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<math> \therefore P(Y>i) = \frac {1}{2^i} </math> | <math> \therefore P(Y>i) = \frac {1}{2^i} </math> | ||

+ | |||

+ | Plugging this result back into the original expression yields | ||

+ | <math> P(Y<X) = \sum_{i=1}^\infty \frac {1}{4^i} = \frac {1}{3} </math> | ||

+ | |||

+ | ==Part D== | ||

+ | *Find <math> P(Y=kX)\ </math> | ||

+ | |||

+ | Noting that X and Y are iid and summing over all possible combinations one arrives at | ||

+ | <math> P(Y=kX) = \sum_{i=1}^\infty i = 1^\infty P(Y=ki, X=i) = \sum_{i=1}^\infty P(Y=ki)P(X=i) </math> | ||

+ | |||

+ | Thus, | ||

+ | <math> P(Y=kX) = \sum_{i=1}^\infty \frac {1}{2^{ki}} \frac {1}{2^i} = \sum_{i=1}^\infty \frac {1}{2^{(k+1)i}} = \frac {1}{2^{(k+1)}-1} </math> |

## Revision as of 11:30, 3 December 2008

## Contents

# Problem 1

X and Y are iid

```
$ P(X=i) = P(Y=i) = \frac {1}{2^i}\ ,i = 1,2,3,... $
```

## Part A

- Find $ P(min(X,Y)=k)\ $

```
Let $ Z = min(X,Y)\ $
```

```
Then finding the pmf of Z uses the fact that X and Y are iid
$ P(Z=k) = P(X \ge k,Y \ge k) = P(X \ge k)P(Y \ge k) = P(X \ge k)^2 $
```

```
$ P(Z=k) = \left ( \sum_{i=k}^N \frac {1}{2^i} \right )^2 = \left ( \frac {1}{2^k} \right )^2 = \frac {1}{4^k} $
```

## Part B

- Find $ P(X=Y)\ $

```
Noting that X and Y are iid and summing across all possible i,
$ P(X=Y) = \sum_{i=1}^\infty P(X=i, Y=i) = \sum_{i=1}^\infty P(X=i)P(Y=i) = \sum_{i=1}^\infty \frac {1}{4^i} = \frac {1}{3} $
```

## Part C

- Find $ P(Y>X)\ $

```
Again noting that X and Y are iid and summing across all possible i,
$ P(Y>X) = \sum_{i=1}^\infty P(Y>i, X=i) = \sum_{i=1}^\infty P(Y>i)P(x=i) $
```

Next, find $ P(Y<i)\ $ $ P(Y>i) = 1 - P(Y \le i) $

```
$ P(Y \le i) = \sum_{i=1}^\infty \frac {1}{2^i} = 1 + \frac {1}{2^i} $
```

```
$ \therefore P(Y>i) = \frac {1}{2^i} $
```

```
Plugging this result back into the original expression yields
$ P(Y<X) = \sum_{i=1}^\infty \frac {1}{4^i} = \frac {1}{3} $
```

## Part D

- Find $ P(Y=kX)\ $

```
Noting that X and Y are iid and summing over all possible combinations one arrives at
$ P(Y=kX) = \sum_{i=1}^\infty i = 1^\infty P(Y=ki, X=i) = \sum_{i=1}^\infty P(Y=ki)P(X=i) $
```

```
Thus,
$ P(Y=kX) = \sum_{i=1}^\infty \frac {1}{2^{ki}} \frac {1}{2^i} = \sum_{i=1}^\infty \frac {1}{2^{(k+1)i}} = \frac {1}{2^{(k+1)}-1} $
```