CUNY 605 Homework Week 8

7.2 11. 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.)

Reference: https://en.wikipedia.org/wiki/Exponential_distribution, https://en.wikipedia.org/wiki/Exponential_decay

The question in the above stem is asking on the expected time (or on average) when the first light bulb will burn out.

The exponential equation can be as shown:

\[R(t) = R_0e^{-t/c}\] Where \(R_0\) is the initial number of light bulbs, which is 100, \(c\) (variable for time constant) is the average lifetime of a lightbulb, which is 1000 hours. We are tasked with finding when the first light bulb will burn out. We need to find when \(R_0 = 100 -> R(t) = 99\). So we can substitute in the variables and solve for \(R(t)\).

\[99 = 100 * e^{-t/1000}\]

# Solve for t
t <- -1 * 1000 * log(99/100)
print(paste0("The expected time for the first light bulb to burn out is: ", round(t,2), " hours."))
## [1] "The expected time for the first light bulb to burn out is: 10.05 hours."

7.2 14. Assume that X1 and X2 are independent random variables, each having an exponential density with parameter λ. Show that \(Z = X_1 − X_2\) has density

\[f_Z(z) = (1/2)\lambda e^{-\lambda |z|}\]

Reference: Introduction to Probability by Grinstead et. al. Page 292, https://faculty.math.illinois.edu/~hildebr/461/jointdistributions.pdf

Looking at our above reference, the exponential probability density function (pdf) is:

\[f_{X1}(x) = f_{X2}(x) = \begin{cases} \lambda e^{-\lambda x} & \text{where } x \geq 0, \\ 0, & \text{otherwise. } \\ \end{cases}\]

Given that \(x_1\) and \(x_2\) has been declared as independent two independent variables, the joint density function can be calculated by:

\[f(x_1, x_2) = f(x_1)f(x_2) = \lambda e^{-\lambda x_1}\lambda e^{-\lambda x_2} = \lambda^2 e^{-\lambda(x_1 + x_2)}\]

Let \(V = X_2\), therefore \(Z = X_1 - X_2 => x_1 = z + v\). We can substitute both v and z into the above equation. This then gives us:

\[g(z,v) = \lambda^2 e^{-\lambda ((z + v) + v)} = \lambda^2 e^{-\lambda (z + 2v)}\]

In order to calculate \(f(x_1, x_2)\), we must integrate the above \(g(x)\) function from \(-\infty\) to \(\infty\).

\[ =\int_{-\infty}^{\infty} \lambda^2 e^{-\lambda (z + 2v)} dv = \lambda /2 * e^{-\lambda z}\]

(To be honest, I’m a little stumped by this problem. I know we have to integrate these two functions after we take the product, but to get from that step to the final answer has been a challenge for me.)

8.2 1. 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.

Reference: Introduction to Probability by Grinstead et. al. Page 316

Chebyshev Inequality: For any positive number \(\epsilon > 0\), we have:

\[P(|X - \mu| \geq \epsilon) \leq \sigma^2/\epsilon^2\] Define upper bound as: \(\frac{\sigma^2}{\epsilon^2}\).

(a) \(P(|X - 10|) \geq 2\)

Here, \(\mu = 2\) and \(\sigma^2 = \frac{100}{3}\) Solve for the upper boundary.

sigma_squared <- 100/3
mu <- 2
upper_bound <- sigma_squared/mu^2
print(paste0("Upper boundary for mu = 2: ", upper_bound))
## [1] "Upper boundary for mu = 2: 8.33333333333333"

(b) \(P(|X - 10|) \geq 5\)

mu <- 5
upper_bound <- sigma_squared/mu^2
print(paste0("Upper boundary for mu = 5: ", upper_bound))
## [1] "Upper boundary for mu = 5: 1.33333333333333"

(c) \(P(|X - 10|) \geq 9\)

mu <- 2
upper_bound <- sigma_squared/mu^9
print(paste0("Upper boundary for mu = 9: ", upper_bound))
## [1] "Upper boundary for mu = 9: 0.0651041666666667"

(d) \(P(|X - 10|) \geq 20\)

mu <- 20
upper_bound <- sigma_squared/mu^2
print(paste0("Upper boundary for mu = 20: ", upper_bound))
## [1] "Upper boundary for mu = 20: 0.0833333333333333"

(Again, despite reading this over and over again, I am not 100% confident in these answers. I will need to review them in class. )