# All of Statistics - Chapter 2 Solutions

## 1.

By Lemma 2.15, . Since is right-continuous, .

## 2.

By Lemma 2.15,

and

## 3.

### 1)

Since is monotone, we can write where is some strictly increasing sequence converging to . Let so that . By continuity of probability, .

### 2)

By additivity, . The desired result follows by moving some terms around.

### 3)

Taking complements, .

### 4)

If is continuous, for all 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 and , it follows that . Next, let . Then,

For ,

## 5.

Suppose and are independent. Then,

To establish the converse, suppose that . For a subset of the support of and a subset of the support of ,

## 6.

Note that

## 7.

Since

it follows that

In particular, when and have the same distribution , .

## 8.

Let . First, note that and . Moreover, for .

## 9.

For , . Therefore, .

## 10.

If and are independent, then

under some lax conditions on and (Borel measurable).

## 11.

### a)

The two variables are dependent because

### b)

The two variables are independent because

is decomposable into the form .

## 12.

If and admit a joint density satisfying , then

The marginal distribution for is where . It follows that . We can similarly define to find that . Moreover, and hence . It follows that , as desired.

## 13.

### a)

Note that

Taking derivatives,

### b)

TODO (Computer Experiment)

## 14.

Let . Then, and hence .

## 15.

For ,

For all ,

## 16.

Note that

Moreover,

As per the hint,

Letting , combining these facts yields

## 17.

First, note that

Therefore,

It follows that

## 18.

TODO (Computer Experiment)

## 19.

Let be strictly increasing with differentiable inverse . Let be a continuous random variable. Then, for ,

and hence . If was instead strictly decreasing, then

and hence . Since a strictly decreasing function has a strictly decreasing inverse, it follows that and hence we can summarize both cases by .

## 20.

Let . Then, and . For , . The region

is either a triangle or a right trapezoid depending on whether or . By covering these case separately, one can derive and , respectively. It follows that

Let . Then, . For , . The region

is either a triangle or a right trapezoid depending on whether or . By covering these cases separately, one can derive and , respectively. It follows that

## 21.

Since

it follows that .