What is the Marginal PMF for X1 and X2 with a Joint PMF of p^2q^x2?

[phiiota]

Homework Statement

Suppose that X1 and X2 have the joint pmf
f(x_{1},x_{2})=p^{2}q^{x_{2}},x_{1}=0,1,2,...,x_{2},x_{2}=0,1,2,...
with
0<p<1,q=1-p

Homework Equations

The Attempt at a Solution

I'm confused because the expression doesn't have x_1 in it. So usually, if I want to find f1x(1), say, I would add up all the x_2 probabilities where x_1=1 to get my marginal (summing over all values of x_2). But without having an x_1 in my pmf, I don't know how to do this. I'm tempted to say the marginal is just the original pmf, but how would I differentiate between p(x_1=1,x_2=2) from p(x_1=0,x_2=2)?

 

[jbunniii]

There must be a typo somewhere. If you sum f(x_1, x_2) over x_1 and x_2, the result will be infinity, not 1. Therefore this is not a valid probability mass function

 

[Ray Vickson]

There must be a typo somewhere. If you sum f(x_1, x_2) over x_1 and x_2, the result will be infinity, not 1. Therefore this is not a valid probability mass function

No, that is not the case: x1 is summed from 0 to x2, and x2 is summed from 0 to ∞.

RGV

 

[jbunniii]

No, that is not the case: x1 is summed from 1 to x2, and x2 is summed from 1 to ∞.

RGV

Oops, you're right. I misread the range for x_1.

 

[phiiota]

okay, so my first thought is that the joint then should be
\sum_{x_{2}=0}^{\infty}\sum_{i=0}^{x_{2}}p^{2}q^{x_{2}}

when I let p=.5, this term does indeed sum to 1, so that's looking alright.
oh, I think I maybe figured this out. So I should have (summing over values in the domain
f_{1}(x_{1})=\sum_{x_{2}=0}^{\infty}p^{2}q^{x_{2}}
f_{2}(x_{2})=\sum_{i=0}^{x_{2}}p^{2}q^{x_{2}}

No, I'm still confused. The f1 sums out to .5 (when I let p=.5). So then I should sum .5 from... 0 to x2 to get 1? But then I'll have x2 in my term, and not 1.
When I integrate f2 from 0 to infinity, I do get one, which I think is good, but I'm still confused by not having an x1 term anywhere.

 

[jbunniii]

okay, so my first thought is that the joint then should be
\sum_{x_{2}=0}^{\infty}\sum_{i=0}^{x_{2}}p^{2}q^{x_{2}}

Yes, that's the sum of the joint pmf.

oh, I think I maybe figured this out. So I should have (summing over values in the domain
f_{1}(x_{1})=\sum_{x_{2}=0}^{\infty}p^{2}q^{x_{2}}

No, you aren't summing over the right range of indices. Try drawing a picture of the region where the joint pmf is nonzero. Then it will be easier to see what the upper and lower limits for x_2 should be in the sum.

f_{2}(x_{2})=\sum_{i=0}^{x_{2}}p^{2}q^{x_{2}}

Yes, this is correct. The summand does not depend on i, so you are simply summing x_2 + 1 copies of the same thing. What is the result?

 

[phiiota]

Okay, so looking at a picture, I know that P(x1,x2)=0 whenever x1>x2 (and x1,x2>0). So I have kind of a triangular set up. So looking at these values, then, I'm thinking that my marginal for x should be
f_{1}(x_{1})=\sum_{i=x_{1}}^{\infty}p^{2}q^{x_{1}}

 

[jbunniii]

Okay, so looking at a picture, I know that P(x1,x2)=0 whenever x1>x2 (and x1,x2>0). So I have kind of a triangular set up. So looking at these values, then, I'm thinking that my marginal for x should be
f_{1}(x_{1})=\sum_{i=x_{1}}^{\infty}p^{2}q^{x_{1}}

Almost, except the exponent for q is wrong: it should be q^i.

You should be able to find a closed form expression for each pmf. For f_{1}(x_{1}), observe that you can pull p^2 outside the sum, and you're left with p^{2}\sum_{i=x_{1}}^{\infty}q^i. How does this sum relate to a geometric series?

