A company buys 100 lightbulbs, each of which has an exponential lifetime of 1000 hours. We are interested in finding the expected time for the first of these bulbs to burn out.
Let’s break this down step by step to understand the expected time for the first of the 100 lightbulbs to burn out.
Exponential Distribution: When we say a lightbulb has an exponential lifetime of 1000 hours, we mean the time until the bulb burns out follows an exponential distribution with a mean (\(\lambda^{-1}\)) of 1000 hours. In the context of the exponential distribution, \(\lambda\) is the rate parameter, which is the reciprocal of the mean. So, \(\lambda = \frac{1}{1000}\) hours\(^{-1}\).
Individual Bulb Lifetime: The probability density function (pdf) of the exponential distribution for a single bulb’s lifetime \(T\) is given by: \[ f_T(t) = \lambda e^{-\lambda t} = \frac{1}{1000} e^{-\frac{1}{1000}t}, \quad t \geq 0. \]
First Bulb to Burn Out: The event we are interested in is the first bulb burning out. This is a minimum of 100 exponential random variables. When you take the minimum of independent exponential random variables, the resulting random variable also follows an exponential distribution, but with a rate parameter that is the sum of the individual rate parameters.
Since all bulbs are identical and independent, the rate parameter for the first bulb to burn out is \(100 \times \lambda = 100 \times \frac{1}{1000} = \frac{1}{10}\) hours\(^{-1}\).
Expected Time for First Bulb to Burn Out: The mean (or expected value) of an exponential distribution with rate parameter \(\lambda\) is \(\frac{1}{\lambda}\). Therefore, the expected time for the first bulb to burn out is: \[ E[T_{\text{first}}] = \frac{1}{100 \times \frac{1}{1000}} = \frac{1}{\frac{1}{10}} = 10 \text{ hours}. \]
So, although each bulb is expected to last 1000 hours on its own, the expected time for the first of these 100 bulbs to fail is only 10 hours, because with more bulbs, there’s a higher chance one of them will fail early on.
Below is the simulation in r.
# Set the number of lightbulbs and simulations
num_bulbs <- 100
num_simulations <- 10000
mean_lifetime <- 1000 # Mean lifetime of each bulb in hours
# Simulate the lifetimes of bulbs
simulated_min_lifetimes <- replicate(num_simulations, {
lifetimes <- rexp(num_bulbs, rate = 1 / mean_lifetime)
min_lifetime <- min(lifetimes)
return(min_lifetime)
})
# Calculate the average minimum lifetime
average_min_lifetime <- mean(simulated_min_lifetimes)
# Print the result
cat("The expected time for the first bulb to burn out (based on simulations) is approximately:", average_min_lifetime, "hours\n")## The expected time for the first bulb to burn out (based on simulations) is approximately: 10.10847 hoursAssume 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 the density
\[ f_Z(z) = \frac{1}{2}\lambda e^{-\lambda|z|} \]
Let \(X_1\) and \(X_2\) be two independent random variables, each with an exponential density with parameter \(\lambda\). We are interested in finding the density function of the difference \(Z = X_1 - X_2\).
The exponential density function for \(X_1\) and \(X_2\) is given by:
\[ f_X(x) = \begin{cases} \lambda e^{-\lambda x}, & \text{if } x \geq 0 \\ 0, & \text{otherwise} \end{cases} \]
We will derive the density function of \(Z = X_1 - X_2\).
Given the independence of \(X_1\) and \(X_2\), the joint density function is the product of their individual densities:
\[ f_{X_1, X_2}(x_1, x_2) = f_{X_1}(x_1) \cdot f_{X_2}(x_2) \]
The density function of \(Z\) can be found by using the convolution formula for the difference of two random variables:
\[ f_Z(z) = \int_{-\infty}^{\infty} f_{X_1}(z + y) f_{X_2}(y) \, dy \]
We consider two cases based on the sign of \(z\).
