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We have $5$ candles each having a lifetime which follows an exponential distribution with parameter $\lambda$. We light up each candle at time $t=0$.

Assume that $Y$ is the time that it takes for the third candle to go off. What is the expectation and variance of $Y$?

My try: First of all, I believe having $5$ candles is irrelevant. We only need to consider one random variable following the exponential distribution, like $X\sim \exp(\lambda)$. It means that on average, it takes $\frac{1}{\lambda}$ for the candle to go off. However, this does not seem like a random variable. It seems like it is a constant. Then, it won't be meaningful to calculate the expectation and variance. Am I right?

Also, we know that at some point, the candle "will" go off. So, does this mean that we cannot predict at which time it will? I am totally confused thinking about these concepts. I appreciate if someone enlightens me.

Note: There is a similar question here. However, the question has not been answered due to the lack of attemps provided by the OP.

Parcly Taxel
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Jared Garreta
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1 Answers1

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The time needed for the first candle to go out is distributed as the minimum of $5$ independent exponential random variables. That minimum is exponentially distributed with rate $5\lambda$ (see here). Since the exponential distribution is memoryless we now start all over; by similar reasoning as before the second-out and third-out times are exponentially distributed with rates $4\lambda$ and $3\lambda$ respectively, and all three new random variables are independent of each other. $Y$ is their sum; since the mean and variance of an exponential distribution with rate $\lambda$ are $\frac1\lambda$ and $\frac1{\lambda^2}$ $$\mathbb E[Y]=\frac1{5\lambda}+\frac1{4\lambda}+\frac1{3\lambda}=\frac{47}{60\lambda}$$ $$\operatorname{Var}(Y)=\frac1{25\lambda^2}+\frac1{16\lambda^2}+\frac1{9\lambda^2}=\frac{769}{3600\lambda^2}$$

Parcly Taxel
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  • Thanks for the answer. Would you please elaborate more the part where $Y$ is the sum of "three" random variables? Why "three"? I don't the the "minimum" part. The question doesn't mention anything about a "minimum". – Jared Garreta May 18 '21 at 11:51
  • But the candles are starting together at $t=0$. Why are you assuming that one of them is going to go off sooner? – Jared Garreta May 18 '21 at 11:59
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    @Jared re your first question: That's from the memory less ness. By the time the first one goes out the clock starts anew – Bananach May 18 '21 at 12:00
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    @JaredGarreta The candles will go off one at a time. This, combined with memorylessness, allows the decomposition of the problem into three subproblems, each waiting for one candle to go out. – Parcly Taxel May 18 '21 at 12:01
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    So the time until it takes for the first (whichever that turns out to be) to go out is a random variable with expectation $1/5\lambda$. The time afterwards for the first of the four remaining ones to go out is $1/4\lambda$... – Bananach May 18 '21 at 12:01