Mean calculation

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

Note by Anurag Choudhary
7 years, 7 months ago

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Comments

You want the expected value of the last order statistic X(3)=max⁡{X1,X2,X3} X_{(3)} = \max\{X_1, X_2, X_3\} of the IID random variables X1,X2,X3∼Exponential(λ) X_1, X_2, X_3 \sim {\rm Exponential}(\lambda) . The cumulative distribution of X(3) X_{(3)} is easy to find: it is simply FX(3)(x)=Pr⁡[max⁡{X1,X2,X3}≤x]=Pr⁡[X1≤x]Pr⁡[X2≤x]Pr⁡[X3≤x]=(1−λe−λx)3. \begin{aligned} 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{aligned} With λ=1 \lambda = 1 , this becomes FX(3)(x)=(1−e−x)3, F_{X_{(3)}}(x) = (1 - e^{-x})^3, so its survival is SX(3)(x)=1−(1−e−x)3. 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 E[X(3)]=∫x=0∞1−(1−e−x)3 dx=116. {\rm E}[X_{(3)}] = \int_{x=0}^\infty 1 - (1-e^{-x})^3 \, dx = \frac{11}{6}. Or you could find the density fX(3)(x)=FX(3)′(x) f_{X_{(3)}}(x) = F'_{X_{(3)}}(x) and compute the integral of xf(x) x f(x) , but that's just extra work.

hero p. - 7 years, 7 months ago
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