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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: Q. 7.49 (page 363)

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

Short Answer

Expert verified

The mean and variance of a lognormal random variable value are $E (X) = e^{m u + 蟽^{2} / 2}$ and

Variancerole="math" localid="1647408572874" $(X) = e^{2 渭 + 蟽^{2}} (e^{蟽^{2}} - 1)$ .

Step by step solution

01

Given Information

Use the normal moment generating function to find the mean and variance of a lognormal random variable.

02

Explanation

Define, $Y = \log X$

Since we know that $X$ is log-normal random variable with parameters $渭$ and $蟽^{2}$ , we have that $Y ~ N (渭, 蟽^{2})$ .

Observe the following equality. For $t 鈮�$ $0$

We have that,

$E (X^{t}) = E ({(e^{Y})}^{t}) = E (e^{t Y}) = M_{Y} (t)$

$= \exp (渭 t + \frac{蟽^{2} t^{2}}{2})$ .

03

Explanation

Now we have that,

$E (X) = \exp (渭 + \frac{蟽^{2}}{2})$ and

$E (X^{2}) = \exp (2 渭 + 2 蟽^{2})$

Which implies, variancerole="math" localid="1647408375853" $(X) = E (X^{2}) - E (X)^{2}$

$= \exp (2 渭 + 2 蟽^{2}) - \exp (2 渭 + 蟽^{2})$

$鈬� Var (X) = e^{2 渭 + 蟽^{2}} (e^{蟽^{2}} - 1)$ .

04

Final answer

The mean and variance of a lognormal random variable value are $E (X) = e^{m u + 蟽^{2} / 2}$ and

variance $(X) = e^{2 渭 + 蟽^{2}} (e^{蟽^{2}} - 1)$ .

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Most popular questions from this chapter

For an event A, let IA equal 1 if A occurs and let it equal 0 if A does not occur. For a random variable X, show that E[X|A] = E[XIA] P(A

In the text, we noted that

$E [{鈭�}_{i = 1}^{鈭�} X_{i}] = {鈭�}_{i = 1}^{鈭�} E [X_{i}]$

when the $X i$ are all nonnegative random variables. Since

an integral is a limit of sums, one might expect that $E [{鈭�}_{0}^{鈭�} X (t) d t] = {鈭�}_{0}^{鈭�} E [X (t)] d t$

whenever $X (t), 0 鈮� t < 鈭�,$ are all nonnegative random

variables; this result is indeed true. Use it to give another proof of the result that for a nonnegative random variable $X$ ,

$E [X) = {鈭�}_{0}^{鈭�} P (X > t} d t$

Hint: Define, for each nonnegative $t$ , the random variable

$X (t)$ by

role="math" localid="1647348183162" $X (t) = 1 i f t < X \ \ 0 i f t 鈮� X$

Now relate $4 q$

${鈭�}_{0}^{鈭�} X (t) d t to X$

Consider $n$ independent flips of a coin having probability $p$ of landing on heads. Say that a changeover occurs whenever an outcome differs from the one preceding it. For instance, if $n = 5$ and the outcome is $H H T H T$ , then there are $3$ changeovers. Find the expected number of changeovers. Hint: Express the number of changeovers as the sum of $n 鈭� 1$ Bernoulli random variables.

Consider a gambler who, at each gamble, either wins or loses her bet with respective probabilities $p$ and $1 - p$ . A popular gambling system known as the Kelley strategy is to always bet the fraction $2 p - 1$ of your current fortune when $p > 12$ . Compute the expected fortune after $n$ gambles of a gambler who starts with $x$ units and employs the Kelley strategy.

A die is rolled twice. Let X equal the sum of the outcomes, and let Y equal the first outcome minus the second.

Compute $C o v (X, Y) .$

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