A company buys 100 lightbulbs, each of which has an exponential lifetime of 1000 hours. What is the expected time for the first of these bulbs to burn out?
\(X_{1}, X_{2}, X_{3},......., X_{100}\) independent random variables.
We are being asked: \(E(X_{i})\)
So, random variable, from the interval of [0,100) with an exponential density with parameter \(\lambda\)
For two independent exponential random variables \(E(X_{i})=1/\lambda_{i}\)
Expected time for the first bulb to burn out => minimum value of \(X_{i}\)
\(\sum_{1}^{100} \lambda_{i} = \lambda = 100/1000\)
\(\lambda_{i}=1/10\) \(E(X_{i})=1/1/10\) = 10 hours
Assume that \(X_{1}\) and \(X_{2}\) are independent random variables, each having an exponential density with parameter \(\lambda\). Show \(Z=X_{1}-X_{2}\)
\(f_{Z}(z)=(1/2)\lambda \exp^{-\lambda z}\)
We have two independend random variables and the question is asking to show the the difference between them as below;
\(Z=X_{1}-X_{2}\) and what is $f{Z}(z)= ? $
The formula for \(X_{1}+X_{2}=W\) \(f{W}(w)=\int_{-\infty}^{\infty}f_{X_{1}}(x)f_{X_{2}}(W-X_{1})dx\)
If we consider \(Z=X_{1}-X_{2}=X_{1}+(-X_{2})\) we can replace the \(X_{2}\) with \(-X_{2}\) in the above calculation.
\(f_{Z}= \int_{-\infty}^{\infty}f_{X_{1}}(x_{1})f_{-X_{2}}(Z-X_{1})dx\)
We also know that \(f_{-X{2}}(z-x)=f_{X_{2}}(x-z)\)
so
\(f_{Z}(z)= \int_{-\infty}^{\infty}f_{X_{1}}(x_{1})f_{X_{2}}(X_{1}-z)dx\)
We need to look at two options. 1) \(z<0\) and 2) \(z\ge 0\)
\(z<0\)
\(=\int_{0}^{\infty}\lambda * e ^{-\lambda * x} * \lambda * e ^{-\lambda(x-z)}dx\)
\(=\lambda * e ^{x*z} \int_{0}^{\infty}\lambda*e^{-2\lambda*x}\)
\(=(\lambda/2)* e^{\lambda*z}\)
\(z\ge 0\)
Distribution of Z is same as distribution of -Z => for example: \(f_{z}(1) = f_{z}(-1)\)
so
\(f_{Z}(z)=f_{Z}(-z)\)
\(=(\lambda/2)*e^{-\lambda*z}\)
\(=(\lambda/2)*e^{-\lambda|z|}\)
Let X be a continuous random variable with mean \(\mu=10\) and variance \(\sigma^2=100/3\) . Using Chebyshev’s Inequality, find an upper bound for the following probabilities.
Chebyshev’s Inequality formula : \(P(|X-\mu|\ge k*\sigma)\le (1/k^2)\)
variance <- sqrt(100/3)
k_a <- 2/variance\(P(|X-10|\ge 2) = (1/k^{2})\)
p_a <- 1/(k_a^2)
p_a## [1] 8.333333k_b <- 5/variance
p_b <- 1/(k_b^2)
p_b## [1] 1.333333k_c <- 9/variance
p_c <- 1/(k_c^2)
p_c## [1] 0.4115226k_d <- 20/variance
p_d <- 1/(k_d^2)
p_d## [1] 0.08333333