\[ u=1000\\ n=100\\ E[x]=\frac{u}{n}=\frac{1000}{100}=10 \]
\[ f_z(Z)=(1/2)\lambda e^{-\lambda |z|} \]
Lets compute the PDF for our x1 and x2
x1:
\[ f(x_1)=(1/2)\lambda e^{-\lambda x_1}\\ f(x_2)=(1/2)\lambda e^{-\lambda x_2} \] Please note that we can assume x1 and x2 to be greater than or equal to zero
http://mathworld.wolfram.com/Convolution.html
Since x1 and x2 are independent, we can compute the joint density function as follows:
\[ f(x_1, x_2)=f(x_1)f(x_2)\\ =(\lambda e^{-\lambda x_1})(\lambda e^{-\lambda x_2})\\ =\lambda^{2}e^{x_1+x_2} \]
We know z=x1+x2 but lets pick some variable h to be x2 such that we have z=x1+h. We can re-work our new equation into x1=z+h. The purpose of this is to use the Jacobian for the transformation of variables.
https://onlinecourses.science.psu.edu/stat414/node/160/
We compute the jacobian as follows:
\[ J=\frac{\delta(x_1, x_2)}{\delta(z,v)}\\ =\begin{vmatrix} 1 & 1 \\ 0 & 1 \end{vmatrix}\\ =1 \]
We can now develop our joint probability distribution with our transformed variables. We will do some substitution a little later.
$$ g(z, v)={-(z+v+v)}\ ={-(z+2v)}
$$
Please note that v>0 and z is bounded from -infinity to infinity
Recall our substitution
\[ z=x_1-v\\ v=x_1-z \] Some additional conditions
\[ v>-z,-\infty <z<\infty\\ v>0,z>0 \]
With our transformed PDF, we can use the conditions we have derived to assemble our integration. Rather than going through every step of the integration, I will highlight the process. Remove the lambda squared since it is a constant, then apply u substitution where u is x+2v. After u substitution, we need to evaluate the resulting improper integral by taking the limit as some arbitrary upper bound goes to infinity.
http://tutorial.math.lamar.edu/Classes/CalcII/ImproperIntegrals.aspx
For z greater than -infinity but less than 0
\[ \int _{ -x }^{ \infty }g(z,v)dv\\ \int _{ -x }^{ \infty }{ \lambda ^{ { 2 } } } { e }^{ -\lambda (x+2v) }dv\\ =\frac{1}{2}e^{\lambda x} \]
for z greater than 0
\[ \int _{ x }^{ \infty }g(z,v)dv\\ \int _{ x }^{ \infty }{ \lambda ^{ { 2 } } } { e }^{ -\lambda (x+2v) }dv\\ =\frac{1}{2}e^{-\lambda x} \]
The probability density function is
\[ g(z) =\frac{1}{2}e^{\lambda x}\ ,-\infty<z<\infty\\ =\frac{1}{2}e^{-\lambda x}\ ,0<z<\infty \]
Now we can combine our equations with our problem statment
\[ f_z(Z)=(1/2)\lambda e^{-\lambda |z|} \\ =(1/2)\lambda e^{-\lambda |x_1+x_2|} \\ \]
Hence proved
Let X be a continuous random variable with mean µ = 10 and variance σ2 = 100/3. Using Chebyshev’s Inequality, find an upper bound for the following probabilities:
\[ P(|X-10|\ge2)=P(|X-10|\ge.2\sqrt3 \frac{10}{\sqrt3})\le\frac{1}{(.2\sqrt3)^{2}}\le8.33 \]
\[ P(|X-10|\ge5)=P(|X-10|\ge.5\sqrt3 \frac{10}{\sqrt3})\le\frac{1}{(.5\sqrt3)^{2}}\le1.33 \]
\[ P(|X-10|\ge9)=P(|X-10|\ge.9\sqrt3 \frac{10}{\sqrt3})\le\frac{1}{(.9\sqrt3)^{2}}\le0.412 \]
\[ P(|X-10|\ge20)=P(|X-10|\ge2\sqrt3 \frac{10}{\sqrt3})\le\frac{1}{(2\sqrt3)^{2}}\le0.083 \]