I have the following, with once again self-fabricated values, question:

Let C(t) be the number of cats to arrive at a cat palace within $t (\geq 0)$ minutes. Suppose that C(t) has a poisson distribution with mean $\mu t$, where arrival rate is $\mu = 15$ arrivals per minute. $(a)$ What is the mean and standard deviation of the time until the first arrival? $(b)$What is the probability that the $15^{th}$ arrival occurs within 1 minute?

My logic:

$(a)$ Both $\mu$ and $\sigma = \lambda$, so they both equal $15$, I think this is wrong because of the mention of "until the first arrival".

$(b)$ This is a discrete poisson problem:

$\lambda = 15,x = 15$: $\frac{e^{-15}15^{15}}{15!} \approx .1024$

I think this one is also wrong, even though the percentage doesn't seem absurd, it does seem a little low based on the mean being equal.

Any hints or other help are greatly appreciated! If you need further clarification please say so. ${ }{}$


1 Answer 1


Note that the inter-arrival times of a Poisson RV is an exponential RV with mean $\frac 1\mu$ and variance $\frac 1{\mu^2}$. This is also true of the time of first arrival.

Secondly, if 15th arrival occurs within 1 minute then we know that the number of arrivals within 1 minute is at least 15. Suppose $X(t)$ is the number of arrivals in time $t$. Then we are interested in $\Pr(X(1)\geq 15)$ where $X(1)$ has a Poisson distribution of parameter $\mu$. Now you have to find the following sum: $$ \Pr(X(1)\geq 15)=\frac{e^{-15}15^{15}}{15!}+\frac{e^{-15}15^{16}}{16!}+\frac{e^{-15}15^{17}}{17!}+\dots\approx 0.534 $$ (here.)

  • $\begingroup$ Wouldn't the mean and variance be $\frac{1}{\mu t}$ and $\frac{1}{(\mu t)^2}$ if it is an exponential RV? Are you sure that series is the correct way to solve the sum? Shouldn't I take 1 - (sum from 1 to 15)? $\endgroup$ Commented Apr 4, 2014 at 0:23
  • $\begingroup$ Let's take a look at it this way. Denote the first arrival time by $T$. What is the probability that the first arrival does not come before $t$? It is $e^{-\mu t}$, which means $\Pr(T\leq t)=1-e^{-\mu t}$. This is an exponential RV with mean $\frac 1\mu$. $\endgroup$
    – Arash
    Commented Apr 4, 2014 at 1:00
  • $\begingroup$ For your second question, I edited the answer! :-) $\endgroup$
    – Arash
    Commented Apr 4, 2014 at 1:05
  • $\begingroup$ @Arash why are we calculating x= 16, x = 17? $\endgroup$
    – user585380
    Commented Sep 9, 2018 at 14:08

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