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Customers line up to be serviced according to a Poisson process at an average rate of five per hour. If the time it takes to serve one customer is a continuous uniform random variable on $[0,4]$, independent of arrival times, what is the probability that the second customer who arrives will have to wait to be served?

I'm really rather lost I think. I'm saying the time $T_n$ that the $n$th customer arrives is a sum of exponential variables with parameter 5, which ends up being a Gamma distribution with parameters $n, 5$. The time that it takes to serve the $n$th customer is $S\sim U(0,4)$. I figure that the second customer waits whenever $T_1+S>T_2$, so that's the probability that I'm looking for. The problem is, I don't know how to find that probability. I tried taking the convolution of $S$ and $T_1$ to get the distribution function of $T_1 +S$ but something seems to be going wrong with that:

$$F_{T_1+S}(a)=\int_{-\infty}^{\infty}F_{T_{1}}(a-y)f_{S}(y)dy=\frac{1}{4}\int_{0}^{4}(1-e^{-5(a-y)})dy \\ =\frac{1}{4}(4-\frac{1}{5}e^{-5a}[e^{20}-1])$$ which is clearly not a CDF (it's negative between $0$ and $4$), so I must have done that wrong. But even if I figure out what I did wrong there, I wouldn't know what to do next.

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  • $\begingroup$ Hint: $T_2=T_1+R$ with $R$ independent of $T_1$ and of known distribution hence $[T_1+S\gt T_2]=[S\gt R]$. $\endgroup$
    – Did
    Commented Aug 13, 2012 at 7:52
  • $\begingroup$ @did ahh, R is just exp(5), yes? So I'm looking for P(R>exp(5)) $\endgroup$
    – crf
    Commented Aug 13, 2012 at 7:58

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Hint: $T_2=T_1+R$ with $R$ independent of $T_1$ and $S$, and $[T_1+S\gt T_2]=[S\gt R]$.

Sub-hint: $\mathrm P(R\gt S)=\int\limits_0^4\mathrm P(R\gt s)\,\frac{\mathrm ds}4$ and $\mathrm P(R\gt s)=\mathrm e^{-5s}$ for every $s\geqslant0$.

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  • $\begingroup$ I don't know why this particular problem has been causing me so much trouble but I think I've got it now. I'm thinking of this as $P(S>R)=P(0<R<S)$ and then if this were a discrete case, that would be something like $\sum_0^4 P(S=s)P(R<s)$ so $\int_0^4 f_S(s)F_R(s)ds$ ought to get me the right thing, which is $\int_{0}^{4}\frac{1}{4}(1-e^{-5s})ds$. The only thing is this reasoning isn't totally right because it's not discrete and I'm not looking at $P(S=s)$. Am I making a good analogy and how can I make this rigorous? Is this better suited to a different question? $\endgroup$
    – crf
    Commented Aug 13, 2012 at 19:23
  • $\begingroup$ As soon as $S$ is continuous with density $f_S$ and $R$ is independent of $S$, the formula $P(R\lt S)=\int f_S(s)F_R(s)ds$ is entirely rigorous. $\endgroup$
    – Did
    Commented Aug 13, 2012 at 20:37
  • $\begingroup$ :) thank you. I'll have to go over proving that, but it makes sense to me. $\endgroup$
    – crf
    Commented Aug 13, 2012 at 20:40

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