Acknowledgements: Thanks to Ben S. for correcting some mistakes.

## 1.

By Lemma 2.15, $$\mathbb{P}(X = x) = F(x) - F(x-)$$. Since $$F$$ is right-continuous, $$F(x) = F(x+)$$.

## 2.

By Lemma 2.15,

$$$\mathbb{P}(2 < X \leq4.8) = F(4.8) - F(2) = 1/10$$$

and

$$$\mathbb{P}(2 \leq X \leq 4.8) = \mathbb{P}(X = 2) + \mathbb{P}(2 < X \leq 4.8) = F(4.8) - F(2-) = 2/10.$$$

## 3.

### 1)

Since $$F$$ is monotone, we can write $$F(x-) = \lim_n F(x_n)$$ where $$(x_n)$$ is some strictly increasing sequence converging to $$x$$. Let $$A_n = \{X \leq x_n\}$$ so that $$\{X < x\} = \cup_n A_n$$. By continuity of probability, $$\mathbb{P}(X < x) = \lim_k \mathbb{P}(A_n) = \lim_n F(x_n)$$.

### 2)

By additivity, $$\mathbb{P}(X \leq x) + \mathbb{P}(x < X \leq y) = \mathbb{P}(X \leq y)$$. The desired result follows by moving some terms around.

### 3)

Taking complements, $$\mathbb{P}(X > x) = 1 - \mathbb{P}(X \leq x) = 1 - F(x)$$.

### 4)

If $$X$$ is continuous, $$\mathbb{P}(X = x) = 0$$ for all $$x$$ by Part 1. The desired result follows from combining this fact with the findings from Part 2.

## 4.

### a)

We can express the CDF using indicator functions:

$F_X(x) = \frac{x}{4} I_{[0, 1)}(x) + \frac{1}{4} I_{[1, \infty)}(x) + \frac{3}{8}\left(x-3\right) I_{[3,5)}(x) + \frac{3}{4} I_{[5, \infty)}(x).$

### b)

Since $$Y = 1/X$$ and $$F_X(0) = 0$$, it follows that $$F_Y(0) = 0$$. For $$y > 0$$,

$$$F_Y(y) = \mathbb{P}(X \geq 1/y) = 1 - \mathbb{P}(X < 1/y) = 1 - F_X(1/y).$$$

## 5.

Suppose $$X$$ and $$Y$$ are independent. Then,

$$$f_{X,Y}(x,y) = \mathbb{P}(X \in \{x\}, Y \in \{y\}) = \mathbb{P}(X \in \{x\}) \mathbb{P}(Y \in \{Y\}) = f_X(x)f_Y(y).$$$

To establish the converse, suppose that $$f_{X,Y} = f_X f_Y$$. For a subset $$A$$ of the support of $$X$$ and a subset $$B$$ of the support of $$Y$$, $$\begin{multline} \mathbb{P}(X\in A,Y\in B) = \sum_{(x,y) \in A \times B} f_{X,Y}(x,y) = \sum_{x \in A} f_X(x) \sum_{y \in B} f_Y(y) \\ = \mathbb{P}(X \in A) \mathbb{P}(Y \in B). \end{multline}$$

## 6.

Note that

$$$F_Y(y) = \begin{cases} 0 & \text{if } y < 0 \\ \mathbb{P}(X \notin A) & \text{if } 0 \leq y < 1 \\ 1 & \text{if } y \geq 1. \end{cases}$$$

## 7.

Since

$$$\mathbb{P}(Z > z) = \mathbb{P}(\min\{X,Y\} > z) = \mathbb{P}(X > z) \mathbb{P}(Y > z) = \left(1 - F_X(z)\right)\left(1 - F_Y(z)\right),$$$

it follows that

$$$F_Z(z) = 1 - (1 - F_X(z))(1 - F_Y(z)) = F_X(z) + F_Y(z) - F_X(z) F_Y(z).$$$

When $$X$$ and $$Y$$ have the same distribution $$F$$, $$F_Z(z) = 2F(z) - F(z)^2$$ and hence $$f_Z(z) = 2f(z) - 2 F(z) f(z)$$. In particular, when $$F$$ is a uniform distribution on $$(0, 1)$$,

$$$f_Z(z) = 2 \left(1 - z\right) I_{(0, 1)}(z).$$$

## 8.

Let $$Y=X^+$$. First, note that $$F_Y(0-)=0$$ and $$F_Y(0)=F_X(0)$$. Moreover, $$F_Y(x)=F_X(x)$$ for $$x > 0$$.