The lower limit of the integral changes to 0 because the exponential density is zero for negative arguments:
\[ f_Z(z) = \int_{0}^{\infty} \lambda e^{-\lambda (z + y)} \lambda e^{-\lambda y} \, dy \]
Simplifying the integral, we get:
\[ f_Z(z) = \lambda^2 e^{-\lambda z} \int_{0}^{\infty} e^{-2\lambda y} \, dy \]
\[ f_Z(z) = \lambda^2 e^{-\lambda z} \left[-\frac{1}{2\lambda} e^{-2\lambda y}\right]_{0}^{\infty} \]
\[ f_Z(z) = \frac{\lambda}{2} e^{-\lambda z} \]
For \(z < 0\), we compute the integral by changing the limits accordingly:
\[ f_Z(z) = \int_{-\infty}^{-z} \lambda e^{-\lambda (y - z)} \lambda e^{-\lambda y} \, dy \]
This is equivalent to:
\[ f_Z(z) = \lambda^2 e^{\lambda z} \int_{-z}^{\infty} e^{-2\lambda y} \, dy \]
Simplifying the integral, we get:
\[ f_Z(z) = \lambda^2 e^{\lambda z} \left[-\frac{1}{2\lambda} e^{-2\lambda y}\right]_{-z}^{\infty} \]
\[ f_Z(z) = \frac{\lambda}{2} e^{\lambda z} \]
Combining both cases, we obtain the density function for \(Z\):
\[ f_Z(z) = \frac{\lambda}{2} e^{-\lambda |z|}, \quad \text{for all } z \]
Through convolution and considering the properties of the exponential distribution, we have shown that the difference of two independent exponential random variables with the same rate \(\lambda\) has the density function:
\[ f_Z(z) = \frac{\lambda}{2} e^{-\lambda |z|} \]
Let \(X\) be a continuous random variable with mean \(\mu = 10\) and variance \(\sigma^2 = \frac{100}{3}\). Using Chebyshev’s Inequality, find an upper bound for the following probabilities. .
Chebyshev’s Inequality states that for any \(k > 0\):
\[ P(|X - \mu| \geq k) \leq \frac{\sigma^2}{k^2} \]
Given a continuous random variable \(X\) with mean \(\mu = 10\) and variance \(\sigma^2 = \frac{100}{3}\), we aim to use Chebyshev’s Inequality to find an upper bound for the following probabilities:
Chebyshev’s Inequality states that for any \(k > 0\):
\[ P(|X - \mu| \geq k\sigma) \leq \frac{1}{k^2} \]
where \(\sigma\) is the standard deviation of \(X\).
First, we calculate the standard deviation \(\sigma\):
\[ \sigma = \sqrt{\sigma^2} = \sqrt{\frac{100}{3}} = \frac{10}{\sqrt{3}} \]
Now, we can calculate the upper bounds for each of the given probabilities:
\[ P(|X - 10| \geq 2) \leq \left(\frac{5}{\sqrt{3}}\right)^2 = \frac{25}{3} \]
\[ P(|X - 10| \geq 5) \leq \left(\frac{2}{\sqrt{3}}\right)^2 = \frac{4}{3} \]
\[ P(|X - 10| \geq 9) \leq \left(\frac{10}{9\sqrt{3}}\right)^2 = \frac{100}{243} \]
\[ P(|X - 10| \geq 20) \leq \left(\frac{1}{2\sqrt{3}}\right)^2 = \frac{1}{12} \]
Below is the ssame solution using r.
# Given values
mu <- 10
variance <- 100 / 3
sigma <- sqrt(variance)
# Chebyshev's Inequality function to calculate the upper bound
chebyshev_upper_bound <- function(deviation) {
k <- deviation / sigma
1 / (k^2)
}
# Deviations for the calculations
deviations <- c(2, 5, 9, 20)
# Calculate upper bounds for each deviation
upper_bounds <- sapply(deviations, chebyshev_upper_bound)
# Print the results
upper_bounds## [1] 8.33333333 1.33333333 0.41152263 0.08333333