Joint Distributions and Covariance Questions and Answers

Q: \(E(X)\) and \(E(Y)\) are, respectively,

Both means are \(3/5\): \(\frac{3}{5}\) and \(\frac{3}{5}\).Integrate over the unit square, splitting each integrand into products of one-variable integrals:\[ \begin{aligned} E(X) &= \tfrac{6}{5}\int\!\!\int x(x + y^{2}) \\ &= \tfrac{6}{5}\left(\tfrac{1}{3} + \tfrac{1}{2}\cdot\tfrac{1}{3}\right) = \tfrac{6}{5}\cdot\tfrac{1}{2} = \tfrac{3}{5} \end{aligned} \]\[ \begin{aligned} E(Y) &= \tfrac{6}{5}\int\!\!\int y(x + y^{2}) \\ &= \tfrac{6}{5}\left(\tfrac{1}{2}\cdot\tfrac{1}{2} + \tfrac{1}{4}\right) = \tfrac{6}{5}\cdot\tfrac{1}{2} = \tfrac{3}{5} \end{aligned} \]The density integrates to \(\frac{6}{5}(\frac{1}{2} + \frac{1}{3}) = 1\), as it must. Both variables lie in [0, 1], so a mean of \(6/5\) is impossible, which rules out two options at once; \(2/5\) for \(E(X)\) is an arithmetic slip.

Q: \(Cov(X, Y)\) is

The covariance is -0.01.First compute \(E(XY)\) over the unit square:\[ \begin{aligned} E(XY) &= \tfrac{6}{5}\int\!\!\int xy(x + y^{2}) \\ &= \tfrac{6}{5}\left(\tfrac{1}{3}\cdot\tfrac{1}{2} + \tfrac{1}{2}\cdot\tfrac{1}{4}\right) \\ &= \tfrac{6}{5}\cdot\tfrac{7}{24} = \tfrac{7}{20} = 0.35 \end{aligned} \]With \(E(X) = E(Y) = 3/5\) from Q.52:\[ Cov(X, Y) = 0.35 - (0.6)(0.6) = 0.35 - 0.36 = -0.01. \]The covariance is not 0, because the density \(x + y^{2}\) does not factor into a function of \(x\) times a function of \(y\), so \(X\) and \(Y\) are not independent. The positive values 0.01 and 0.02 have the wrong sign.

Q: Let \(X\) and \(Y\) be two random variables having the joint probability density function\[ f(x,y) = \begin{cases} 2 & \text{if } 0 \lt x \lt y \lt 1 \\ 0 & \text{otherwise.} \end{cases} \]Then the conditional probability \(P\left(X \le \frac{2}{3} \mid Y = \frac{3}{4}\right)\) is equal to

The conditional probability is \(\frac{8}{9}\).The marginal density of Y is \(f_{Y}(y) = \int_{0}^{y} 2\,dx = 2y\) for \(0 \lt y \lt 1\). So the conditional density of X given \(Y = y\) is\[ \begin{aligned} f_{X \mid Y}(x \mid y) &= \frac{2}{2y} = \frac{1}{y}, \\ &0 \lt x \lt y \end{aligned} \]That is, given \(Y = \frac{3}{4}\), X is uniform on \((0, \frac{3}{4})\):\[ \begin{aligned} P\left(X \le \tfrac{2}{3} \mid Y = \tfrac{3}{4}\right) &= \frac{2/3}{3/4} \\ &= \frac{8}{9} \end{aligned} \]\(\frac{2}{3}\) would be the answer if X were uniform on \((0, 1)\), ignoring the conditioning.

Q: Let \((X, Y)\) be a random vector with \(Var(X) = Var(Y) = 1\) and \(Cov(X, Y) = -0.5\). For which one of the following values of \(b\), the random variables \(U = bX + Y\) and \(V = X + bY\) are uncorrelated?

