Exercises

$\sum_{i=1}^2 \sum_{j=1}^3 c(i+j) = c(2+3+4+3+4+5) = 21c = 1$, so $c = 1/21$.

$P_{X}(1) = (2+3+4)/21 = 9/21 = 3/7$, $P_{X}(2) = (3+4+5)/21 = 12/21 = 4/7$.

$P_{Y}(1) = (2+3)/21 = 5/21$, $P_{Y}(2) = (3+4)/21 = 7/21 = 1/3$, $P_{Y}(3) = (4+5)/21 = 9/21 = 3/7$.

$P_{X,Y}(1,1) = 2/21$, but $P_{X}(1)P_{Y}(1) = (3/7)(5/21) = 15/147 = 5/49 \ne 2/21$. Therefore $X$ and $Y$ are not independent.

$f_{X,Y}(x,y) = 1/\pi$ for $x^2 + y^2 \le 1$.

$f_{X}(x) = \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \frac{1}{\pi}\,dy = \frac{2\sqrt{1-x^2}}{\pi}, \quad |x| \le 1.$$

$\mathbb{P}(X > Y) = \int_0^\infty e^{-y} \cdot 2e^{-2y}\,dy = 2\int_0^\infty e^{-3y}\,dy = \frac{2}{3}.$$

$f_{X}(x) = \int_x^\infty e^{-y}\,dy = e^{-x}$ for $x \ge 0$.

$f(y \mid x) = e^{-y}/e^{-x} = e^{-(y-x)}$ for $y \ge x$. Given $X = x$, $Y - x \sim \text{Exp}(1)$.

$\mathbb{E}[Y \mid X = x] = x + 1$ (shift by $x$, plus mean 1).

$\mathbb{E}[\text{Var}(Y \mid X)] = \mathbb{E}[X^2/12] = 2/12 = 1/6$.

$\text{Var}(\mathbb{E}[Y \mid X]) = \text{Var}(X/2) = \text{Var}(X)/4 = 1/4$.

$\text{Var}(Y) = 1/6 + 1/4 = 5/12$.

Verification: $\mathbb{E}[Y^2] = \mathbb{E}[\mathbb{E}[Y^2 \mid X]] = \mathbb{E}[X^2/3] = 2/3$. $\mathbb{E}[Y]^2 = (1/2)^2 = 1/4$. $\text{Var}(Y) = 2/3 - 1/4 = 5/12$. Confirmed.

$F_{Z}(z) = \int_0^1 \int_0^{zy} dx\,dy = \int_0^1 zy\,dy = z/2.$$

$F_{Z}(z) = \int_0^{1/z} zy\,dy + \int_{1/z}^1 1\,dy = \frac{z}{2z^2} + 1 - \frac{1}{z} = 1 - \frac{1}{2z}.$$

$f_{Z}(z) = \begin{cases} 1/2 & 0 < z \le 1 \\ 1/(2z^2) & z > 1 \end{cases}.$$

$X_{(k)} \sim \text{Beta}(k, n-k+1)$, so $\mathbb{E}[X_{(k)}] = k/(n+1)$.

$J = \det\begin{pmatrix} 1/2 & 1/2 \\ 1/2 & -1/2 \end{pmatrix} = -1/2$, so $|J| = 1/2$.

$f_{U,V}(u,v) = \frac{1}{2\pi}\exp\!\left(-\frac{(u+v)^2 + (u-v)^2}{8}\right) \cdot \frac{1}{2} = \frac{1}{4\pi}\exp\!\left(-\frac{u^2+v^2}{4}\right).$$This factors as$f_{U}(u) \cdot f_{V}(v)$where$U \sim \mathcal{N}(0, 2)$and$V \sim \mathcal{N}(0, 2)$. So$U$and$V$ are independent.

From the Rayleigh result, $R$ has PDF $f_{R}(r) = (r/\sigma^2)e^{-r^2/(2\sigma^2)}$. Let $W = R^2$, so $R = \sqrt{W}$ and $dR/dW = 1/(2\sqrt{W})$.

$f_{W}(w) = f_{R}(\sqrt{w}) \cdot \frac{1}{2\sqrt{w}} = \frac{1}{2\sigma^2}\exp\!\left(-\frac{w}{2\sigma^2}\right), \quad w \ge 0.$$This is$\text{Exp}(1/(2\sigma^2))$, or equivalently$\text{Gamma}(1, 1/(2\sigma^2))$. When$\sigma^2 = 1$, this is$\chi^2(2)$ — the chi-squared distribution with 2 degrees of freedom.

For $z \ge 0$: $$f_{Z}(z) = \int_0^z \lambda_1 e^{-\lambda_1 u} \cdot \lambda_2 e^{-\lambda_2(z-u)}\,du = \lambda_1\lambda_2 e^{-\lambda_2 z}\int_0^z e^{-(\lambda_1 - \lambda_2)u}\,du.$$

$= \frac{\lambda_1\lambda_2}{\lambda_1 - \lambda_2}\bigl(e^{-\lambda_2 z} - e^{-\lambda_1 z}\bigr), \quad z \ge 0.$$

This is a hypoexponential distribution — the difference of two exponential survival functions, weighted by a constant.

