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# ISI MStat PSB 2006 Problem 6 | Counting Principle & Expectations

This is a very beautiful sample problem from ISI MStat PSB 2006 Problem 6 based on counting principle . Let's give it a try !!

## Problem- ISI MStat PSB 2006 Problem 6

Let $$Y_{1}, Y_{2}, Y_{3}$$ be i.i.d. continuous random variables. For i=1,2, define $$U_{i}$$ as

$$U_{i}= \begin{cases} 1 , & Y_{i+1}>Y_{i} \\ 0 , & \text{otherwise} \end{cases}$$
Find the mean and variance of $$U_{1}+U_{2}$$ .

### Prerequisites

Basic Counting Principle

Probability

Continuous random variable

## Solution :

$$E(U_1+U_2)=E(U_1)+E(U_2)$$

Now $$E(U_1)=1 \times P(Y_2>Y_1) = \frac{1}{2}$$, as there are only two cases either $$Y_2>Y_1$$ or $$Y_2<Y_1$$.

Similarly , $$E(U_2)= \frac{1}{2}$$

So, $$E(U_1+U_2)= 1$$

$$Var(U_1+U_2)=Var(U_1)+Var(U_2)+2Cov(U_1,U_2)$$ .

$$Var(U_1)=E({U_1}^2)-{E(U_1)}^2=1^2 \times P(Y_2>Y_1) - {\frac{1}{2}}^2 = \frac{1}{2}-\frac{1}{4}=\frac{1}{4}$$

Similarly ,$$Var(U_2)=\frac{1}{4}$$

$$Cov(U_1,U_2)=E(U_1 U_2)-E(U_1)E(U_2)=1 \times P(Y_3>Y_2>Y_1) - {\frac{1}{2}}^2 = \frac{1}{3!}-{\frac{1}{2}}^2$$ ( as there are 3! possible arrangements of $$Y_i's$$ keeping inequalities fixed .

Therefore , $$Var(U_1+U_2) = 2 \times \frac{1}{4} + 2 \times (\frac{1}{6}- \frac{1}{4}) = \frac{1}{3}$$

## Food For Thought

Find the same under the condition that $$Y_i's$$ are iid poission random variables .