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

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

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

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