Joint Distributions and Covariance Questions and Answers

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.

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.

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.

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 \(4 - 2.5 = 1.5 \neq 0\).