Solution. \(f_{Y\mid X}(y\mid 0)=\) conditional distribution of \(Y\) given \(X=0\) \[f_{Y\mid X}(y\mid 0)=\frac {f_{X,Y}(0,y)}{f_X(0)}=\frac {P(X=0,Y=y)}{P(X=0)}=\frac {P(X=0,Y=y)}{0.3}.\] We have that \begin {align*} f_{Y\mid X}(y\mid 0) & = \frac {P(X=0,Y=0)}{0.3}, \quad y = 0\\ & = \frac {P(X=0,Y=1)}{0.3}\, , \quad y = 1\\ & = \frac {P(X=0,Y=2)}{0.3}\, , \quad y = 2\\ & = \frac {P(X=0,Y=3)}{0.3}\, , \quad y = 3. \end {align*}

Therefore

Solution. We first obtain the conditional distribution of \(X\) given that \(Y = y\). \[f_{X\mid Y}(x\mid y) = \frac {f(x,y)}{f_Y(y)}\] Then marginal \(f_Y(y)\) is \[f_Y(y) = \frac {1}{y}e^{-y}\int ^{\infty }_0 e^{-x/y}\, dx = \frac {1}{y}e^{-y}\left [-ye^{-x/y}\right ]^{\infty }_0 = e^{-y}.\] Hence, \[f_{X\mid Y}(x\mid y = 1) = \frac {e^{-x}\, e^{-1}}{e^{-1}} = e^{-x}.\] Therefore \[P(X > 1\mid Y = 1) = \int ^{\infty }_1e^{-x}\, dx = \left [-e^{-x}\right ]^{\infty }_1 = e^{-1}.\] □

Proof. In the independent case \[f_{X\mid Y}(x\mid y^*) = \frac {f_{X,Y}(x,y)}{f_Y(y^*)}=\frac {f_{X,Y}(X=x,Y=y^*)}{f_Y(y^*)} = \frac {f_X(x)\cdot f_Y(y^*)}{f_Y(y^*)} = f_X(x).\] □

Proof. Suppose that \(X\) and \(Y\) are continuous independent random variables. Then the joint p.d.f can be written as \(f_{X,Y}(x,y) = f_X(x)\, f_Y(y)\). Thus \begin {align*} E(XY) & = \int _x\int _y xy\, f_{X,Y}(x,y)\, dy\, dx\\ & = \int _x\int _y xy\, f_X(x)\, f_Y(y)\, dy \, dx\\ & = \left [\int _x x\, f_X(x)\, dx\right ]\, \left [\int _yy\,f_Y(y)\, dy\right ]\\ & = E(X)\, E(Y). \end {align*} □

Answer. Yes only if \(X\) and \(Y\) have joint normal distribution. Otherwise no! □

Proof. Let \[Q(t)=E\{[(X-\mu _X) + (Y-\mu _Y)t]^2\},\] since \([(X-\mu _X) + (Y-\mu _Y)t]^2\geq 0\) \[\implies E\{[(X-\mu _X)+(Y-\mu _Y)t]^2\}\geq 0\] i.e. \(\quad Q(t)\geq 0\) \begin {align*} Q(t) & = E\left \{\left (X-\mu _X\right )^2+2\left (X-\mu _X\right )\left (Y-\mu _Y\right )t+t^2\left (Y-\mu _Y\right )^2\right \}\\ & = E\left (X-\mu _X\right )^2+2tE\left [\left (X-\mu _X\right )\left (Y-\mu _Y\right )\right ]+t^2E\left (Y-\mu _Y\right )^2\\ & = \operatorname {Var}(X)+2\operatorname {Cov}(X,Y)t+t^2\operatorname {Var}(Y). \end {align*}

Therefore \[Q(t)= \operatorname {Var}(Y)t^2+2\operatorname {Cov}(X,Y)t + \operatorname {Var}(X).\] \(Q(t)\) is a quadratic in \(t\), since \(Q(t)\geq 0\) implies its discriminant must be less than or equal to zero.i.e. \[\left [2\operatorname {Cov}(X,Y)\right ]^2-4\operatorname {Var}(Y)\,\operatorname {Var}(X)\leq 0\] \[4\left [\operatorname {Cov}(X,Y)\right ]^2\leq 4\operatorname {Var}(X)\,\operatorname {Var}(Y)\] \[\implies \frac {\left [\operatorname {Cov}(X,Y)\right ]^2}{\operatorname {Var}(X)\,\operatorname {Var}(Y)}\leq 1\] \[\implies \left [\frac {\operatorname {Cov}(X,Y)}{\sqrt {\operatorname {Var}(X)\,\operatorname {Var}(Y)}}\right ]^2\leq 1\] \[\rho ^2_{XY}\leq 1\] Therefore, \(-1\leq \rho _{XY}\leq 1\). □

Solution. Since \(Y = X^2\) \begin {align*} \operatorname {Cov}(X,Y) & = E(XY) - E(X)\, E(Y)\\ & = E(X^3) - E(X)\, E(X^2). \end {align*}

Since \(X\) has the probability density function \[f_X(x) = \begin {cases} \frac {1}{2}\, & \text {if}\,\, -1\leq x\leq 1,\\ 0 & \text {otherwise}, \end {cases}\] we find that \[E(X) = \int ^1_{-1}x\, \frac {1}{2}\, dx = 0\] and \[E(X^3) = \int ^1_{-1}x^3\, \frac {1}{2}\, dx = 0.\] Therefore \[\operatorname {Cov}(X,Y) = 0.\] Hence \(\rho = 0\) although \(X\) and \(Y\) are clearly dependent: \(Y\) is a function of \(X\). For instance \(P(X>\frac 12, Y<\frac 14) = 0\), while \(P(X>\frac 12)P(Y<\frac 14) = \frac 14\cdot \frac 12>0\). □

