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# Probability Theory | ISI MStat 2015 PSB Problem B5

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 $$\Rightarrow Z = X + Y$$ ~ $$N ( 0 , 3)$$.

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

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

$$= \frac{36}{\sqrt{6}\pi}$$.