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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}.

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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