$\text{Var}(2X - 3Y + 5) = 4\text{Var}(X) + 9\text{Var}(Y) - 12\text{Cov}(X,Y) = 16 + 81 - 36 = 61$.

$\rho_{X,Y} = 3/\sqrt{4 \cdot 9} = 3/6 = 0.5$.

$\mathbb{E}[X] = \mathbb{E}[\mathbb{E}[X \mid N]] = \mathbb{E}[Np] = \lambda p$.

$\mathbb{E}[\text{Var}(X \mid N)] = \mathbb{E}[Np(1-p)] = \lambda p(1-p)$.

$\text{Var}(\mathbb{E}[X \mid N]) = \text{Var}(Np) = p^2\text{Var}(N) = \lambda p^2$.

$\text{Var}(X) = \lambda p(1-p) + \lambda p^2 = \lambda p$.

The result $\text{Var}(X) = \lambda p = \mathbb{E}[X]$ shows that $X \sim \text{Poisson}(\lambda p)$ — consistent with the Poisson splitting property.

$F_{W}(w) = \int_0^w \int_0^1 dy\,dx + \int_w^1 \int_0^{w/x} dy\,dx = w + \int_w^1 \frac{w}{x}\,dx = w + w\ln(1/w) = w(1 - \ln w).$$

$f_{W}(w) = \frac{d}{dw}[w(1 - \ln w)] = 1 - \ln w - 1 = -\ln w, \quad 0 < w < 1.$$

For $x < y$, we need exactly 1 RV at $x$, $n-2$ RVs in $(x,y)$, and 1 RV at $y$. The number of assignments is $n(n-1)$:

$$f_{X_{(1)},X_{(n)}}(x,y) = n(n-1)\,[F_{X}(y) - F_{X}(x)]^{n-2}\,f_{X}(x)\,f_{X}(y)$$

for $x < y$, and zero otherwise.

Let $U = X/Y$, $V = Y$. Inverse: $X = UV$, $Y = V$. Jacobian: $|J| = |v|$.

$f_{U,V}(u,v) = \frac{1}{2\pi}e^{-(u^2v^2 + v^2)/2}|v| = \frac{|v|}{2\pi}e^{-v^2(1+u^2)/2}.$$

$f_{U}(u) = \int_{-\infty}^{\infty} \frac{|v|}{2\pi}e^{-v^2(1+u^2)/2}\,dv = \frac{2}{2\pi}\int_0^{\infty} v\,e^{-v^2(1+u^2)/2}\,dv = \frac{1}{\pi(1+u^2)}.$$

This is the Cauchy distribution — the ratio of two independent standard Gaussians. The Cauchy has no finite mean or variance, illustrating that ratios of RVs can have much heavier tails than the original variables.

$P_{Z}(k) = \sum_{j=0}^2 P_{X}(k-j)\,P_{Y}(j)$ for $k = 0, 1, 2, 3, 4$.

$P_{Z}(0) = 1/9$, $P_{Z}(1) = 2/9$, $P_{Z}(2) = 3/9$, $P_{Z}(3) = 2/9$, $P_{Z}(4) = 1/9$.

This is a triangular distribution on $\{0, 1, 2, 3, 4\}$.

$\text{Cov}(X+Y, X-Y) = \text{Cov}(X,X) - \text{Cov}(X,Y) + \text{Cov}(Y,X) - \text{Cov}(Y,Y) = \text{Var}(X) - \text{Var}(Y).$$

$Y = \rho X + \sqrt{1-\rho^2}\,Z$ with $Z \perp X$. Then

$$\mathbb{E}[|XY|] = \rho\,\mathbb{E}[X|X|] + \sqrt{1-\rho^2}\,\mathbb{E}[|X|]\mathbb{E}[|Z|].$$

But $\mathbb{E}[X|X|] = \mathbb{E}[X^2\,\text{sgn}(X)] = 0$ by symmetry (odd function), so this direct route does not simplify nicely. Instead, use polar integration.

Transform to polar coordinates in the $(X, Z)$ plane and integrate $|X(\rho X + \sqrt{1-\rho^2}\,Z)|$ over the bivariate standard Gaussian. After careful computation (splitting into quadrants), the result is

$$\mathbb{E}[|XY|] = \frac{2}{\pi}\bigl(\sqrt{1-\rho^2} + \rho\arcsin\rho\bigr).$$

At $\rho = 0$: $\mathbb{E}[|XY|] = 2/\pi = \mathbb{E}[|X|]^2$. At $\rho = 1$: $\mathbb{E}[|X^2|] = 1$.