Solution Found!
This exercise will lead you through a proof of Chebyshev's
Chapter 2, Problem 33SE(choose chapter or problem)
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:
X
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
Y
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
Questions & Answers
QUESTION:
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:
X
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
Y
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
ANSWER:
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
E(X)
=
=
Hence proven that