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Let $\{N_t\}$ be a Poisson process of rate $\mu>0.$ If at time $t=10$ we have $N_{10}=4$ arrivals, what is the distribution of the time $T$ of the $3$rd arrival?

Attempt. If we were not given the total number of arrivals, the distribution of time $T$ would be $Gamma(3,\mu).$ Given the total number $N$ of arrivals though, the total times are uniformly equidistributed on interval $[0,T]$. Does this lead me somewhere?

Thank you in advance for the help.

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    $\begingroup$ What you say is correct, so you are asking for the third order statistic of four independent uniform random variables on the interval $[0,10]$. $\endgroup$
    – angryavian
    Commented Jul 4, 2018 at 18:03

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Continuing my initial statement and @angryavians's comment, i am posting a solution:

Let $U_{(i)},~i=1,2,3,4$ be the uniformly equidistributed arrival times on $[0,10]$ ($U_{(i)}$ means the $i$-th arrival time $U_i,~i=1,2,3,4$ with respect to the function ordering). Then $U_{(3)}$ takes values on $[0,10]$ and $U_{(3)}\leq u$ if and only if:

either $U_1,~U_2,~U_3,~U_4\leq u$, with probability $(u/10)^4$,

either exactly one of the $U_i'$s is $>u$, with probability $4\,(u/10)^3\,(1-u/10)$.

So the desired distribution is given as: $$F_{(3)}(u)=P(U_{(3)}\leq u)=(u/10)^4+4\,(u/10)^3\,(1-u/10)$$ for $u\in [0,1]]$, as $0$ for $u<0$ and as $1$ for $u>10.$

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