\((11)\) 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.)

Answer
\[E(X) = \mu/n = 1000/100 = 10 hours\]

\(Z(14)\) Assume that X1 and X2 are independent random variables, each having an exponential density with parameter \(\lambda\). Show that Z = X1 - X2 has density
\[fz(z) = (1/2)\lambda e ^ {-\lambda|z|}\]

Answer
The convolution formula of W = X + Y, we know the probability density would be: \[fw(W) = \int_{-\infty}^{\infty}fx(x)f{y}(w-x)dx\]

X1 and X2 are independent random variables.
Z = X1 - X2 or X1 + (-X2)
X2 = Z - X1
(To keep it simple, X1 = X, X2 = -Y)

\[fz(z) = \int_{-\infty}^{\infty}fx(x)f(-y)(z - x)dx\] \[fz(z) = \int_{-\infty}^{\infty}fx(x)f(y)(x - z)dx\] For Z < 0 \[fz(z) = \int_{0}^{\infty}\lambda e ^ {-\lambda x} \lambda e ^ {-\lambda(x - z)}dx\] \[fz(z) = \lambda e ^ {-\lambda z} \int_{0}^{\infty} \lambda e ^ {-2\lambda x} dx = (\lambda/2) e ^ {\lambda z}\] Similarly, for z >= 0, we will have to take the negative the exponent \[fz(z) = (\lambda / 2) e ^ {-\lambda z}\] Hence, combing both limit Z<0 and Z >= 0 we can write:
\[fz = (\frac{1}{2}) \lambda e ^ {-\lambda |z|}\]

Reference: https://www.youtube.com/watch?v=f8Nli1AfygM

\((1)\) 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.
Answer
\[P(|X - \mu| >= k\sigma ) =< \frac {\sigma ^ 2}{k^2 \sigma ^ 2} = \frac {1}{k^2}\]

(a) P(|X - 10| >= 2)

mu <- 10
variance <- 100/3
sigma <- sqrt(variance)
# k * sigma = 2
k <- 2 / sigma
1 / k ^ 2
## [1] 8.333333

Probability cannot be greater than 1 so, upper bound is 1.
(b) P(|X - 10| >= 5)

k <- 5 / sigma
1 / k ^ 2
## [1] 1.333333

Probability cannot be greater than 1 so, upper bound is 1.
(c) P(|X - 10| >= 9)

k <- 9 / sigma
1 / k ^ 2
## [1] 0.4115226

Upper bound is : 0.4115226
(d) P(|X - 10| >= 20)

k <- 20 / sigma
1 / k ^ 2
## [1] 0.08333333

Upper bound is: 0.08333333