## 9.

For $$x>0$$, $$F_X(x) = \int_0^x \lambda e^{-\lambda t} dt = 1 - e^{-\lambda x}$$. Therefore, $$F^{-1}(q) = -\ln(1 - q) / \lambda$$.

## 10.

If $$X$$ and $$Y$$ are independent, then

$\begin{multline} \mathbb{P}(g(X) \in A, h(Y) \in B) = \mathbb{P}(X \in g^{-1}(A), Y \in h^{-1}(B)) \\ = \mathbb{P}(X \in g^{-1}(A)) \mathbb{P}(Y \in h^{-1}(B)) = \mathbb{P}(g(X) \in A) \mathbb{P}(h(Y) \in B) \end{multline}$

under some lax conditions on $$g$$ and $$h$$ (Borel measurable).

## 11.

### a)

The two variables are dependent because

$$$\mathbb{P}(X = 1, Y = 0) = 0 \neq p (1 - p) = \mathbb{P}(X = 1) \mathbb(Y = 0).$$$

### b)

The two variables are independent because

$$$\mathbb{P}(X=i,Y=j) =\frac{\lambda^{i+j}e^{-\lambda}}{\left(i+j\right)!}\binom{i+j}{i}p^{i}\left(1-p\right)^{j} =e^{-\lambda}\frac{\lambda^{i}p^{i}}{i!}\frac{\lambda^{j}\left(1-p\right)^{j}}{j!}$$$

is decomposable into the form $$g(i)h(j)$$.

## 12.

If $$X$$ and $$Y$$ admit a joint density satisfying $$f(x,y) = g(x)h(y)$$, then

$$$\mathbb{P}(X\leq x,Y\leq y) =\int_{-\infty}^x\int_{-\infty}^yf(s,t)dtds =\int_{-\infty}^xg(s)ds\int_{-\infty}^yh(t)dt.$$$

The marginal distribution for $$X$$ is $$\mathbb{P}(X\leq x)=c_h\int_{-\infty}^xg(s)ds$$ where $$c_h=\int_{-\infty}^{\infty}h(t)dt$$. It follows that $$f_X = h c_h$$. We can similarly define $$c_g$$ to find that $$f_Y = g c_g$$. Moreover, $$c_h c_g=1$$ and hence $$c_g = 1 / c_h$$. It follows that $$f_{X,Y} = f_X f_Y$$, as desired.

## 13.

### a)

Note that

$$$F_Y(y) = \mathbb{P}(e^X \leq y) = \mathbb{P}(X \leq \ln y) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\ln y} \exp\left(-\frac{x^2}{2}\right) dy.$$$

Taking derivatives,

$$$f_Y(y) = \frac{1}{y\sqrt{2\pi}}\exp\left(-\frac{\left(\ln y\right)^2}{2}\right).$$$

### b)

TODO (Computer Experiment)

## 14.

Let $$0 < r < 1$$. Then, $$F_R(r) = \pi r^2 / \pi = r^2$$ and hence $$f_R(r) = 2r$$.

## 15.

For $$0\leq y\leq1$$,

$$$F_Y(y) = \mathbb{P}(F(X) \leq y) = \mathbb{P}(X \leq F^{-1}(y)) = F(F^{-1}(y)) = y.$$$

For all $$x$$,

$$$F_X(x) = \mathbb{P}(F^{-1}(U) \leq x) = \mathbb{P}(U\leq F(x)) = F(x).$$$

## 16.

Note that

$$$\mathbb{P}(X=x\mid X+Y=n)=\frac{\mathbb{P}(X=x,Y=n-x)}{\mathbb{P}(X+Y=n)}.$$$

Moreover,

$$$\mathbb{P}(X=x,Y=n-x)=\frac{e^{-\lambda}\lambda^{x}}{x!}\frac{e^{-\mu}\mu^{n-x}}{\left(n-x\right)!}.$$$

