Sums, maximums, and minimums of exponential random variables

Question

Let $y_1$, $y_2$,...,$y_{10}$ be independent exponential random variables with mean $1$. (a) Find the distribution of their sum.

(b) Suppose the random variables above represent the lifetime (in thousands of hours) of $10$ light bulbs in a house. Derive the density function of the lifetime of the first light bulb to burn out and the density function of the last light bulb to burn out.

I got the answer to part (a) of the question, which is the distribution of their sum is a gamma distribution with $\alpha = 10$ and $\beta= 1$. However, I have no idea where to start with part (b). Please help!

Answer 1

To keep the notation simple, I will show how to work a similar problem with three bulbs instead of ten. I hope this gives the 'start' that you need.

Minimum. Suppose $X_1, X_2, X_3$ are independently exponential with mean 1 (thus rate 1). Let $V = X_{(1)}$ be the distribution of the minimum of the $X_i$ (first to burn out), and let $W = X_{(3)}$ be the distribution of the maximum of the $X_i$ (last to burn out). Then for $t > 0,$ $$1 - F_v(t) = P(V > t) = P(X_1 > t, X_2 > t, X_3 > t)\\ = P(X_1 > t)P(X_2 > t)P(X_3 > t) = (e^{-t})^3 = e^{-3t},$$ so that $F_V(t) = 1 - e^{-3t},$ which is the CDF of an exponential distribution with rate $\lambda = 3$ and mean $\mu = 1/3.$ You can get the density function of $V$ by differentiating this CDF.

Maximum. Also, $$F_W(t) = P(W \le t) = P(X_1 \le t, X_2 \le t, X_3 \le t)\\ = P(X_1 \le t)P(X_2 \le t)P(X_3 \le t) = (1 - e^{-t})^3.$$ which is not the CDF of an exponential distribution. [Intuitively, $W$ cannot have the exponential no-memory property. If some time has passed, one or two of the bulbs may already have burned out, so the expected time until the last of the three burns out is smaller than it was at the start.]

The PDF of $W$ can be found by differentiating its CDF, but the resulting formula is a little messy for three bulbs (and messier still for ten). Because you are not asked to use the density function for anything, perhaps it is sufficient to say that the density function of $W$ is $$f_W(t) = F_W^\prime(t) = \frac{d}{dt}(1-e^{-t})^n,$$ where $n$ is the number of bulbs. Also, questions about the probability that the last bulb burns out before time $t_0$ can be answered directly in terms of the CDF.

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Note: Some approximate results for $n = 10$ are easily simulated: $E(V) = SD(V) = 0.10$ is closely approximated by simulation; $E(W) \approx 2.93,\, SD(W) \approx 1.24.$ Because all individual bulbs have the same distribution, intuitive arguments can be made that $E(W) = \frac{1}{10} + \frac 1 9 + \cdots + 1$ $= 2.928968$ and $SD(W) = \sqrt{\frac{1}{10^2} + \frac{1}{9^2} +\cdots + 1} = 1.244897.$

v = replicate( 10^6, min(rexp(10)) ) 
mean(v);  sd(v)
## 0.09999971
## 0.09979044

w = replicate( 10^6, max(rexp(10)) ) 
mean(w);  sd(w)
## 2.928545
## 1.244993

Histograms of the simulated distributions of $V$ and $W$ are shown below. The exact density function of $V$ is superimposed on the histogram in the left panel. The density curve shown for $W$ was obtained from simulated values of $W$ using a kernel density estimator. (It is a right-skewed curve, but neither exponential nor gamma.)