Pg. 303, Ex. 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 Ex. 10.)

lambda <- 0.001
mu <- 1 / lambda
k = 100
expected <- mu / k

Expected First Failure: \(10\) hours.

Pg. 303, Ex. 14:

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

I’m having a lot of difficulty with this one, but here’s what I’ve done:

\(\int_{0}^{z} = \lambda e^{-\lambda(z + x_2)}\lambda e^{-\lambda x_2} dx_2\)

\(\int_{0}^{z} = \lambda^2 e^{-\lambda(z + 2x_2)} dx_2\)

\(\lambda^2(z + 2x_2) e^{-\lambda(z + 2x_2)}\)

And if I hardcode some numbers to test this out, I can confirm that my fZz variable calculation matches the result of the book’s calculation, a variable which I named check.

lambda <- 0.1
X1 <- 7
X2 <- 4
Z <- X1 - X2

fZz <- round(lambda^2 * (Z + 2 * X2) * exp(-1 * lambda * (Z + 2 * X2)), 3)
check <- round((1 / 2 * lambda) * (exp(-1 * lambda * Z)), 3)

print(fZz)
## [1] 0.037
print(check)
## [1] 0.037
(fZz == check)
## [1] TRUE

But my method and the book’s method don’t evaluate the same for all values of \(X_1\), \(X_2\), and \(lambda\), so there is probably something I’m missing. I can’t really see how to get from my calculation to the book’s calculation unfortunately.

Pp. 320-321, Ex. 1:

1 Let X be a continuous random variable with mean μ = 10 and variance σ2 = 100/3. Using Chebyshev’s Inequality, find an upper bound for the following probabilities.

mu = 10
var = 100 / 3
sd = var^0.5
ka = 2
kb = 5
kc = 9
kd = 20

a <- 1 / ka^2
au <- ka * sd + mu
b <- 1 / kb^2
bu <- kb * sd + mu
c <- 1 / kc^2
cu <- kc * sd + mu
d <- 1 / kd^2
du <- kd * sd + mu

\((a) P (|X − 10|≥ 2) \le 0.25\). So there’s a 75% chance X is below 21.5470054.

\((b) P (|X − 10|≥ 5) \le 0.04\). So there’s a 96% chance X is below 38.8675135.

\((c) P (|X − 10|≥ 9) \le 0.0123457\). So there’s a 98.8% chance X is below 61.9615242.

\((d) P (|X − 10|≥ 20) \le 0.0025\). So there’s a 99.8% chance X is below 125.4700538.

I apologize for this late submission, and I understand if credit’s not possible. I just wanted to finish the work and not get any further behind.