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 .

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