This is a detailed solution based on Probability Theory of ISI MStat 2015 PSB Problem B5, with the prerequisites mentioned explicitly. Stay tuned for more.

## Problem

Suppose that \(X\) and \(Y\) are random variables such that

\(E(X+Y) = E(X-Y)=0 \)

Var\((X+Y)=3 \)

Var\((X-Y)=1\)

(a) Evaluate Cov\((X, Y)\).

(b) Show that \(E|X+Y| \leq \sqrt{3}\).

(c) If in addition, it is given that \((X, Y)\) is bivariate normal, calculate E\((|X+Y|^{3})\).

### Prerequisites

**Basic Probability theory ( Expectation, Variance, and Covariance )****Normal Distribution****Gamma Integral**

## Solution

**(a)**

\(Var(aX + bY) = a^2\text{Var}(X) +2ab\text{Cov}(X,Y) + b^2\text{Var}(Y) \rightarrow (*) \)

Using \((*)\), we use Var\((X+Y)\) – Var\((X-Y)\) = \(4 \text{Cov}(X,Y) = 2\)

\(\Rightarrow \text{Cov}(X,Y) = \frac{1}{2} \).

**(b)**

Say \(Z = (X + Y)\)

\({\text{E}(Z)} = 0\).

\( \text{Var}(Z) = \text{E}(Z^2) – {\text{E}(Z)}^2 = \text{E}(Z^2) \).

Do you remember the Cauchy – Schwartz Inequality?

\(3 = \text{Var}(Z) = \text{E}(Z^2) = \text{E}(|Z|^2) \overset{ Cauchy – Schwartz Inequality }{ \geq } {\text{E}(|Z|)}^2 \). Hence, \( {\text{E}(|Z|)} \leq \sqrt{3} \).

**(c)**

\(Z = (X + Y)\) and \( \text{E}(X) = \text{E}(Y) = 0\).

\((X, Y)\) is bivariate normal \( \Rightarrow Z = X + Y \) ~ \(N ( 0 , 3)\).

\( \text{E}(|Z|^3) \overset{\text{ Z is symmetric around 0}}{=}\)

\( 2 \times \frac{1}{\sqrt{6}\pi} \times \int_{0}^{\infty} z^3 e^{ – \frac{z^2}{6}} dz \overset{u = \frac{z^2}{6}}{=} \frac{1}{\sqrt{6}\pi} \times 36 \times \int_{0}^{\infty} u^{2 – 1}.e^{ – u} du = \frac{1}{\sqrt{6}\pi} \times 36\Gamma(2) \)

\(= \frac{36}{\sqrt{6}\pi}\).

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