Moment generating function and expectation

BookMark440
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Homework Statement


Let X denote a random variable with the following probability mass function:
P(j)= 2^(-j), j=1,2,3,...
(a) Compute the moment generating function of X.
(b) Use your answer to part (a) to compute the expectation of X.

Homework Equations


m.g.f of X is M (t) = E[e^tX]


The Attempt at a Solution



I computed the MGF (X) to be: e^t/(2-e^t). I need a suggestion about the next step. I'm confused about the relationship between the MGF and expectation of X.

Thanks!
 
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That link was very helpful. I completed the problem but I am a little uncertain about my strategy of finding the derivative by parts and the answer:

X has p.d.f : p(j) = 2^(-j), j=1,2,3,...

I computed the MGF for X = e^t/(2-e^t)

Then E[X] = d/dt (e^t/(2-e^t)) [evaluated for t= 0] =

numerator = (2-e^t)d/dt(e^t) - (e^t)d/dt(2-e^t)
denominator = (2-e^t)^2

Evaluating for t=0, the final answer is:
E[X] = 2

Is this the correct strategy and answer?

Thanks!
 
Hi BookMark440! :smile:

(try using the X2 tag just above the Reply box :wink:)
BookMark440 said:
Then E[X] = d/dt (e^t/(2-e^t)) [evaluated for t= 0] =

numerator = (2-e^t)d/dt(e^t) - (e^t)d/dt(2-e^t)
denominator = (2-e^t)^2

Evaluating for t=0, the final answer is:
E[X] = 2

Is this the correct strategy and answer?!

Looks good! :biggrin:
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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