 

[phiiota]

okay, so finally, we have that
f_{1}(x_{1})=\sum_{i=x_{1}}^{\infty}p^{2}q^{i}=p^{2}\sum_{i=x_{1}}^{\infty}q^{i}
so \sum_{i=0}^{\infty}q^{i}=\dfrac{1}{1-(1-p)}=\dfrac{1}{p}. \sum_{i=0}^{x_{1}}q^{i}=\dfrac{1-q^{x_{1}-1}}{1-q}=\dfrac{1-q^{x_{1}-1}}{p}, so taking the difference, we have
f_{1}(x)=p^{2}\left(\dfrac{q^{x_{1}-1}}{p}\right)=pq^{x_{1}-1}.

for f_2(x_2), we just have
f_{2}(x_{2})=\sum_{i=0}^{x_{2}}p^{2}q^{x_{2}}=(x_{2}+1)p^{2}q^{x_{2}}

is this right? thanks for your help.

 

[jbunniii]

okay, so finally, we have that
f_{1}(x_{1})=\sum_{i=x_{1}}^{\infty}p^{2}q^{i}=p^{2}\sum_{i=x_{1}}^{\infty}q^{i}
so \sum_{i=0}^{\infty}q^{i}=\dfrac{1}{1-(1-p)}=\dfrac{1}{p}.

Yes.

\sum_{i=0}^{x_{1}}q^{i}=\dfrac{1-q^{x_{1}-1}}{1-q}=\dfrac{1-q^{x_{1}-1}}{p},

The upper limit on your sum should be x_1 - 1, not x_1. Otherwise you will lose the x_1 term when you take the difference.

for f_2(x_2), we just have
f_{2}(x_{2})=\sum_{i=0}^{x_{2}}p^{2}q^{x_{2}}=(x_{2}+1)p^{2}q^{x_{2}}

Yes, but you need to specify the range of values of x_2 for which this formula is valid. For example, if x_2 is negative, this formula isn't right.

 

[Ray Vickson]

okay, so finally, we have that
f_{1}(x_{1})=\sum_{i=x_{1}}^{\infty}p^{2}q^{i}=p^{2}\sum_{i=x_{1}}^{\infty}q^{i}
so \sum_{i=0}^{\infty}q^{i}=\dfrac{1}{1-(1-p)}=\dfrac{1}{p}. \sum_{i=0}^{x_{1}}q^{i}=\dfrac{1-q^{x_{1}-1}}{1-q}=\dfrac{1-q^{x_{1}-1}}{p}, so taking the difference, we have
f_{1}(x)=p^{2}\left(\dfrac{q^{x_{1}-1}}{p}\right)=pq^{x_{1}-1}.

for f_2(x_2), we just have
f_{2}(x_{2})=\sum_{i=0}^{x_{2}}p^{2}q^{x_{2}}=(x_{2}+1)p^{2}q^{x_{2}}

is this right? thanks for your help.

What you have written makes no sense: your summation gives the wrong answer
\sum_{i=0}^{x_{2}}p^{2}q^{x_{2}}= (x_2+1) p^2 q^{x_2}; do you see why?
Maybe you intended to say
f_2(x_2) = \sum_{i=0}^{x_2}p^2 q^i,
in which case that is what you should have written.

BTW: it will save you a lot of time and trouble if you use only the needed brackets in LaTeX. When the subscript or superscript is a single symbol, you do not need curly brackets, so it is OK to write f_2(x_2) and q^i, for example, but you DO need brackets when you have things like \sum_{i=0}^{x_2}, because the sub- or super-scripts have more than one symbol. However, you do NOT need to write ^{x_{2}}.

RGV

 

[jbunniii]

What you have written makes no sense: your summation gives the wrong answer
\sum_{i=0}^{x_{2}}p^{2}q^{x_{2}}= (x_2+1) p^2 q^{x_2}; do you see why?

Assuming the joint pmf is correctly stated in the original post, this is in fact a valid expression for the pmf of x_2. What makes you say that (x_2+1) p^2 q^{x_2} (for x_2 \geq 0, otherwise 0) is the wrong answer? It's certainly a valid pmf: it takes on only nonnegative values and it sums to 1.

 

[Ray Vickson]

Assuming the joint pmf is correctly stated in the original post, this is in fact a valid expression for the pmf of x_2. What makes you say that (x_2+1) p^2 q^{x_2} (for x_2 \geq 0, otherwise 0) is the wrong answer? It's certainly a valid pmf: it takes on only nonnegative values and it sums to 1.

OK, right! I should not post anything before my first cup of coffee!

RGV

 

[phiiota]

Thanks. As far as the latex goes, I usually type it in an editor, and then paste the code in here.