CUNY SPS DATA 605 - Assignment 8

Problem 1

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?

The distribution of the minimum value of n independent expontially distributed variables with mean \(\mu\) would also be exponential with mean \(\mu / n\). So in this case the expected value would be 1000 hrs/100 or 10 hours.

Problem 2

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

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

Math

\[ \begin{align} f_Z(z) = f_{x_1} * f_{-x_2}(z) &= \int_{-\infty}^{\infty} f_{-x_2}(z-x_1)\; f_{x_1}(x_1)\; dx_1 \\ f_Z(z) &= \int_{-\infty}^{\infty} f_{x_2}(x_1-z)\; f_{x_1}(x_1)\; dx_1 \end{align} \]

The propbability density function for an exponential distribution is:

\[ f_X(x) = \lambda e^{-\lambda x} \]

So plugging that into the formula above gives us:

\[ \begin{align} f_Z(z) &= \int_{-\infty}^{\infty} \lambda e^{-\lambda(x_1-z)} \lambda e^{-\lambda x_1} dx_1 \\ &= \int_{-\infty}^{\infty} \lambda^2 e^{-2\lambda x_1+\lambda z}dx_1 \\ &= \int_{-\infty}^{\infty} \lambda^2 e^{\lambda(-2 x_1+ z)}dx_1 \\ &= \int_{-\infty}^{\infty} \lambda^2 e^{\lambda(z-2 x_1)}dx_1 \\ \end{align} \]

This is as far as I could get. I don’t remember integration well enough to solve this, so I tried putting it into Wolfram Alpha to integrate and I got this:

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

When I should be getting this:

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

Simulation

Just out of cuiosity I wanted to see what the distrubution would look like so I ran a simulation and plotted it. Pretty coool that \(X - Y\) is a mirror image of \(Y - X\). Makes sense now that I think about it, but I hadn’t really anticipated that…

x <- rexp(400)
y <- rexp(400)
hist(x)

hist(y)

hist(x-y)

hist(y-x)

Problem 3

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.

Write a function

chebyshev <- function(e, var){
    pX = var / e^2
    return(pX)
}

(a) \(P(|X−10| \geq 2)\)

var <- 100/3
e <- 2
p <- chebyshev(e, var)
p

## [1] 8.333333

Since we can’t have a probability greater than 1…

\[P(|X−10| \geq 2) \leq 1\]

(b) \(P(|X−10| \geq 5)\)

e <- 5
p <- chebyshev(e, var)
p

## [1] 1.333333