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The number of people queuing at a supermarket check-out during some time interval of length $t$ minutes follows a Poisson process with mean $\displaystyle\frac{t}{2}$. The variable $Y$ denotes the time in minutes from when the supermarket opens until the first customer arrives at the check-out.

(a) Find the p.d.f. of the variable $Y$.

(b) On a certain day the person running the check-out arrives 5 minutes late (5 minutes after opening), but no customers have yet arrived at the check-out. Find the distribution of the time that the check-out attendant will have to wait for the first customer to arrive at the check-out.

I found the p.d.f. in part (a) to be, in terms of $t$

\begin{array}{@{}c@{\quad}l@{}} 0.5e^{-0.5t} & \text{if $t$ is positive}\\ 0 & \text{otherwise}\\ \end{array} So my question is whether the answer to (b) is

\begin{array}{@{}c@{\quad}l@{}} 0.5e^{-0.5(5-t)} & \text{if $t>5$}\\ 0 & \text{otherwise}\\ \end{array}

However, I'm not sure whether this is correct. If this is an incorrect answer, could someone please explain why?

Any help much appreciated, as always.

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  • $\begingroup$ Does your answer to (b) give a pdf which integrates to $1$? $\endgroup$ – Henry Oct 16 '16 at 23:21
  • $\begingroup$ I would have thought memorylessness might be involved $\endgroup$ – Henry Oct 16 '16 at 23:22
  • $\begingroup$ Do you know what the correct answer should be? $\endgroup$ – wrb98 Oct 17 '16 at 10:05
  • $\begingroup$ Is it possibly a conditional probability question? $\endgroup$ – wrb98 Oct 17 '16 at 10:41
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Let's do some conditional probabilities!

We want to compute the conditional probability $$ P(Y-5 = t | Y > 5)=\frac{P(Y=t+5)}{P(Y>5)} = \frac{\frac{e^{-\frac{t+5}{2}}}{2}}{1-\int_0^5\frac{e^{-\frac{t}{2}}}{2}dt} = \frac{\frac{e^{-\frac{t+5}{2}}}{2}}{e^{5/2}} = \frac{e^{-\frac{t}{2}}}{2} $$

So the final distribution is exactly the same!

This is because the distribution was an exponential distribution, and exponential distributions are memoryless.

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