The value is \(b = 2 - \sqrt{3}\).Expand the covariance using bilinearity:\[ \begin{aligned} Cov(U, V) &= b\,Var(X) + (b^{2} + 1)Cov(X, Y) + b\,Var(Y) \\ &= 2b - 0.5(b^{2} + 1) \end{aligned} \]Setting this to zero gives \(b^{2} - 4b + 1 = 0\), so \(b = 2 \pm \sqrt{3}\). Of these only \(2 - \sqrt{3}\) is offered.\(\sqrt{3} - 2\) is its negative and gives \(Cov(U,V) = 2(\sqrt{3} - 2) - 0.5(8 - 4\sqrt{3}) = 4\sqrt{3} - 8 \neq 0\). Substituting \(b = \sqrt{3}\) gives \(2\sqrt{3} - 2 \neq 0\), and \(b = 2\) gives \(4 - 2.5 = 1.5 \neq 0\).

Q: Let X and Y be random variables with joint distribution :\[ \Pr(X = i, Y = j) = \begin{cases} 1/6 & \text{for } i = 5,6 \text{ and } j = 3,6,9 \\ 0 & \text{otherwise} \end{cases} \]Then the mutual information I (X, Y) = .................. .

The mutual information is 0.The joint distribution puts probability 1/6 on each of the 2 × 3 = 6 pairs. The marginals are therefore uniform:\[ \begin{aligned} P(X = i) &= 3 \times \tfrac{1}{6} = \tfrac{1}{2} \\ P(Y = j) &= 2 \times \tfrac{1}{6} = \tfrac{1}{3} \end{aligned} \]Since \(\tfrac{1}{2} \times \tfrac{1}{3} = \tfrac{1}{6}\) for every pair, \(P(X, Y) = P(X)P(Y)\): X and Y are independent, and independent variables share no information.\[ \begin{aligned} I(X, Y) &= H(X) + H(Y) - H(X, Y) \\ &= \log 2 + \log 3 - \log 6 = 0 \end{aligned} \]The distractors are the separate entropies: log(2) is H(X), log(3) is H(Y), and log(6) is the joint entropy H(X, Y).

Q: Let X and Y be random variables with joint distribution :\[ \Pr(X = i, Y = j) = \begin{cases} 1/6 & \text{for } i = 5,6 \text{ and } j = 3,6,9 \\ 0 & \text{otherwise} \end{cases} \]Then the conditional entropy H(X|Y) = .................. .

The conditional entropy is log(2).The joint distribution spreads probability 1/6 evenly over the six pairs with \(i \in \{5, 6\}\) and \(j \in \{3, 6, 9\}\). So X is uniform on two values and Y uniform on three, and every joint probability is the product of the marginals:\[ \begin{aligned} P(X=i) &= \tfrac{1}{2}, \quad P(Y=j) = \tfrac{1}{3} \\ P(X=i, Y=j) &= \tfrac{1}{2} \cdot \tfrac{1}{3} = \tfrac{1}{6} \end{aligned} \]X and Y are therefore independent, and knowing Y tells us nothing about X:\[ H(X \mid Y) = H(X) = \log 2 \]0 would mean Y determines X completely. log(3) is H(Y), and log(6) is the joint entropy H(X, Y) = log 2 + log 3.

Direction for [ Question No: 1 To 2 ] :

The joint probability density function of two random variables X and Y is given as

\[ f(x, y) = \begin{cases} \frac{6}{5}(x + y^{2}), & 0 \le x \le 1, 0 \le y \le 1 \\ 0, & \text{elsewhere} \end{cases} \]

\(E(X)\) and \(E(Y)\) are, respectively,

A.\(\frac{2}{5}\) and \(\frac{3}{5}\)
B.\(\frac{3}{5}\) and \(\frac{3}{5}\)
C.\(\frac{3}{5}\) and \(\frac{6}{5}\)
D.\(\frac{4}{5}\) and \(\frac{6}{5}\)

Exam Year(s)

GATE Mathematics - 2012,

Answer: B

Both means are \(3/5\): \(\frac{3}{5}\) and \(\frac{3}{5}\).

