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

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The discussion focuses on solving the equation f(x) = ce^-x for x = 1, 2, 3... Participants seek to find the value of c, the moment generating function, and the probability generating function of X, while verifying the expected value E(x). The correct value of c is derived as e - 1, using the properties of geometric series. Participants also discuss the moment generating function, with guidance provided on how to continue the calculations involving infinite series. The thread emphasizes the importance of showing work in probability and statistics problems.
JoanneTan
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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.
 
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JoanneTan said:
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.
 
Ok.. For the first question which is to find value of c.
ImageUploadedByPhysics Forums1398268199.770761.jpg

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