As per the hint,

$$$\mathbb{P}(X+Y=n)=e^{-\lambda-\mu}\frac{\left(\lambda+\mu\right)^{n}}{n!}.$$$

Letting $$\pi = \lambda / (\lambda + \mu)$$, combining these facts yields

$$$\mathbb{P}(X=x\mid X+Y=n)=\binom{n}{x}\pi^{x}\left(1-\pi\right)^{n-x}.$$$

## 17.

First, note that

$$$f_Y(1/2) = \int_0^{1}f(x,1/2)dx = c\int_0^{1}\left(x+\frac{1}{4}\right)dx = \frac{3}{4}c.$$$

Therefore,

$$$f_{X\mid Y}(x\mid1/2) = \frac{f_{X,Y}(x,1/2)}{f_Y(1/2)} = \frac{4}{3}\left(x+\frac{1}{4}\right)I_{(0,1)}(x).$$$

It follows that

$$$\mathbb{P}(X<1/2\mid Y=1/2) = \frac{4}{3}\int_0^{1/2}\left(x+\frac{1}{4}\right)dx = \frac{1}{3}.$$$

## 18.

TODO (Computer Experiment)

## 19.

Let $$r$$ be strictly increasing with differentiable inverse $$s$$. Let $$X$$ be an (absolutely) continuous random variable. Then, for $$Y=r(X)$$,

$$$F_Y(y) = \mathbb{P}(r(X) \leq y) = \mathbb{P}(X \leq s(y)) = F_X(s(y))$$$

and hence $$f_Y(y) = f_X(s(y))s^{\prime}(y)$$. If $$r$$ was instead strictly decreasing, then

$$$F_Y(y) = \mathbb{P}(X\geq s(y)) = 1 - F_X(s(y))$$$

and hence $$f_Y(y)=-f_X(s(y))s^{\prime}(y)$$. Since a strictly decreasing function has a strictly decreasing inverse, it follows that $$s^{\prime}(y) < 0$$ and hence we can summarize both cases by $$f_Y = (f_X\circ s) |s^{\prime}|$$.

## 20.

Let $$W=X-Y$$. Then, $$F_{W}(-1)=0$$ and $$F_{W}(1)=1$$. For $$-1 < w <1$$, $$F_{W}(w)=\mathbb{P}(Y\geq X-w)$$. The region

$$$\left\{ \left(x,y\right)\colon y\geq x-w,0\leq x,y\leq1\right\}$$$

is either a triangle or a right trapezoid depending on whether $$-1 < w < 0$$ or $$0 < w < 1$$:

By covering these case separately, one can derive $$F_{W}(w)=(1+w)^{2}/2$$ and $$F_{W}(w)=-w^2/2+w+1/2$$, respectively. It follows that

$$$f_{W}(w) = \begin{cases} 1+w & \text{if }-1 < w < 0\\ 1-w & \text{if }0 < w < 1\\ 0 & \text{otherwise}. \end{cases}$$$

Let $$V=X/Y$$. Then, $$F_{V}(0)=0$$. For $$v>0$$, $$F_{V}(v)=\mathbb{P}(Y\geq X/v)$$. The region

$$$\left\{ (x,y)\colon y\geq\frac{x}{v},0\leq x,y\leq1\right\}$$$

is either a triangle or a rectangle plus a right trapezoid depending on whether $$0 < v < 1$$ or $$v > 1$$. By covering these cases separately, one can derive $$F_{V}(v)=2/v$$ and $$F_{V}(v)=1/(2v)+(1-1/v)$$, respectively. It follows that

$$$f_{V}(v) = \begin{cases} 1/2 & \text{if }0<v<1\\ 1/(2v^{2}) & \text{if }v>1\\ 0 & \text{otherwise}. \end{cases}$$$

## 21.

Since

$$$F_Y(y) = \mathbb{P}(\max\{X_{1},\ldots,X_{n}\}\leq y) = \mathbb{P}(X_{1}\leq y)^{n} = \left(1-e^{-\beta y}\right)^{n},$$$

it follows that $$f_Y(y) = \beta ne^{-\beta y}(1-e^{-\beta y})^{n-1}$$.