[Math] Light bulbs with exponential life

exponential functionprobability

Five light bulbs are controlled simultaneously. Each has an exponential lifespan of parameters $λ_i$ with i = 1 .. 5, independent.
I have to find:

1) expected time of a failure and distribution

2) expected time of two failures and distribution

3) expected time and breakdown of all light bulbs

These are my partial answers:

1) $ Y = \min (X_1 , X_2 , X_3, X_4, X_5) \sim exp(\sum_{i = 1}^{5}\lambda_i)$

$E[Y] = \frac{1}{\sum_{i = 1}^{5}\lambda_i}$

$F_Y(x) = {\begin{cases}1-e^{-(\sum_{i = 1}^{5}\lambda_i) \cdot x}&x\geq 0,\\0&x<0\end{cases}}$

2) Y = ?

$E[Y] = \frac{\lambda_1}{\lambda_1 + … + \lambda_5}\left(\frac1{\sum_{i = 1}^{5}\lambda_i} + \frac1{\sum_{i = 2}^{5}\lambda_i}\right) + \frac{\lambda_2}{\lambda_1 + … + \lambda_5}\left(\frac1{\sum_{i = 1}^{5}\lambda_i} + \frac1{\sum_{i = 1_{i \ne 2}}^{5}\lambda_i}\right)
+ \frac{\lambda_3}{\lambda_1 + … + \lambda_5}\left(\frac1{\sum_{i = 1}^{5}\lambda_i} + \frac1{\sum_{i = 1_{i \ne 3}}^{5}\lambda_i}\right)
+ \frac{\lambda_4}{\lambda_1 + … + \lambda_5}\left(\frac1{\sum_{i = 1}^{5}\lambda_i} + \frac1{\sum_{i = 1_{i \ne 4}}^{5}\lambda_i}\right)+ \frac{\lambda_5}{\lambda_1 + … + \lambda_5}\left(\frac1{\sum_{i = 1}^{5}\lambda_i} + \frac1{\sum_{i = 1}^{4}\lambda_i}\right)$

(Using the suggestion of the answer of @Misha Lavrov)

$F_Y(x) = $ ?

3) $ Y = \max (X_1 , X_2 , X_3, X_4, X_5)$

$E[Y] =$ ?

$F_Y(x) = {\begin{cases}P(Y \le x) = P(X_1 \le x, \dots, X_5 \le x) \\
= \prod_{i=1}^5 P(X_i \le x) = \prod_{i=1}^5 (1 – e^{-\lambda_i x})&x\geq 0,\\0&x<0\end{cases}}$

(Using the suggestion of the answer of @BruceET)

Best Answer

You already have an excellent hint from @MishaLavrov.(+1) Here are two additional suggestions toward a solution.

First, let the five rates all be $\lambda$ and $W$ be the waiting time for the last bulb to die. Then, by the 'no-memory' property of exponential random variables: $$E(W) = \frac{1}{5\lambda} + \frac{1}{4\lambda} + \frac{1}{3\lambda} + \frac{1}{2\lambda} + \frac{1}{\lambda}.$$

Also, the CDF of $W$ is $F_W(t) = (1 - e^{-\lambda t})^5.$

These solutions account for results of the following simulation in R statistical software:

set.seed(1222); m = 10^6; lam = 2
w = replicate(m, max(rexp(5, lam)))
mean(w); 2*sd(w)/sqrt(m)
## 1.140875          # aprx E(W)
## 0.001206653       # 95% margin of simulation error
(1/lam)*sum(1/(1:5))
## 1.141667          # exact E(W)

mean(w <= 1)
## 0.483739          # aprx P(W < 1)
(1-exp(-lam))^5
## 0.4833244         # exact P(W < 1)

Second, generalizing the method above for $E(W)$ can be done by extending @MishaLavrov's Answer, but you seem to be finding that approach difficult for five bulbs. (Maybe not so difficult after you've thought about the equal-rate case.)

Anyhow, getting the CDF $F_W(t) = \prod_{i=1}^5 (1 - e^{-\lambda_i t}),\, t > 0$ is not difficult. Getting $E(W)$ from $F_W$ may be a bit tedious, but that approach uses straightforward algebra and calculus.

Note: If one of the five $\lambda_i$ is very much smaller than the others (one light bulb is very much more durable), then $E(W)$ depends heavily on that one failure rate.

Note: In response to question in Comment. Let $W$ be the waiting time until the fifth bulb fails. Thus $W = \max\{X_1, \dots X_5\}$ and $$F_W(t) = P(W \le t) = P(X_1 \le t, \dots, X_5 \le t) \\ = \prod_{i=1}^5 P(X_i \le t) = \prod_{i=1}^5 (1 - e^{-\lambda_i t}),$$ where independence is used to get from the first line to the second. To finish the general case, multiply the five terms (collecting terms to minimize the mess), differentiate to get the PDF of $W$ and then integrate as needed to get $E(W).$ The last integration may require some integration by parts.

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