Chapter 5: Q 5.5 (page 215)

Use the result that for a nonnegative random variable $Y$ ,
$E [Y] = \int_{0}^{\infty} P \{Y > t\} d t$
to show that for a nonnegative random variable $X$ ,
$E [X^{n}] = \int_{0}^{\infty} n x^{n - 1} P \{X > x\} d x$
Hint: Start with
$E [X^{n}] = \int_{0}^{\infty} P \{X^{n} > t\} d t$
and make the change of variables $t = x^{n}$ .

Short Answer

Expert verified

Therefore, we’ve shown $E [X^{n}] = \int_{0}^{\infty} n x^{n - 1} P \{X > x\} d x = \int_{0}^{\infty} n x^{n - 1} P \{X > t\} d t$ as required.

Step by step solution

Given information:

Use the result that for a non-negative random variable $Y$ ,

$E [Y] = \int_{0}^{\infty} P \{Y > t\} d t$

Hint: Start with

$E [X^{n}] = \int_{0}^{\infty} P \{X^{n} > t\} d t$

and make the change of variables $t = x^{n}$

Explanation:

By letting $Y = X^{n}$ we know that

$E [X^{n}] = \int_{0}^{\infty} P \{X^{n} > t\} d t$

Making the change of variables $t = x^{n}$ and $d t = n x^{n - 1} d x$ , we get that

$E [X^{n}] = \int_{0}^{\infty} n x^{n - 1} P \{X^{n} > x^{n}\} d x$

Explanation:

Since $u \mapsto u^{n}$ is an increasing function of $u \geq 0$ the events $\{X^{n} > x^{n}\}$ and $\{X > x\}$ are identical. So we’ve shown

$E [X^{n}] = \int_{0}^{\infty} n x^{n - 1} P \{X > x\} d x = \int_{0}^{\infty} n x^{n - 1} P \{X > t\} d t$ as required.

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Short Answer

Step by step solution

Given information:

Explanation:

Explanation:

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Geometry

Probability and Statistics

Theoretical and Mathematical Physics

Calculus

Applied Mathematics

Logic and Functions

Study anywhere. Anytime. Across all devices.

Use the result that for a nonnegative random variableY,EY=∫0∞PY>tdtto show that for a nonnegative random variableX,EXn=∫0∞nxn-1PX>xdxHint: Start withEXn=∫0∞PXn>tdtand make the change of variablest=xn.

Short Answer

Step by step solution

Given information:

Explanation:

Explanation:

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Geometry

Probability and Statistics

Theoretical and Mathematical Physics

Calculus

Applied Mathematics

Logic and Functions

Study anywhere. Anytime. Across all devices.