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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.

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    $\begingroup$ The task makes most sense if "the third candle" doesn't refer to a predetermined numbering of the candles, but the third of the five candles to "go off"? In other words, the task is asking of the time until any three of the five candles have gone off. $\endgroup$ – Troposphere May 18 at 11:59
  • $\begingroup$ @Troposphere I understand but this is the question I am given. I am not sure about what it means $\endgroup$ – Jared Garreta May 18 at 12:00
  • $\begingroup$ In other words I think "the time it takes for the third candle to go off" is an attempt to say "the time it takes until the the third time one of the candles to out". $\endgroup$ – Troposphere May 18 at 12:03
  • $\begingroup$ @Troposphere The problem is trivial otherwise. $\endgroup$ – Parcly Taxel May 18 at 12:05
  • $\begingroup$ @ParclyTaxel: Indeed. However, when the OP wrote "I believe having 5 candles is irrelevant" it sounded like he was misunderstanding the problem. $\endgroup$ – Troposphere May 18 at 12:09
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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 $\frac1{\lambda^2}$ $$\mathbb E[Y]=\frac1{5\lambda}+\frac1{4\lambda}+\frac1{3\lambda}=\frac{47}{60\lambda}$$ $$\operatorname{Var}(Y)=\frac1{25\lambda^2}+\frac1{16\lambda^2}+\frac1{9\lambda^2}=\frac{769}{3600\lambda^2}$$

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  • $\begingroup$ 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". $\endgroup$ – Jared Garreta May 18 at 11:51
  • $\begingroup$ But the candles are starting together at $t=0$. Why are you assuming that one of them is going to go off sooner? $\endgroup$ – Jared Garreta May 18 at 11:59
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    $\begingroup$ @Jared re your first question: That's from the memory less ness. By the time the first one goes out the clock starts anew $\endgroup$ – Bananach May 18 at 12:00
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    $\begingroup$ @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. $\endgroup$ – Parcly Taxel May 18 at 12:01
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    $\begingroup$ 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 $1/4\lambda$... $\endgroup$ – Bananach May 18 at 12:01

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