Let $N(t)$ be a Poisson process with rate $\lambda$, if $\tau_1$ and $\tau_2$ are the times of the first and second arrivals. I need to find $E[\tau_1 \vert N(1) = 1]$ and $E[\tau_2 \vert N(1)=1]$.
The arrival times are defined $\tau_n = T_1 + T_2 + \cdots + T_n$, with $T_i \sim \exp(\lambda)$, then conditioning we have: $$ E[\tau_1 \vert N(1)=1] = E[\tau_1 \vert \tau_1 < 1, \tau_2 > 1 ] = E[T_1 \vert T_1 < 1 , T_1 + T_2 >1].$$
But I don't know how to use it or if I'm conditioning correctly. I would appreciate any hint.
I'm going to calculate $E[\tau_1 \vert \tau_1 \leq 1, \tau_2 > 1 ] = E[T_1 \vert T_1 \leq 1 , T_1 + T_2 >1]$. First of all, note that $$P[T_1\leq t \vert T_1 \leq 1 , T_1 + T_2 >1]=\begin{cases} 1&\text{if }t\ge1,\\\frac{P[T_1\leq t , T_1 + T_2 >1]}{P[T_1\leq 1 , T_1 + T_2 >1]}&\text{if }0\leq t<1.\end{cases}$$ We konw that the distribution of $(T_1,T_2)$ has density $f_{T_1,T_2}(t_1,t_2)=\lambda^2e^{-\lambda(t_1+t_2)}$, $t_1,t_2\ge0$. So the distribution of $(T_1,T_1+T_2)$ has density $f_{T_1,T_1+T_2}(t_1,s)=\lambda^2e^{-\lambda s}$, $s\ge t_1\ge0$. Hence for $0\leq t\leq1$, $$P[T_1\leq 1 , T_1 + T_2 >1]=\int_0^t\int_1^\infty\lambda^2e^{-\lambda s}\,ds\,dt_1=t\lambda e^{-\lambda},$$ and then $$P[T_1\leq t \vert T_1 \leq 1 , T_1 + T_2 >1]=\frac{P[T_1\leq t , T_1 + T_2 >1]}{P[T_1\leq 1 , T_1 + T_2 >1]}=\frac{t\lambda e^{-\lambda}}{\lambda e^{-\lambda}}=t.$$ Therefore $$E[T_1 \vert T_1 \leq 1 , T_1 + T_2 >1]=\int_0^1 t\,dt=\frac12.$$