Question

Problem 3. (5 pts) Discrete Random Variables (a) (3 pts) Let A = {1,2,3,4}. Pick a subset B C A uniformly among the 24 subsets (i.e. the power set of A) and let X be its size. Then likewise pick a subset C CB uniformly from the power set of B and let y be its size. Give the joint p.m.f of (X,Y) and compute E(X – Y). Note: X,Y can take value 0 if you pick the empty set. You can either write down a table or a compact expression of the form P(X = 1, Y = j). х (b) (2 pts) Verify that for j> 1: р n-j + P(Bin(n, p) = j) X P(Binan, p) = -1) 1-P 3 Now use this fact to determine the value of j for which P(Bin(n,p) = j) is highest. ("--j+1)

Answer 1

a)

$P[X=x,Y=y]=\left\{\begin{matrix} P[X=x]P[Y=y|X=x] & x=0,1,2,3,4; y=0,1,..,x\\ 0&, otherwise \end{matrix}\right.\newline =\left\{\begin{matrix} \frac{\binom{4}{x}}{2^{4}}\frac{\binom{x}{y}}{2^{x}}& x=0,1,2,3,4; y=0,1,..,x\\ 0&, otherwise \end{matrix}\right.$

Clearly,$X \sim Bin(4,\frac{1}{2})$ and $Y|X=x \sim Bin(x,\frac{1}{2})$

$E(X-Y)=E_{X}E_{Y|X}(X-Y)=E_{X}(X-E_{Y|X}(Y))=E_{X}(X-\frac{X}{2})\newline =E_{X}(\frac{X}{2})=\frac{1}{2}\times 4\times \frac{1}{2}=1$

b)

$P[Bin(n,p)=j]=\binom{n}{j}p^{j}(1-p)^{n-j}=\frac{n!}{(n-j)!j!}p^{j}(1-p)^{n-j}\newline =\frac{p}{(1-p)}\frac{n!}{(n-j)!j!}p^{j-1}(1-p)^{n-j+1}\newline =\frac{p}{(1-p)}\frac{(n-j+1)}{j}\frac{n!}{(n-j+1)!(j-1)!}p^{j-1}(1-p)^{n-j+1}\newline =\frac{p}{(1-p)}\frac{(n-j+1)}{j}P[Bin(n,p)=(j-1)]$