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? (See Exercise 10.)
For any of the bulbs, $X_i = $ its independent random variable. Let \(\Sigma X_i = min(x_1, x_2, ... x_{100})\). Let \(n=100\). From the problem, we know that \(X_i\) is exponentially distributed.
The expected lifetime of bulb \(i\) is 1000. \[ E[X_i] = \frac{1}{\lambda_i} = 1000 \rightarrow \lambda_i = \frac{1}{1000} \]
We have \(\Sigma X_i = min(x_1, x_2, ... x_{100})\). Since \(X_i \sim Exponential(\lambda_i)\), \(\Sigma X_i \sim Exponential(\Sigma_{i=1}^{100} \lambda_i)\).
\[ \Sigma \lambda_i = \Sigma_{i=1}^{100} \frac{1}{1000} = \frac{1}{10} \\ E[min(x_1, x_2, ... x_{100})] = \frac{1}{\lambda} = \frac{1}{1/10} = 10 \] Therefore, the expected time for the first of these bulbs to burn out is 10.
# number of bulbs
n <- 100
# expected lifetime of any bulb
ex <- 1000
# lambda of i
lambda_i <- 1 / ex
# calculate expected time for first of bulbs to burn out
lambda_min <- rep(lambda_i, n)
for (i in 1:n) {
lambda_min[i] <- lambda_i * (1 - pexp(1, rate = lambda_min[i]))
}
expected_time_first_burn_out <- 1 / sum(lambda_min)
print(expected_time_first_burn_out)## [1] 10.01001Assume that \(X_1\) and \(X_2\) are independent random variables, each having an exponential density with parameter \(\lambda\). Show that \(Z = X_1 − X_2\) has density \(f_z(z) = (1/2)\lambda e^{-\lambda |z|}\).
\[ f(x_1) = \lambda e^{-\lambda x_1}\\ f(x_2) = \lambda e^{-\lambda x_2} \]
For \(Z<0\): \[ Z = X_1 X_2 \rightarrow X_2 = X_1 - Z\\ f_z(z) = \int_{-\infty}^{\infty} f_{x_2}(x_1 - z) f_{x_1}(x_1) dx_1 \\ = \int_{0}^{\infty} \lambda e^{-\lambda (x_1 - z)} \lambda e^{-\lambda x_1}dx_1 = \lambda^2 \int_0^{\infty} e^{-\lambda (x_1 - z)} e^{-\lambda x_1}dx_1\\ = \lambda^2 \int_0^{\infty} e^{-\lambda x_1} e^{\lambda z} e^{-\lambda x}dx_1 = \lambda^2 e^{\lambda z} \int_0^{\infty} e^{-2 \lambda x_1} dx_1\\ = \frac{\lambda ^2 e^{\lambda z}}{-2\lambda} [e^{-2 \lambda x_1}]_0^{\infty} = (-1/2)\lambda e^{\lambda z} [0-1] = (-1)((-1/2)\lambda e^{\lambda z}) \\ f_z(z) = (1/2)\lambda e^{\lambda |z|} \]
For \(Z>0\): \[ Z= X_1 - X_2 \rightarrow X_1 = Z + X_2\\ f_z(z) = \int_{-\infty}^{\infty} f_{x_1} (x_2+z) f_{x_2} (x_2)dx_2\\ = \int_0^{\infty}\lambda e^{-\lambda (x_2 + z)} \lambda e^{-\lambda x_2}dx_2 = \lambda^2 \int_0^{\infty} e ^{-\lambda x_2 - \lambda z} e^{-\lambda x_2}dx_2\\ = \lambda^2 \int_0^{\infty} e^{-\lambda x_2}e^{-\lambda z} e^{-\lambda x_2}dx_2 = \lambda ^2 e^{-\lambda z} \int_0^{\infty} e^{-2\lambda x_2}dx_2\\ = \frac{\lambda^2 e^{-\lambda z}}{-2\lambda}[e^{-2\lambda x_2}]_0^{\infty} = \frac{\lambda^2 e^{-\lambda z}}{-2\lambda} [0 -1] \\ = (1/2) \lambda e^{-\lambda |z|} \]
So, in conclusion:
\[ f_z(Z) = \begin{cases} (1/2) \lambda e^{-\lambda |z|} \text{ for } Z>0\\ (1/2)\lambda e^{\lambda |z|} \text{ for } Z<0\\ \end{cases} \]
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.
Chebychev’s Inequality: \[ P(|X-E[X]| > k\sigma)\leq \frac{1}{k^2} \text{ for } k>0 \]
# variance
var <- 100/3
# mean
mean <- 10
# standard deviation
sd <- sqrt(var)\[ k\sigma = 2\\ k\sqrt{\frac{100}{3}} = 2 \rightarrow k = 0.3464\\ \text{Upper Bound } = \frac{1}{k^2} = \frac{1}{0.3464^2} = \frac{25}{3} \approx 8.333 \]
k_a <- 2/sd
upper_bound_a <- 1/k_a^2
print(upper_bound_a)## [1] 8.333333\[ k\sigma = 5 \rightarrow k = \frac{5}{\sqrt{\frac{100}{3}}} \approx 0.866025\\ \frac{1}{k^2} = \frac{1}{0.866025^2} \approx 1.333 = \frac{4}{3} \]
k_b <- 5/sd
upper_bound_b <- 1/k_b^2
print(upper_bound_b)## [1] 1.333333k_c <- 9/sd
upper_bound_c <- 1/k_c^2
print(upper_bound_c)## [1] 0.4115226\[ k\sigma = 20 \rightarrow k=\frac{20}{\sqrt{\frac{100}{3}}} \approx 3.4641\\ \frac{1}{k^2} = \frac{1}{3.4641^2} \approx 0.08333 = \frac{1}{12} \]
k_d <- 20/sd
upper_bound_d <- 1/k_d^2
print(upper_bound_d)## [1] 0.08333333