Communication, Networking, Signal and Image Processing (CS)

Question 1: Probability and Random Processes

August 2015

## Contents

### Solution 1

First of all, the conditional distribution can be written as:

$P(X=x|X+Y=n) =\frac{P(X=x, X+Y=n)}{P(X+Y=n)} =\frac{P(X=x, Y=n-x)}{P(X+Y=n)}$

And

$P(X=x, Y=n-x) =P(X=x)P(Y=n-x)\\ =\frac{e^{-\lambda_1}\lambda^x}{x!}\times \frac{e^{-\lambda_2}\lambda^(n-x)}{(n-x)!}\\ =\frac{e^{-(\lambda_1+\lambda_2)}}{x!} \left( \begin{array}{c} n\\x \end{array} \right) \lambda_1^x\lambda_2^{n-x}$

Also

$P(X+Y=n) ={\sum_{k=0}^{k=n}P(X=k,Y=n-k)} ={\sum_{k=0}^{k=n}P(X=k)P(Y=n-k)}\\ =\frac{e^{-(\lambda_1+\lambda_2)}}{n!}\sum_{k=0}^{k=n} \left( \begin{array}{c} n\\k \end{array} \right) \lambda_1^k\lambda_2^{n-k}\\ =\frac{e^{-(\lambda_1+\lambda_2)}}{n!}(\lambda_1+\lambda_2)^n$

So, we get $P(X=x|X+Y=n) = \left( \begin{array}{c} n\\k \end{array} \right) (\frac{\lambda_1}{\lambda_1+\lambda_2})^x(\frac{\lambda_2}{\lambda_1+\lambda_2})^{n-x}$

### Solution 2

Let $Z=X+Y$,

$P_Z(k)=P_Z(z=k)=P_Z(x+y=k)\\ =\sum_{i=0}^{k}P(x=i)(y=k-i)=\sum_{i=0}^{k}\frac{\lambda_1^i e^{\lambda_1}}{i!}\cdot\frac{\lambda_2^{k-i}e^{-\lambda_2}}{(k-i)!} =e^{-\lambda_1-\lambda_2}\cdot \frac{(\lambda_1+\lambda_2)^k}{k!}$

Using $(a+b)^k=\sum_{i=0}^{k}a^ib^{(k-i)}\cdot \frac{k!}{i!(k-i)!}$

Therefore,

$P_{X|Z}(x=k|z=n)=\frac{P(x=k,z=n)}{P(z=n)}=\frac{P(x=k,x+y=n)}{P(z=n)}=\frac{P(x=k,y=n-k)}{P(z=n)}\\ =\frac{\lambda_1^k e^{-\lambda_1}}{k!}\frac{\lambda_2^{n-k} e^{-\lambda_2}}{(n-k)!}\frac{n!}{e^{-\lambda_1}e^{-\lambda_2}(\lambda_1+\lambda_2)^n} = \frac{\lambda_1^k \lambda_2^{n-k} }{(\lambda_1+\lambda_2)^n}\frac{n!}{k!(n-k)!}$

### Solution 3

We will view this problem through the lens of Bayes' Theorem. As such, we can write the conditional distribution as

$P(X = x | X+Y = n) = \frac{P(X = x, X+Y = n)}{P(X+Y = n)} = \frac{P(X = x, Y = n - X)}{\sum^n_{k = 0}P(X = k, Y = n - k)}$.

Since $X$ and $Y$ are independent, we can further write

$P(X = x | X+Y = n) = \frac{P(X = x)P(Y = n - X)}{\sum^n_{k = 0}(P(X=k)P(Y = n-k))}$.

Now let us separate the above expression into numerator and denominator. Recalling that $X$ and $Y$ are independent Poisson r.v.s, the numerator is given by

$P(X = x)P(Y = n - X) = \frac{e^{-\lambda_1}\lambda_1^x}{x!}\cdot\frac{e^{-\lambda_2}\lambda_2^{n-x}}{(n-x)!}$.

Multiplying the above by $\frac{n!}{n!}$ gives

$P(X = x)P(Y = n - X) = \frac{e^{-\lambda_1 + \lambda_2}}{n!}{n\choose x}\lambda_1^x\lambda_2^{n-x}$.

Now let us examine the denominator. Again, we make use of the fact that $X$ and $Y$ are independent Poisson r.v.s to write

$\sum^n_{k = 0}(P(X=k)P(Y = n-k)) = \sum^n_{k = 0}\left(\frac{e^{-\lambda_1}\lambda_1^k}{k!}\frac{e^{-\lambda_2}\lambda_2^{n-k}}{(n-k)!}\right)$.

Again, we multiply by $\frac{n!}{n!}$ to obtain

$\sum^n_{k = 0}(P(X=k)P(Y = n-k)) = \frac{e^{-\lambda_1 + \lambda_2}}{n!}\sum^n_{k = 0}{n\choose k}\lambda_1^k\lambda_2^{n-k}$.

We can make use of the binomial formula to simplify this expression. Recall that the binomial formula is given by

$(a+b)^n = \sum^n_{k = 0}{n\choose k}a^k b^{n - k}$.

We use this to write

$. \sum^n_{k = 0}(P(X=k)P(Y = n-k)) = \frac{e^{-\lambda_1 + \lambda_2}}{n!}\cdot(\lambda_1 + \lambda_2)^n$

Putting this all together, we can finally write

$P(X = x | X+Y = n) = \frac{\frac{e^{-\lambda_1 + \lambda_2}}{n!}{n\choose x}\lambda_1^x\lambda_2^{n-x}}{\frac{e^{-\lambda_1 + \lambda_2}}{n!}\cdot(\lambda_1 + \lambda_2)^n} = {n\choose x}\frac{\lambda_1^x\lambda_2^{n-x}}{(\lambda_1 + \lambda_2)^n}$

and we are done.

### Similar Problem

If $X$ and $Y$ are independent binomial random variables with success probabilities $p$ and $q$ respectively, find the probability mass function of $X$ when $X + Y = k$. In addition, investigate what happens to this p.m.f. when $p = q$.