Integrate over the unit square, splitting each integrand into products of one-variable integrals:

\[ \begin{aligned} E(X) &= \tfrac{6}{5}\int\!\!\int x(x + y^{2}) \\ &= \tfrac{6}{5}\left(\tfrac{1}{3} + \tfrac{1}{2}\cdot\tfrac{1}{3}\right) = \tfrac{6}{5}\cdot\tfrac{1}{2} = \tfrac{3}{5} \end{aligned} \]

\[ \begin{aligned} E(Y) &= \tfrac{6}{5}\int\!\!\int y(x + y^{2}) \\ &= \tfrac{6}{5}\left(\tfrac{1}{2}\cdot\tfrac{1}{2} + \tfrac{1}{4}\right) = \tfrac{6}{5}\cdot\tfrac{1}{2} = \tfrac{3}{5} \end{aligned} \]

The density integrates to \(\frac{6}{5}(\frac{1}{2} + \frac{1}{3}) = 1\), as it must. Both variables lie in [0, 1], so a mean of \(6/5\) is impossible, which rules out two options at once; \(2/5\) for \(E(X)\) is an arithmetic slip.

Discuss in Forum

\(Cov(X, Y)\) is

A.-0.01
B.0
C.0.01
D.0.02

Exam Year(s)

GATE Mathematics - 2012,

Answer: A

The covariance is -0.01.

First compute \(E(XY)\) over the unit square:

\[ \begin{aligned} E(XY) &= \tfrac{6}{5}\int\!\!\int xy(x + y^{2}) \\ &= \tfrac{6}{5}\left(\tfrac{1}{3}\cdot\tfrac{1}{2} + \tfrac{1}{2}\cdot\tfrac{1}{4}\right) \\ &= \tfrac{6}{5}\cdot\tfrac{7}{24} = \tfrac{7}{20} = 0.35 \end{aligned} \]

With \(E(X) = E(Y) = 3/5\) from Q.52:

\[ Cov(X, Y) = 0.35 - (0.6)(0.6) = 0.35 - 0.36 = -0.01. \]

The covariance is not 0, because the density \(x + y^{2}\) does not factor into a function of \(x\) times a function of \(y\), so \(X\) and \(Y\) are not independent. The positive values 0.01 and 0.02 have the wrong sign.

Discuss in Forum

Let \(X\) and \(Y\) be two random variables having the joint probability density function

\[ f(x,y) = \begin{cases} 2 & \text{if } 0 \lt x \lt y \lt 1 \\ 0 & \text{otherwise.} \end{cases} \]

Then the conditional probability \(P\left(X \le \frac{2}{3} \mid Y = \frac{3}{4}\right)\) is equal to

A.\(\frac{5}{9}\)
B.\(\frac{2}{3}\)
C.\(\frac{7}{9}\)
D.\(\frac{8}{9}\)

Exam Year(s)

GATE Mathematics - 2015,

Answer: D

The conditional probability is \(\frac{8}{9}\).

The marginal density of Y is \(f_{Y}(y) = \int_{0}^{y} 2\,dx = 2y\) for \(0 \lt y \lt 1\). So the conditional density of X given \(Y = y\) is

\[ \begin{aligned} f_{X \mid Y}(x \mid y) &= \frac{2}{2y} = \frac{1}{y}, \\ &0 \lt x \lt y \end{aligned} \]

That is, given \(Y = \frac{3}{4}\), X is uniform on \((0, \frac{3}{4})\):

\[ \begin{aligned} P\left(X \le \tfrac{2}{3} \mid Y = \tfrac{3}{4}\right) &= \frac{2/3}{3/4} \\ &= \frac{8}{9} \end{aligned} \]

\(\frac{2}{3}\) would be the answer if X were uniform on \((0, 1)\), ignoring the conditioning.

Discuss in Forum

Let \((X, Y)\) be a random vector with \(Var(X) = Var(Y) = 1\) and \(Cov(X, Y) = -0.5\). For which one of the following values of \(b\), the random variables \(U = bX + Y\) and \(V = X + bY\) are uncorrelated?

A.\(2 - \sqrt{3}\)
B.\(\sqrt{3} - 2\)
C.\(\sqrt{3}\)
D.\(2\)

Exam Year(s)

IIT JAM - 2026,

Answer: A

The value is \(b = 2 - \sqrt{3}\).

