There are 3 bulbs in a room. If we switched on all of them. What is the total expected time till the room remains lit? Assume the "on" time for each bulb is an exponential random variable with λ=1 hour

You want the expected value of the last order statistic \( X_{(3)} = \max\{X_1, X_2, X_3\} \) of the IID random variables \( X_1, X_2, X_3 \sim {\rm Exponential}(\lambda) \). The cumulative distribution of \( X_{(3)} \) is easy to find: it is simply \[ \begin{align*} F_{X_{(3)}}(x) &= \Pr[\max\{X_1, X_2, X_3\} \le x] \\ &= \Pr[X_1 \le x]\Pr[X_2 \le x]\Pr[X_3 \le x] \\ &= (1 - \lambda e^{-\lambda x})^3. \end{align*} \] With \( \lambda = 1 \), this becomes \[ F_{X_{(3)}}(x) = (1 - e^{-x})^3, \] so its survival is \[ S_{X_{(3)}}(x) = 1 - (1 - e^{-x})^3. \] Now recalling that the expected value of a continuous random variable whose support is positive is simply the integral of its survival function, we immediately obtain \[ {\rm E}[X_{(3)}] = \int_{x=0}^\infty 1 - (1-e^{-x})^3 \, dx = \frac{11}{6}. \] Or you could find the density \( f_{X_{(3)}}(x) = F'_{X_{(3)}}(x) \) and compute the integral of \( x f(x) \), but that's just extra work.

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestYou want the expected value of the last order statistic \( X_{(3)} = \max\{X_1, X_2, X_3\} \) of the IID random variables \( X_1, X_2, X_3 \sim {\rm Exponential}(\lambda) \). The cumulative distribution of \( X_{(3)} \) is easy to find: it is simply \[ \begin{align*} F_{X_{(3)}}(x) &= \Pr[\max\{X_1, X_2, X_3\} \le x] \\ &= \Pr[X_1 \le x]\Pr[X_2 \le x]\Pr[X_3 \le x] \\ &= (1 - \lambda e^{-\lambda x})^3. \end{align*} \] With \( \lambda = 1 \), this becomes \[ F_{X_{(3)}}(x) = (1 - e^{-x})^3, \] so its survival is \[ S_{X_{(3)}}(x) = 1 - (1 - e^{-x})^3. \] Now recalling that the expected value of a continuous random variable whose support is positive is simply the integral of its survival function, we immediately obtain \[ {\rm E}[X_{(3)}] = \int_{x=0}^\infty 1 - (1-e^{-x})^3 \, dx = \frac{11}{6}. \] Or you could find the density \( f_{X_{(3)}}(x) = F'_{X_{(3)}}(x) \) and compute the integral of \( x f(x) \), but that's just extra work.

Log in to reply