Difference between revisions of "ECE PhD QE CNSIP 2015 Problem1.1" - Rhea

Latest revision as of 22:04, 31 January 2016

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 $.

Back to QE CS question 1, August 2015

Back to ECE Qualifying Exams (QE) page