Question

(a) (2 pts) An urn contains 3 red and 5 green balls. At each step of this game, we pick one ball at random, note its color and return the ball to the urn together with anoter ball of the same color. Prove by induction that the probability that the ball we pick a red ball at the n-th step is 3/8. (b) (2pts) Consider any two random variables X, Y of any distirbution and not necesarily independent. Given that P(X = a) = P1, P(max(X,Y) = a) = P2, and P(min(X,Y)= a) = P3, find P(Y = a) in terms of P1, P2 and 23. Hint:Use the first Bayes theorem. Choose a suitable event-complement pair to partition the probability space by comparing X and Y. EXEY) (c) (2 pts) Show that, if X and Y are two uncorrelated (i.e. EXY Bernoulli (Indicator) random variables then they are independent.

Answer 1

a)

Let
n
denote the colour of the ball drawn at the $n^{th}$ step.

For n=2,

$P[X_{2}=red]=P[X_{1}=red]P[X_{2}=red|X_{1}=red]+\newline P[X_{1}=green]P[X_{2}=red|X_{1}=green]\newline =\frac{3}{8} \frac{4}{9}+\frac{5}{8}\frac{3}{9}=\frac{3}{8}$

Assuming the probability of drawing a red ball at n=k-1 step is 3/8.

For n=k,

$P[X_{k}=red]=\newline P[X_{k-1}=red]P[X_{k}=red|X_{k-1}=red]+\newline P[X_{k-1}=green]P[X_{k}=red|X_{k-1}=green]\newline =\frac{3}{8}\times P[X_{k}=red|X_{k-1}=red]+\frac{5}{8}\times P[X_{k}=red|X_{k-1}=green]\newline =\frac{3}{8}\frac{4}{9}+\frac{5}{8}\frac{3}{9}=\frac{3}{8}$

Hence, by mathematical induction the probability of drawig a red ball at the $n^{th}$ is 3/8.

b)

P2 = P[mat(X,Y) = a) = P[X = a] P[mat(X,Y) = a X = a) + P[X + a] P[max(X,Y) = a X +a] = P[X = a]PLY <a|X = a] + P[X #a]PLY = a Xa] = piPLY <a X = a1 + (1 - pı)P[Y = ax + a)....(*)

$p_{3}=P[min(X,Y)=a]=P[X=a]P[min(X,Y)=a|X=a]+P[X\neq a]P[min(X,Y)=a|X\neq a]\newline =P[X=a]P[Y\geq a|X=a]+P[X\neq a]P[Y=a|X\neq a]\newline =p_{1}P[Y\geq a|X=a]+(1-p_{1})P[Y=a|X\neq a]....(**)$

adding (*) and (**), we have

$p_{2}+p_{3}=(P[X= a]P[Y\leq a|X=a]+P[X = a]P[Y>a|X=a])+P[X=a]P[Y=a|X=a]+P[X\neq a]P[Y=a|X\neq a]=p_{1}+P[Y=a]\newline =>P[Y=a]=p_{2}+p_{3}-p_{1}$

c)

$E(XY)=P[X=1,Y=1]$

$E(X)E(Y)=P[X=1]P[Y=1]$

Now as X and Y are uncorrelated we have,

$P[X=1,Y=1]=P[X=1]P[Y=1]\newline =>P[X=1]P[Y=1|X=1]=P[X=1]P[Y=1]\newline =>P[Y=1|X=1]=P[Y=1]=>1-P[Y=1|X=1]=1-P[Y=1]\newline =>P[Y=0|X=1]=P[Y=0]$

Therefore, the conditional distribution of Y given is the unconditional distribution of Y.

Hence X and Y are independent.