# Thread: Question mostly on sum transformation than on actual marginal dsitribution

1. ## Question mostly on sum transformation than on actual marginal dsitribution

Hi,

So I have a problem given as example in class, but with a whole bunch of steps skipped (don't know if intentionally...):

For two RVs, $X_1, X_2$, the joint pmf is given by:

$p(m, n) = {{n}\choose{m}}p_1(1-p_1)^{n}p_2^{m}(1-p_2)^{n-m}$

Now, marginal distrib. of $X_1$ is defined as:

$p_{X_1}(m) = \sum_{\text{all } n} p(m,n) = \sum_{n = m}^{\infty}{{n}\choose{m}}p_1(1-p_1)^{n}p_2^{m}(1-p_2)^{n-m} =$

$p_1p_2^{m}(1-p_1)^{m} \sum_{n = m}^{\infty}{{n}\choose{m}}(1-p_1)^{n-m}(1-p_2)^{n-m} =$
.
. steps missing which I can't figure out!!!
.
$= p_1p_2^{m}(1-p_1)^{m} [ 1- (1-p_{1})(1-p_2)]^{-(m+1)}$

I would really appreciate if you could help we understand the transforms,
Thanks

2. ## Re: Question mostly on sum transformation than on actual marginal dsitribution

Originally Posted by dokrbb
Hi,

So I have a problem given as example in class, but with a whole bunch of steps skipped (don't know if intentionally...):

For two RVs, $X_1, X_2$, the joint pmf is given by:

$p(m, n) = {{n}\choose{m}}p_1(1-p_1)^{n}p_2^{m}(1-p_2)^{n-m}$

Now, marginal distrib. of $X_1$ is defined as:

$p_{X_1}(m) = \sum_{\text{all } n} p(m,n) = \sum_{n = m}^{\infty}{{n}\choose{m}}p_1(1-p_1)^{n}p_2^{m}(1-p_2)^{n-m} =$

$p_1p_2^{m}(1-p_1)^{m} \sum_{n = m}^{\infty}{{n}\choose{m}}(1-p_1)^{n-m}(1-p_2)^{n-m} =$
You know, I am sure, that $(1- p_1)^{n-m}= (1- p_1)^n(1- p_1)^{-m}$ and similarly for $(1- p_2)^{n-m}$. So the first thing I would do is factor $(1- p_1)^{-m}$ and $(1- p_2)^{-m}$ to give
$p_1p_2^{m}(1-p_2)^{-m} \sum_{n = m}^{\infty}{{n}\choose{m}}(1-p_1)^{n}(1-p_2)^{n}$
Now try using the "binomial theorem"- $(a- b)^n= \sum_{m=0}^n \begin{pmatrix}n \\ m\end{pmatrix}a^mb^{n-m}$

steps missing which I can't figure out!!!
.
$= p_1p_2^{m}(1-p_1)^{m} [ 1- (1-p_{1})(1-p_2)]^{-(m+1)}$

I would really appreciate if you could help we understand the transforms,
Thanks

3. ## Re: Question mostly on sum transformation than on actual marginal dsitribution

OMG, Thank you, very much!

4. ## Re: Question mostly on sum transformation than on actual marginal dsitribution

Originally Posted by HallsofIvy
You know, I am sure, that $(1- p_1)^{n-m}= (1- p_1)^n(1- p_1)^{-m}$ and similarly for $(1- p_2)^{n-m}$. So the first thing I would do is factor $(1- p_1)^{-m}$ and $(1- p_2)^{-m}$ to give
$p_1p_2^{m}(1-p_2)^{-m} \sum_{n = m}^{\infty}{{n}\choose{m}}(1-p_1)^{n}(1-p_2)^{n}$
Now try using the "binomial theorem"- $(a- b)^n= \sum_{m=0}^n \begin{pmatrix}n \\ m\end{pmatrix}a^mb^{n-m}$
So, I thought I got it, but...

$p_1p_2^{m}(1-p_2)^{-m} \sum_{n = m}^{\infty}{{n}\choose{m}}[(1-p_1)(1-p_2)]^{n}$

now, by the binomial we have $(a + b)^n= \sum_{m=0}^n \begin{pmatrix}n \\ m\end{pmatrix}a^mb^{n-m}$

so, how he is getting the $(1 - (term^n)^{-m-1})$...