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Probability Mass Function/Moment Generating Function

  • Thread starter tangodirt
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1. Homework Statement

The pmf of a random variable X is given by f(x) = π(1 − π)x for x = 0, 1, ..., ∞, and 0 ≤ π ≤ 1.
a) Show that this function actually is a pmf.
b) Find E(X).
c) Find the moment generating function of X, MX(t) = E(etX).

2. The attempt at a solution
My solution was done numerically in MATLAB, but I suppose that there is probably an analytical solution as well. My biggest issue is the interpretation of π.

[PLAIN]http://img401.imageshack.us/img401/2015/proofox.png [Broken]

For the second part, I also did this numerically, by solving the series:

[tex]\sum_{x}x \cdot p(x) = \sum_{x}x \cdot f(X=x)[/tex]
Which evaluates, also through MATLAB, to be: E(X) = 1, but I suspect there is probably an analytical method for this as well?

The part I am struggling the most with is the last bit. I can get my MGF down to:

[tex]\sum_{\forall x} e^{tx} \pi (1-\pi)^{x}[/tex]
But I am not sure how to get rid of the infinite summation. I tried an infinite geometric series, but it only holds true for:

[tex]|e^{t}(1-\pi)| < 1[/tex]
Which means that E(X) cannot be found with the MGF.

Any ideas?
 
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Ray Vickson

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1. Homework Statement

The pmf of a random variable X is given by f(x) = π(1 − π)x for x = 0, 1, ..., ∞, and 0 ≤ π ≤ 1.
a) Show that this function actually is a pmf.
b) Find E(X).
c) Find the moment generating function of X, MX(t) = E(etX).

2. The attempt at a solution
My solution was done numerically in MATLAB, but I suppose that there is probably an analytical solution as well. My biggest issue is the interpretation of π.

[PLAIN]http://img401.imageshack.us/img401/2015/proofox.png [Broken]

For the second part, I also did this numerically, by solving the series:

[tex]\sum_{x}x \cdot p(x) = \sum_{x}x \cdot f(X=x)[/tex]
Which evaluates, also through MATLAB, to be: E(X) = 1, but I suspect there is probably an analytical method for this as well?

The part I am struggling the most with is the last bit. I can get my MGF down to:

[tex]\sum_{\forall x} e^{tx} \pi (1-\pi)^{x}[/tex]
But I am not sure how to get rid of the infinite summation. I tried an infinite geometric series, but it only holds true for:

[tex]|e^{t}(1-\pi)| < 1[/tex]
Which means that E(X) cannot be found with the MGF.

Any ideas?
You should not need Matlab to do part (a); in fact, relying on software instead of thinking means you will _never_ learn the material. You need to show that sum{(1-p)^k, k=0..infinity} is 1/p. Are you really sure you have never, ever, seen this type of material before? It is just elementary algebra.

Part (b) is just a little bit trickier, but again, involves only standard results that are widely available If q = 1-p, you want to evaluate S = 0 + 1*q + 2*q^2 + 3*q^3 + ... . You can write this as q + q^2 + q^3 + ... + [q^2 + 2*q^3 + 3*q^4 + ... ], and [...] = q*(q + 2*q^2 + 3*q^3 +...) = q*S. The first summation is q*(1 + q + q^2 + ...) = q/(1-q), so we have S = q/(1-q) + q*S. This is an equation for S that you can solve. (There are many other ways of getting the result, but the above is the most "elementary".)

You can get part (c) similarly: just write out the first few terms of what you are supposed to evaluate, then look for a pattern.

RGV
 
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54
1
You should not need Matlab to do part (a); in fact, relying on software instead of thinking means you will _never_ learn the material. You need to show that sum{(1-p)^k, k=0..infinity} is 1/p. Are you really sure you have never, ever, seen this type of material before? It is just elementary algebra.
To be honest, I cannot recall if I've seen the material or not. It seems like I have, but it's been so long that I cannot remember.

Part (b) is just a little bit trickier, but again, involves only standard results that are widely available If q = 1-p, you want to evaluate S = 0 + 1*q + 2*q^2 + 3*q^3 + ... . You can write this as q + q^2 + q^3 + ... + [q^2 + 2*q^3 + 3*q^4 + ... ], and [...] = q*(q + 2*q^2 + 3*q^3 +...) = q*S. The first summation is q*(1 + q + q^2 + ...) = q/(1-q), so we have S = q/(1-q) + q*S. This is an equation for S that you can solve. (There are many other ways of getting the result, but the above is the most "elementary".)
To be honest, this really confuses me. Also, when solving your final equation, I get:

[tex]S = \frac{q}{1-q} + qS[/tex]
[tex]S = \frac{q}{(q-1)^2}[/tex]

Which should return 1 for any value of q such that 0 <= q <= 1, no? For q = 0.5, it evaluates to 2, for q = 0.75, it evaluates to 12, etc.

You can get part (c) similarly: just write out the first few terms of what you are supposed to evaluate, then look for a pattern.

RGV
Regardless, even if I have learned about series (however long ago), I've never done series manipulation like this, as far as I am aware.
 

Ray Vickson

Science Advisor
Homework Helper
Dearly Missed
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To be honest, I cannot recall if I've seen the material or not. It seems like I have, but it's been so long that I cannot remember.



To be honest, this really confuses me. Also, when solving your final equation, I get:

[tex]S = \frac{q}{1-q} + qS[/tex]
[tex]S = \frac{q}{(q-1)^2}[/tex]

Which should return 1 for any value of q such that 0 <= q <= 1, no? For q = 0.5, it evaluates to 2, for q = 0.75, it evaluates to 12, etc.



Regardless, even if I have learned about series (however long ago), I've never done series manipulation like this, as far as I am aware.
Sorry: I should have said S = p*(0 + q + 2*q^2 + 3*q^3+...), which is p times what I wrote before, so should have gotten S = p*q/(1-q)^2 = p*q/p^2 = q/p. Here, p = 1-q. There is no reason to expect S to be 1, because S is NOT the sum of the probabilities (which would be sum p(x)); it is the expected value of X, which is sum x*p(x).

RGV
 

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