Exercise 7.2.11 (p. 303)

From exercise 10, we are given that the density for a random variable \(M\) which is the minimum of a set of random variables \(X_1, X_2, \dots, X_n\) following an exponential distribution is \(\mu/n\), \(\mu\) is the mean of the random \(X_j\)

In the case of our company, we can take 1000 hours to be the mean value for the exponential lifetime density function. Plugging into our formula, we get:

\[\begin{aligned} E(M) = \frac{\mu}{n} = 1000 / 100 = 10 \end{aligned}\]

Where \(M\) in this case is the lifetime of the shortest-burning bulb, or the minimum value of the random variable corresponding to the lifetime of each bulb. In this case, the company can expect the first bulb to burn out in 10 hours

Exercise 7.2.14 (p. 303)

We know based on the definition of a random variable \(X\) having an exponential distribution that the density function looks like:

\[\begin{aligned} f_X(x) = \begin{cases} \lambda e^{-\lambda x} & \text{if } x \ge 0\\ 0 &otherwise \end{cases} \end{aligned}\]

Combining these two, we can take their convolution to get the following:

\[\begin{aligned} f_{x_1, x_2}(x_1, x_2) = \lambda^{2} e^{-\lambda x_1} e^{-\lambda x_2} \end{aligned}\]

Assuming both \(x_1, x_2\) share the same rate parameter \(\lambda\).

where we can take \(x_1, x_2\) to be the values taken by random variables \(X_1, X_2\). We can do some manipulation within our probability distribution, where \(z\) is the value taken by \(Z\). We can find our cumulative distribution function, \(F_Z(z)\). then differentiate to get the probability distribution for \(Z\)

\[\begin{aligned} F_Z(z) = P(Z \le z) \newline = P(X_1 - X_2 \le z) \newline = P(X_2 \ge X_1 - z) \newline \end{aligned}\]

We’ll need to split this problem over our domain of \(-\infty \le z \le \infty\). Let’s take the case where \(z \le 0\) first:

\[\begin{aligned} F_Z(z) = \int_{0}^{\infty} \int_{x_1 - z}^{\infty} \lambda^2 e^{-\lambda x_1} e^{-\lambda x_2} \,dx_2\,dx_1 \newline = -\lambda \int_{0}^{\infty} e^{\lambda z - 2 x_1 \lambda} \,dx1 \newline = \frac{1}{2}e^{\lambda z} \end{aligned}\]

The integral bounds we’re using here reflect the fact that we care about the case where \(x_2 > x_1\), since we bounded \(z \le 0\), that means \(0 \le x_1 \le \infty\)

Now we can look at our case when \(z > 0\):

\[\begin{aligned} F_Z(z) = 1 - P(Z \le z) \newline = 1 - P(X_1 - X_2 \le z) \newline = 1 - P(X_2 \ge X_1 - z) \newline = 1 - \int_{z}^{\infty} \int_{0}^{x_1 - z} \lambda^2 e^{-\lambda x_1} e^{-\lambda x_2} \,dx_2\,dx_1 \newline = 1 + \lambda \int_{z}^{\infty} e^{-\lambda (2x_1 - z)} \,dx1 \newline = \frac{1}{2}e^{\lambda z} \end{aligned}\]

We can then differentiate our \(F_Z(z)\) with respect to \(z\) to find our density \(f_Z(z)\)

\[\begin{equation} f_Z(z)= \begin{cases} \frac{\lambda}{2}e^{-\lambda z} & \text{if } z > 0\\ \frac{\lambda}{2}e^{\lambda z} & \text{if } z \le 0 \end{cases} \end{equation}\]

In this case, the sign of \(z\) matters in terms of where we are in our domain \([-\infty, \infty]\). In both of the cases above, the exponential argument evaluates to a value less than or equal to 0, keeping the probability density function bounded. When \(z \ne 0\), the above cases evaluate to the same value. Because of this, we can simpliify our density function to:

\[\begin{aligned} f_Z(z)= \frac{\lambda}{2}e^{-\lambda |z|} \end{aligned}\]

Exercise 8.2.1 (p. 320)

Chebyshev’s inequality tells us that for a continuous random variable with density \(f(x)\), with finite expected value and variance, for any positive number \(\epsilon > 0\) we have:

\[\begin{aligned} P(|X - \mu| \ge \epsilon) \le \frac{\sigma^2}{\epsilon^2} \end{aligned}\]

We can define an R function to take \(\epsilon\) as our input to help us calculate these upper bounds numerically:

chebyshev_upper <- function(sigma, epsilon){
  # Calculates upper bound for Chebyshev's inequality based on random var variance sigma and a positive number epsilon
  ((sigma ** 2) / (epsilon ** 2))
  
}

# test call of our function
4 == chebyshev_upper(10, 5)
## [1] TRUE

In our case, we can plug in our mean value \(\mu = 10\) and variance \(\sigma^2 = 100 / 3\) to find our upper bound

  1. \[\begin{aligned} P(|X - 10| \ge 2) \le \frac{(100 / 3)^2}{2^2} = \frac{111.\overline{11}}{4} = 277.\overline{777} \end{aligned}\]

Let’s test with our R function

problem_sigma = 100 / 3 
chebyshev_upper(problem_sigma, 2)
## [1] 277.7778
  1. \[\begin{aligned} P(|X - 10| \ge 5) \le \frac{(100 / 3)^2}{5^2} = \frac{1111.111}{25} = 44.\overline{444} \end{aligned}\] We can check our calculation using our function
chebyshev_upper(problem_sigma, 5)
## [1] 44.44444
  1. \[\begin{aligned} P(|X - 10| \ge 9) \le \frac{(100 / 3)^2}{9^2} = \frac{1111.111}{81} = 13.717 \end{aligned}\]
chebyshev_upper(problem_sigma, 9)
## [1] 13.71742
  1. \[\begin{aligned} P(|X - 10| \ge 20) \le \frac{(100 / 3)^2}{20^2} = \frac{1111.111}{400} = 2.\overline{777} \end{aligned}\] checking our last upper bound programmatically:
chebyshev_upper(problem_sigma, 20)
## [1] 2.777778

In fact, we could complete these problems in a for loop, iterating over each value of epsilon

library(glue)
epsilons <- c(2, 5, 9, 20)

for (e in epsilons){
  upper_bound <- chebyshev_upper(problem_sigma, e)
  print(glue("Epsilon: {e}, upper bound: {upper_bound}"))
}
## Epsilon: 2, upper bound: 277.777777777778
## Epsilon: 5, upper bound: 44.4444444444444
## Epsilon: 9, upper bound: 13.7174211248285
## Epsilon: 20, upper bound: 2.77777777777778