Solution. To find \[\rho _{XY}=\frac {\operatorname {Cov}(X,Y)}{\sqrt {\operatorname {Var}(X)\cdot \operatorname {Var}(Y)}},\] we need to find \[\operatorname {Cov}(X,Y)=E(XY)-E(X)\,E(Y),\] \[\operatorname {Var}(X)=E(X^2)-[E(X)]^2\] and \[\operatorname {Var}(Y)=E(Y^2)-[E(Y)]^2.\] The marginals of \(X\) and \(Y\) are

and

. Then \[E(X)=\sum ^2_{x=0}\,xf_X(x)\, =\, \frac {15}{28}+\frac {6}{28}=\frac {21}{28}=\frac {3}{4},\]

\[E(X^2)=\sum ^2_{x=0}\, x^2f_X(x)\, =\frac {15}{28}+\frac {12}{28}=\frac {27}{28},\] Hence \[\operatorname {Var}(X)=\frac {27}{28}-\frac {9}{16}=\frac {45}{112}=\frac {9}{4}\cdot \frac {5}{28}.\] Also \[E(Y)=\sum ^2_{y=0}yf_Y(y)\, =\frac {12}{28}+\frac {2}{28}=\frac {14}{28}=\, \frac {1}{2},\]

\[E(Y^2)=\sum ^2_{y=0}y^2f_Y(y)=\frac {12}{28}+\frac {4}{28}=\frac {16}{28}=\,\frac {4}{7}\]

\[\operatorname {Var}(Y)=\frac {4}{7}-\frac {1}{4}=\frac {16-7}{28}=\, \frac {9}{28}.\]

\[E(XY)=1\times \frac {3}{14}=\frac {3}{14}.\] The covariance \[\operatorname {Cov}(X,Y)= \frac {3}{14}-\frac {3}{4}\cdot \frac {1}{2} = \frac {12 - 21}{56} = -\frac {9}{56}.\] Therefore \[\rho _{XY} =\frac {-\frac {9}{56}}{\sqrt {\frac {45}{112}\cdot \frac {9}{28}}} = \frac {-\frac {9}{56}}{\frac {9\sqrt 5}{56}} = -\frac {1}{\sqrt {5}} = -0.447 .\] □

Proof. Complete the square in \(v\): \(u^2 - 2\rho uv + v^2 = (v - \rho u)^2 + (1 - \rho ^2)u^2\). So \begin {align*} f_{X,Y}(x,y) &= \frac {1}{\sigma _X\sqrt {2\pi }}e^{-u^2/2}\\ &\quad \cdot \frac {1}{\sigma _Y\sqrt {2\pi (1 - \rho ^2)}}\exp \left \{-\frac {(v - \rho u)^2}{2(1 - \rho ^2)}\right \}. \end {align*}

In terms of \(y\), \[v - \rho u = \frac {1}{\sigma _Y}\left [y - \mu _Y - \rho \frac {\sigma _Y}{\sigma _X}(x - \mu _X)\right ],\] so the second factor is the density at \(y\) of a normal distribution with mean \(m(x) = \mu _Y + \rho \frac {\sigma _Y}{\sigma _X}(x - \mu _X)\) and variance \(\sigma _Y^2(1 - \rho ^2)\).

(i)

Integrating over \(y\), the second factor integrates to \(1\), leaving \(f_X(x) = \frac {1}{\sigma _X\sqrt {2\pi }}e^{-u^2/2}\), the \(N(\mu _X, \sigma _X^2)\) density. By symmetry \(Y\sim N(\mu _Y, \sigma _Y^2)\).

(ii)

\(f_{Y\mid X}(y\mid x) = \frac {f_{X,Y}(x,y)}{f_X(x)}\) is exactly the second factor. The statement for \(X\) given \(Y\) follows by exchanging the roles of the variables.

(iii)

By (ii), \(E(Y\mid X) = m(X)\), so \begin {align*} E\left [(X - \mu _X)(Y - \mu _Y)\right ] &= E\left [(X - \mu _X)\left (m(X) - \mu _Y\right )\right ]\\ &= \rho \frac {\sigma _Y}{\sigma _X}E(X - \mu _X)^2 = \rho \sigma _X\sigma _Y, \end {align*}

and the correlation is \(\rho \). If \(\rho = 0\) the joint density factorises into \(f_X(x)f_Y(y)\), so \(X\) and \(Y\) are independent; conversely independent variables have zero correlation.

Solution.

(a)

By theorem 3.3.22, \(Y\mid X = 186\) is normal with mean \(175 + 0.5\cdot \frac 87(186 - 172) = 175 + 8 = 183\) and variance \(64(1 - 0.25) = 48\), so standard deviation \(6.93\).

(b)

\(P(Y>186\mid X = 186) = P\left (Z>\frac {186 - 183}{6.93}\right ) = P(Z>0.43) = 0.333\).

(c)

Unconditionally \(Y\sim N(175, 64)\), so \(P(Y>186) = P\left (Z>\frac {11}{8}\right ) = P(Z>1.375) = 0.085\).

The son of a tall father is expected to be tall, \(183\) cm against the average of \(175\), but less exceptional than his father: the father is \(2\) standard deviations above the mean of fathers, and the son’s expected height is only \(1\) standard deviation above the mean of sons. This is Galton’s regression to the mean, and it is a consequence of \(|\rho |<1\), not of any biological mechanism.