PROB_Midsem_PartB

Vasudeva S N

Problem 3

Question

Suppose it is given in a probability space that at least one, but no more than three, of the events $A_r$, $1\le r\le n$, occur, where $n\ge 3$; the probability of at least two occurring is $1/2$. Further if $P(A_r)=p$, $P(A_r\cap A_s)=q$, $r\ne s$ and $P(A_r\cap A_s\cap A_t)=x$, $r<s<t$, then show that $p\ge \frac{3}{2n}$ and $q\le \frac{4}{n}$.

We know that at least one of the $n$ events occur, so $P\left(\bigcup A_i\right)=1$. Next,

$$\begin{array}{rl}P\left(\bigcup _{i=1}^n{A}_i\right)& =\sum _{i}P(A_i)-\sum _{i\ne j}P(A_i\cap A_j)+\sum _{i\ne j\ne k}P(A_i\cap A_j\cap A_k)\\ ⟹1& =np-\left(\genfrac{}{}{0ex}{}{n}{2}\right)q+\left(\genfrac{}{}{0ex}{}{n}{3}\right)x(1)\end{array}$$

since no more than three events can occur concurrently. The probability of at least two occurring is given by

$$\begin{array}{rl}P\left(\bigcup _{i\ne j}(A_i\cap A_j)\right)& =\sum _{i\ne j}P(A_i\cap A_j)-3\sum _{i\ne j\ne k}P(A_i\cap A_j\cap A_k)+\sum _{i\ne j\ne k}P(A_i\cap A_j\cap A_k)\\ ⟹\frac{1}{2}& =\left(\genfrac{}{}{0ex}{}{n}{2}\right)q-2\left(\genfrac{}{}{0ex}{}{n}{3}\right)x(2)\end{array}$$

Now, $(1)+(2)$ gives us

$$\begin{array}{rl}& \frac{3}{2}=np-\left(\genfrac{}{}{0ex}{}{n}{3}\right)x\\ & p=\frac{3}{2n}+\left(\genfrac{}{}{0ex}{}{n}{3}\right)\frac{x}{n}\\ & p\ge \frac{3}{2n}\end{array}$$

From $(2)$, we know that $q\left(\genfrac{}{}{0ex}{}{n}{2}\right)=\frac{1}{2}+2\left(\genfrac{}{}{0ex}{}{n}{3}\right)x$. Thus, we need a bound for $x$ in terms of $q$. Observe that $P(\text{exactly two events occur})=$ $P(\text{at least two events occur})-$ $P(\text{exactly three events occur})$. The probability that exactly three events occur is $\left(\genfrac{}{}{0ex}{}{n}{3}\right)x$. Thus,

$$\begin{array}{rl}P(\text{exactly two events occur})& =\left(\genfrac{}{}{0ex}{}{n}{2}\right)q-3\left(\genfrac{}{}{0ex}{}{n}{3}\right)x\ge 0\\ & 2\left(\genfrac{}{}{0ex}{}{n}{3}\right)x\le \frac{2}{3}\left(\genfrac{}{}{0ex}{}{n}{2}\right)q\end{array}$$

Thus,

$$\begin{array}{rl}q\left(\genfrac{}{}{0ex}{}{n}{2}\right)& \le \frac{1}{2}+\frac{2}{3}\left(\genfrac{}{}{0ex}{}{n}{2}\right)q\\ \left(\genfrac{}{}{0ex}{}{n}{2}\right)q& \le \frac{3}{2}\\ q& \le \frac{3}{n(n-1)}\le \frac{4}{n}\end{array}$$

Problem 4

Question

Let $X$ and $Y$ be discrete random variables with mean $0$, variance $1$ and covariance $c$. Prove that $E(\text{Max}(X^2,Y^2))\le 1+\sqrt{1-c^2}$.

Note that since $EX=EY=0$ and $\text{Var}(X)=\text{Var}(Y)=1$, $E(X^2)=E(Y^2)=1$, and $E(XY)=c$.

$$\begin{array}{rl}E(\text{Max}(X^2,Y^2))& =E\left(\frac{X^2+Y^2}{2}+\left|\frac{X^2-Y^2}{2}\right|\right)\\ & =\frac{E(X^2)}{2}+\frac{E(Y^2)}{2}+\frac{E(|X-Y|(X+Y))}{2}\\ & \le 1+\frac{1}{2}\sqrt{E|X-Y{|}^{2}E(X+Y{)}^2}\\ & \le 1+\frac{1}{2}\sqrt{(2-2E(XY))(2+2E(XY))}\\ & \le 1+\sqrt{1-E(XY{)}^2}\\ & \le 1+\sqrt{1-c^2}\end{array}$$

Problem 5

Question

Define the conditional variance of $Y$ given $X$ by $\text{Var}(Y|X)=E((Y-E(Y|X){)}^2|X)$. Show that $\text{Var}(Y)=E(\text{Var}(Y|X))+\text{Var}(E(Y|X))$.

$$\begin{array}{rl}& E(\text{Var}(Y|X))+\text{Var}(E(Y|X))\\ & =E(Y-E(Y|X){)}^2+E((E(Y|X){)}^2)-(E(E(Y|X)){)}^2\\ & =E(Y^2+(E(Y|X){)}^2-2YE(Y|X))+E((E(Y|X){)}^2)-(EY{)}^2\\ & =E(Y^2)-(EY{)}^2+2E(E(Y|X)E(Y|X))-2E(YE(Y|X))\end{array}$$

Now, note that $E(Y|X)$ is a function of $X$, so $E(Y|X)E(Y|X)=E(YE(Y|X)|X)$. Thus, $E(E(Y|X)E(Y|X))=E(YE(Y|X))$. Therefore, $E(\text{Var}(Y|X))+\text{Var}(E(Y|X))=E(Y^2)-(EY{)}^2=\text{Var}(Y)$.