==> Problem 11 on page 303 of probability text.
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?
According to formula
\[E(v) = E_0 e ^ {-\lambda v} \]
\[\lambda = \frac{1}{1000}\]
\[\ n.\lambda = \frac{100}{1000} = \frac{1}{10}\] Probability of fail \[P(minV1,V2,...,Vn<=v)=1 - e ^ {\frac{-1}{10}v}\]
Calculate Expected value \[E[minXi]=\frac{1}{1/10}= 10\] Hence, the expected time for the first of these bulbs to burn out is 10Hrs
==> Problem 14 on page 303 of probability text.
Assume that X1 and X2 are independent random variables, each having an exponential density with parameter \(\lambda\) . Show that \[Z = X_1 − X_2\] has density \[ f_z(Z) = (1/2) \lambda e ^ {- \lambda{|z|} } \]
Given, \(Z = X_1 − X_2\) so, \(X_1 = X_2 + Z\)
Use the convolution formula, we can write:
\[f_Z(z) = \int_{-\infty}^{0}f_{X_1}(x_1)f_{X_2}(x_2)dx_2\] \[f_Z(z) = \int_{-\infty}^{0}f_{X_1}(z+x_2)f_{X_2}(x_2)dx_2\]
\[f_{X_1}(z+x_2) = \lambda e^{-\lambda (z+x_2)}\] \[f_{X_2}(x_2)= \lambda e^{-2\lambda x_2}\]
For \(X_2 \geq X_1\), \(-\infty \; to \; 0\) \[f_Z(z) = \int_{-\infty}^0\lambda e^{-\lambda (z+x_2)} \lambda e^{-\lambda x_2}dx_2\] \[f_Z(z) = \lambda^2 e^{-\lambda z}(\frac{ -1}{2\lambda})\] \[f_Z(z) = \frac{-\lambda e^{-\lambda z}}{2}\]
for \(X_1 \geq X_2\), \(0 \; to \; \infty\) \[f_Z(z) = \int_0^{\infty}\lambda e^{-\lambda (z+x_2)} \lambda e^{-\lambda x_2}dx_2\] \[f_Z(z) = \lambda^2 e^{-\lambda z}(\frac{ -1}{2\lambda})\] \[f_Z(z) = \frac{\lambda e^{-\lambda z}}{2}\]
Taken together, we have \[ f_z(Z) = (1/2) \lambda e ^ {- \lambda{|z|} } \]
==> Problem 1 on page 321 of probability text.
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 is calculated as :-
\(P(|X−μ|≤kσ)≤1/k ^ 2\)
We are given:-
mu <- 10
sigma <- sqrt(100/3)
Cheby_In <- function(k) { # Function to calculate Chebyshev's Inequality
1/k^2
print(paste("Chebyshev's Inequality Calculated upper bound", round((1/k^2),4)))
}- P(|X - 10| >= 2).
## [1] "Chebyshev's Inequality Calculated upper bound 8.3333"- P(|X - 10| >= 5).
## [1] "Chebyshev's Inequality Calculated upper bound 1.3333"- P(|X - 10| >= 9).
## [1] "Chebyshev's Inequality Calculated upper bound 0.4115"- P(|X - 10| >= 20).
## [1] "Chebyshev's Inequality Calculated upper bound 0.0833"