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.
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\((|X+Y|^{3})\).
(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.
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\((|X+Y|^{3})\).
(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 \( \Rightarrow Z = X + Y \) ~ \(N ( 0 , 3)\).
\( \text{E}(|Z|^3) \overset{\text{ Z is symmetric around 0}}{=}\)
\( 2 \times \frac{1}{\sqrt{6}\pi} \times \int_{0}^{\infty} z^3 e^{ - \frac{z^2}{6}} dz \overset{u = \frac{z^2}{6}}{=} \frac{1}{\sqrt{6}\pi} \times 36 \times \int_{0}^{\infty} u^{2 - 1}.e^{ - u} du = \frac{1}{\sqrt{6}\pi} \times 36\Gamma(2) \)
\(= \frac{36}{\sqrt{6}\pi}\).
