}+ \cdot\cdot\cdot= 1$

You can factor out $ce^{-1}$ and have $ce^{-1}(1+ e^{-1}+ e^{-2}+ \cdot\cdot\cdot)$

That is, as you say, a geometric series with common factor $e^{-1}$ so is equal to $\frac{ce^{-1}}{1- e^{-1}}= 1$. That, you have. Now multiply both numerator and denominator by $e$:
$\frac{c}{e- 1}= 1$.

 

[JoanneTan]

Oh! Ok.. I get u now.. Thanks a lot! But the second question, moment generating function,
As u can see from the photo, I done until half, can help me to continue? Cause dono how to substitute.

 

[Ray Vickson]

Oh! Ok.. I get u now.. Thanks a lot! But the second question, moment generating function,
As u can see from the photo, I done until half, can help me to continue? Cause dono how to substitute.

You need to calculate the sum $$\sum_{n=1}^{\infty} e^{-n} e^{kn}<br /> = \sum_{n=1}^{\infty} r^n, \text{ where } r = e^{k-1}$$
You have already seen how to do such summations; look at part (a)!