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 .

Similar Problems and Solutions

ISI MStat PSB 2008 Problem 10 — Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

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 .

Similar Problems and Solutions

ISI MStat PSB 2008 Problem 10 — Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

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