Moment generating functions help

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Homework Statement


[/B]
Let X be a random variable with support on the positive integers (1, 2, 3, . . .) and PMF f(x) = C2 ^(-x) .

(a) For what value(s) of C is f a valid PMF?
(b) Show that the moment generating function of X is m(t) = Ce^t/(2− e^t) , and determine the interval for t for which it is valid. (You may use your value for C calculated in question 1, if you would like).
(c) Using the MGF, calculate the expected value and the variance of X.

The Attempt at a Solution


[/B]
a)

sum from -∞ to ∞ of C/(2^x)=1

C(1/2+1/4...)=1
C=1 as it converges

b)

m(t)=E[e^tx]=integral from -∞ to ∞ of ((e^tx)*(2^(-x)))
=integral from -∞ to ∞ of (e^tx)/(2^x)

is this the right way to go about calculating it?
 
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Mark53 said:

Homework Statement


[/B]
Let X be a random variable with support on the positive integers (1, 2, 3, . . .) and PMF f(x) = C2 ^(-x) .

(a) For what value(s) of C is f a valid PMF?
(b) Show that the moment generating function of X is m(t) = Ce^t/(2− e^t) , and determine the interval for t for which it is valid. (You may use your value for C calculated in question 1, if you would like).
(c) Using the MGF, calculate the expected value and the variance of X.

The Attempt at a Solution


[/B]
a)

sum from -∞ to ∞ of C/(2^x)=1

C(1/2+1/4...)=1
C=1 as it converges

b)

m(t)=E[e^tx]=integral from -∞ to ∞ of ((e^tx)*(2^(-x)))
=integral from -∞ to ∞ of (e^tx)/(2^x)

is this the right way to go about calculating it?
No. For a discrete random variable we have
$$E f(X) = \sum_x p(x) f(x), $$
so involves summation, not integration.
 
Ray Vickson said:
No. For a discrete random variable we have
$$E f(X) = \sum_x p(x) f(x), $$
so involves summation, not integration.

when calculating the sum I get:

the sum of x=0 to ∞ of (e^tx)/(2^x)=e^t/2

which is wrong am I still missing something?
 
Mark53 said:
when calculating the sum I get:

the sum of x=0 to ∞ of (e^tx)/(2^x)=e^t/2

which is wrong am I still missing something?
Yes: you are basically saying that ##\sum_{k=0}^{\infty} r^k = r, ## which is wrong.
 
Ray Vickson said:
Yes: you are basically saying that ##\sum_{k=0}^{\infty} r^k = r, ## which is wrong.
Do I need to see if the series converges or find the partial sum?

not sure how to start solving it
 
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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