## 1.

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

By Lemma 2.15,

and

## 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 $% $. By continuity of probability, $% $.

### 2)

By additivity, $% $. 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:

### b)

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

For $y > 0$,

## 5.

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

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 $Y$, $\begin{multline} \mathbb{P}(X\in A,Y\in B) = \sum_{(x,y) \in A \times B} f_{X,Y}(x,y) = \sum_{x \in A} f_X(x) \sum_{y \in B} f_Y(y) \\ = \mathbb{P}(X \in A) \mathbb{P}(Y \in B). \end{multline}$

Note that

## 7.

Since

it follows that

In particular, when $X$ and $Y$ have the same distribution $F$, $F_Z(z) = 2F(z) - F(z)^2$.

## 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

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

## 11.

### a)

The two variables are dependent because

### b)

The two variables are independent because

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

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

Taking derivatives,

### b)

TODO (Computer Experiment)

## 14.

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

## 15.

For $0\leq y\leq1$,

For all $x$,

## 16.

Note that

Moreover,

As per the hint,

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

First, note that

Therefore,

It follows that

## 18.

TODO (Computer Experiment)

## 19.

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

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

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 $% $ 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 $% $, $F_{W}(w)=\mathbb{P}(Y\geq X-w)$. The region

is either a triangle or a right trapezoid depending on whether $% $ or $% $. 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

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

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

## 21.

Since

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