Expand the covariance using bilinearity:

\[ \begin{aligned} Cov(U, V) &= b\,Var(X) + (b^{2} + 1)Cov(X, Y) + b\,Var(Y) \\ &= 2b - 0.5(b^{2} + 1) \end{aligned} \]

Setting this to zero gives \(b^{2} - 4b + 1 = 0\), so \(b = 2 \pm \sqrt{3}\). Of these only \(2 - \sqrt{3}\) is offered.

\(\sqrt{3} - 2\) is its negative and gives \(Cov(U,V) = 2(\sqrt{3} - 2) - 0.5(8 - 4\sqrt{3}) = 4\sqrt{3} - 8 \neq 0\). Substituting \(b = \sqrt{3}\) gives \(2\sqrt{3} - 2 \neq 0\), and \(b = 2\) gives \(4 - 2.5 = 1.5 \neq 0\).

Discuss in Forum

Let X and Y be random variables with joint distribution :

\[ \Pr(X = i, Y = j) = \begin{cases} 1/6 & \text{for } i = 5,6 \text{ and } j = 3,6,9 \\ 0 & \text{otherwise} \end{cases} \]

Then the mutual information I (X, Y) = .................. .

A.0
B.log(2)
C.log(3)
D.log(6)

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: A

The mutual information is 0.

The joint distribution puts probability 1/6 on each of the 2 × 3 = 6 pairs. The marginals are therefore uniform:

\[ \begin{aligned} P(X = i) &= 3 \times \tfrac{1}{6} = \tfrac{1}{2} \\ P(Y = j) &= 2 \times \tfrac{1}{6} = \tfrac{1}{3} \end{aligned} \]

Since \(\tfrac{1}{2} \times \tfrac{1}{3} = \tfrac{1}{6}\) for every pair, \(P(X, Y) = P(X)P(Y)\): X and Y are independent, and independent variables share no information.

\[ \begin{aligned} I(X, Y) &= H(X) + H(Y) - H(X, Y) \\ &= \log 2 + \log 3 - \log 6 = 0 \end{aligned} \]

The distractors are the separate entropies: log(2) is H(X), log(3) is H(Y), and log(6) is the joint entropy H(X, Y).

Discuss in Forum

Let X and Y be random variables with joint distribution :

\[ \Pr(X = i, Y = j) = \begin{cases} 1/6 & \text{for } i = 5,6 \text{ and } j = 3,6,9 \\ 0 & \text{otherwise} \end{cases} \]

Then the conditional entropy H(X|Y) = .................. .

A.0
B.log(2)
C.log(3)
D.log(6)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: B

The conditional entropy is log(2).

The joint distribution spreads probability 1/6 evenly over the six pairs with \(i \in \{5, 6\}\) and \(j \in \{3, 6, 9\}\). So X is uniform on two values and Y uniform on three, and every joint probability is the product of the marginals:

\[ \begin{aligned} P(X=i) &= \tfrac{1}{2}, \quad P(Y=j) = \tfrac{1}{3} \\ P(X=i, Y=j) &= \tfrac{1}{2} \cdot \tfrac{1}{3} = \tfrac{1}{6} \end{aligned} \]

X and Y are therefore independent, and knowing Y tells us nothing about X:

\[ H(X \mid Y) = H(X) = \log 2 \]

0 would mean Y determines X completely. log(3) is H(Y), and log(6) is the joint entropy H(X, Y) = log 2 + log 3.

Discuss in Forum

Loading…

Topics

Other Subjects

Tags: Joint Distributions and Covariance Questions and Answers || Joint Distributions and Covariance MCQ Questions and Answers || Joint Distributions and Covariance GK Questions and Answers || Joint Distributions and Covariance GK MCQ Questions || Joint Distributions and Covariance Multiple Choice Questions and Answers || Joint Distributions and Covariance GK || GK on Joint Distributions and Covariance || Probability and Statistics Questions and Answers || Probability and Statistics MCQ Questions and Answers || Probability and Statistics GK Questions and Answers || GK on Probability and Statistics

Joint Distributions and Covariance Questions and Answers

Direction for [ Question No: 1 To 2 ] :

Topics

Other Subjects

About Us

Stay informed about all important Dates & Notifications