This exercise will lead you through a proof of Chebyshev's

Statistics for Engineers and Scientists | 4th Edition | ISBN: 9780073401331 | Authors: William Navidi

ISBN: 9780073401331 38

Solution for problem 33SE Chapter 2

Statistics for Engineers and Scientists | 4th Edition

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Statistics for Engineers and Scientists | 4th Edition

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Problem 33SE

This exercise will lead you through a proof of Chebyshev's inequality. Let $X$ be a continuous random variable with probability density function $f(x)$. Suppose that $P(X<0)=0$, so $f(x)=0$ for $x \leq 0$.

a. Show that $\mu_{X}=\int_{0}^{\infty} x f(x) d x$.

b. Let $k>0$ be a constant. Show that $\mu_{X} \geq \int_{k}^{\infty} k f(x) d x=k P(X \geq k)$.

c. Use part (b) to show that $P(X \geq k) \leq \mu_{X} / k$. This is called Markov's inequality. It is true for discrete as well as for continuous random variables.

d. Let $Y$ be any random variable with mean $\mu_{Y}$ and variance $\sigma_{Y}^{2}$. Let $X=\left(Y-\mu_{Y}\right)^{2}$. Show that $\mu_{X}=\sigma_{Y}^{2}$.

e. Let be $k>0$ a constant. Show that $P\left(\left|Y-\mu_{Y}\right| \geq k \sigma_{Y}\right)=P\left(X \geq k^{2} \sigma_{Y}^{2}\right)$

f. Use part (e) along with Markov's inequality to prove Chebyshev's inequality: $P\left(\left|Y-\mu_{Y}\right| \geq k \sigma_{Y}\right) \leq 1 / k^{2}$

Equation Transcription:

$X$

$f(x)$

$P(X<0)=0$

$f(x)=0$

$x\leq{}0$

${\mu{}}_{X}=\int\limits_{0}^{\infty{}}xf(x)dx$

${\mu{}}_{X}\geq{}$ $\int\limits_{k}^{\infty{}}kf(x)dx=kP(X\geq{}k)$

$P(X\geq{}k)\leq{}{\mu{}}_{X}/k$

$Y$

${\mu{}}_{Y}$

${\sigma{}}_{Y}^{2}$

$X={\left({Y-{\mu{}}_{Y}}\right)}^{2}$

${\mu{}}_{X}={\sigma{}}_{Y}^{2}$

$P(\left|{Y-{\mu{}}_{Y}}\right|\geq{}$ $k{\sigma{}}_{Y})=P\left({X\geq{}{k}^{2}{\sigma{}}_{Y}^{2}}\right)$

$P(\left|{Y-{\mu{}}_{Y}}\right|\geq{}k{\sigma{}}_{Y})\leq{}1/{k}^{2}$

Text Transcription:

f(x)

P(X<0)=0

f(x)=0

x leq 0

muX=integral_0^ infinity xf(x)dx

mu_X geq integral_k^ infinity kf(x)dx=kP(X qec k)

P(X gec k)lec mu_X/k

mu_Y

sigma_Y^2

X=(Y-mu_Y)^2

mu_X=sigma_Y^2

P(|Y-mu_Y| gec k sigma_Y)=P(X gec k^2 sigma_Y^2)

P(|Y-mu_Y|gec k sigma_Y) lec 1/k^2

Step-by-Step Solution:

Answer

Step 1 of 6

a) Let x is a continuous random variable with probability density function f(x)

Here we have to show that $\mu{}{\ }_{x}=$ $\int\limits_{0}^{\infty{}}x\ f(x)\ dx$

$\mu{}{\ }_{x}=$ E(X)

= $\sum\limits_{x=0}^{\infty{}}$ $x\ f(x)dx$

= $\int\limits_{0}^{\infty{}}x\ f(x)\ dx$

Hence proven that $\mu{}{\ }_{x}=$ $\int\limits_{0}^{\infty{}}x\ f(x)\ dx$

Step 2 of 6

Chapter 2, Problem 33SE is Solved

Step 3 of 6

Textbook: Statistics for Engineers and Scientists

Edition: 4

Author: William Navidi

ISBN: 9780073401331

This textbook survival guide was created for the textbook: Statistics for Engineers and Scientists , edition: 4. This full solution covers the following key subjects: show, let, inequality, random, Continuous. This expansive textbook survival guide covers 153 chapters, and 2440 solutions. Statistics for Engineers and Scientists was written by and is associated to the ISBN: 9780073401331. The full step-by-step solution to problem: 33SE from chapter: 2 was answered by , our top Statistics solution expert on 06/28/17, 11:15AM. The answer to “?This exercise will lead you through a proof of Chebyshev's inequality. Let $X$ be a continuous random variable with probability density function $f(x)$. Suppose that $P(X<0)=0$, so $f(x)=0$ for $x \leq 0$.a. Show that $\mu_{X}=\int_{0}^{\infty} x f(x) d x$.b. Let $k>0$ be a constant. Show that $\mu_{X} \geq \int_{k}^{\infty} k f(x) d x=k P(X \geq k)$.c. Use part (b) to show that $P(X \geq k) \leq \mu_{X} / k$. This is called Markov's inequality. It is true for discrete as well as for continuous random variables.d. Let $Y$ be any random variable with mean $\mu_{Y}$ and variance $\sigma_{Y}^{2}$. Let $X=\left(Y-\mu_{Y}\right)^{2}$. Show that $\mu_{X}=\sigma_{Y}^{2}$.e. Let be $k>0$ a constant. Show that $P\left(\left|Y-\mu_{Y}\right| \geq k \sigma_{Y}\right)=P\left(X \geq k^{2} \sigma_{Y}^{2}\right)$f. Use part (e) along with Markov's inequality to prove Chebyshev's inequality: $P\left(\left|Y-\mu_{Y}\right| \geq k \sigma_{Y}\right) \leq 1 / k^{2}$Equation Transcription: Text Transcription: Xf(x)P(X<0)=0f(x)=0x leq 0muX=integral_0^ infinity xf(x)dxmu_X geq integral_k^ infinity kf(x)dx=kP(X qec k) P(X gec k)lec mu_X/kYmu_Ysigma_Y^2X=(Y-mu_Y)^2mu_X=sigma_Y^2P(|Y-mu_Y| gec k sigma_Y)=P(X gec k^2 sigma_Y^2)P(|Y-mu_Y|gec k sigma_Y) lec 1/k^2” is broken down into a number of easy to follow steps, and 164 words. Since the solution to 33SE from 2 chapter was answered, more than 388 students have viewed the full step-by-step answer.