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A First Course in Probability

Sheldon M. Ross

$Math Studyset 91影视 Explanations$ Math

9th Edition 路 432 pages

ISBN: 9780321794772

Chapter 7: Q7.34 (page 361)

7.34. For another approach to Theoretical Exercise 7.33, let Tr denote the number of flips required to obtain a run of r consecutive heads. (a) Determine E[Tr|Tr鈭�1]. (b) Determine in terms of 贰摆罢谤鈭�1闭. (c) What is E[T1]? (d) What is E[Tr]?

Short Answer

Expert verified

$= \frac{1}{p} [E [T_{0}] = 0]$

Step by step solution

01

Given Information

Let be the probability that a coin lands on heads. Let Tr denote the number of flips required to obtain a run of r consecutive heads.

We have to find

$E [T r | T r 鈭� 1] .$

$E [T r 鈭� 1]$

E[T1]

$E [T r]$

02

Explanation Of a

Let $p$ be the probability that a coin lands on heads. Let $E [T_{r}]$ denote the number of flips required to obtain a run of $r$ consecutive heads.

Determine $E [T_{r} 鈭� T_{r - 1}] .$

$E [T_{r} 鈭� T_{r - 1}] = T_{r - 1} + 1 + (1 - p) E [T_{r}]$

03

Explanation Of b

Determine $E [T_{f}]$

Taking expectations on both sides of (a) yields,

$E [T_{r}] = E [T_{r - 1}] + 1 + (1 - p) E [T_{r}]$

$= \frac{1}{p} + \frac{1}{p} E [T_{r - 1}]$

04

Explanation Of c

DetermineSubstitute for r in part (b)

05

Explanation Of d

Determine $E [T_{r}]$

$E [T_{r}] = \frac{1}{p} + \frac{1}{p} E [T_{r - 1}]$

$= \frac{1}{p} + \frac{1}{p} [\frac{1}{p} + \frac{1}{p} E [T_{r - 1}]]$

$= (\frac{1}{p}) + {(\frac{1}{p})}^{2} + {(\frac{1}{p})}^{2} E [T_{r - 2}]$

$= (\frac{1}{p}) + {(\frac{1}{p})}^{2} + {(\frac{1}{p})}^{3} + {(\frac{1}{p})}^{3} E [T_{r - 3}]$

$= {鈭�}_{i = 1}^{p} {(\frac{1}{p})}^{i} + {(\frac{1}{p})}^{r} E [T_{0}]$

$= {鈭�}_{i = 1}^{r} {(\frac{1}{p})}^{i} [E [T_{0}] = 0]$

06

Final Answer

$E [T_{r} 鈭� T_{r - 1}] = T_{r - 1} + 1 + (1 - p) E [T_{r}]$

贰摆罢谤鈭�1闭. $= \frac{1}{p} + \frac{1}{p} E [T_{r - 1}]$

E[T1]

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Most popular questions from this chapter

There are n items in a box labeled H and m in a box labeled T. A coin that comes up heads with probability p and tails with probability 1 鈭� p is flipped. Each time it comes up heads, an item is removed from the H box, and each time it comes up tails, an item is removed from the T box. (If a box is empty and its outcome occurs, then no items are removed.) Find the expected number of coin flips needed for both boxes to become empty. Hint: Condition on the number of heads in the first n + m flips.

Let $U 1, U 2, . . .$ be a sequence of independent uniform $(0, 1)$ random variables. In Example $5 i$ , we showed that for $0 鈮� x 鈮� 1, E [N (x)] = e^{x}$ , where

$N (x) = min \{n : {鈭�}_{i = 1}^{n} 鈥� U_{i} > x\}$

This problem gives another approach to establishing that result.

(a) Show by induction on n that for $0 < x 鈮� 1$ 0 and all $n 鈮� 0$

$P {N (x) 鈮� n + 1} = \frac{x^{n}}{n!}$

Hint: First condition on $U 1$ and then use the induction hypothesis.

use part (a) to conclude that

$E [N (x)] = e^{x}$

If $Y = a X + b$ where a and b are constants, express the moment generating function of $Y$ in terms of the moment generating function of $X$ .

A group of 20 people consisting of 10 men and 10 women is randomly arranged into 10 pairs of 2 each. Compute the expectation and variance of the number of pairs that consist of a man and a woman. Now suppose the 20 people consist of 10 married couples. Compute the mean and variance of the number of married couples that are paired together.

The positive random variable $X$ is said to be a lognormal random variable with parameters $渭$ and $蟽^{2}$ if $\log (X)$ is a normal random variable with mean $渭$ and variance role="math" localid="1647407606488" $蟽^{2}$ . Use the normal moment generating function to find the mean and variance of a lognormal random variable

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