I am trying to solve a problem like the following.
Q) The event that a man arrives at a bank $\sim Poisson(\lambda)$. If two men visited the bank between 9:30 AM and 10:30 AM, what is the probability that both entered before 10:00 AM?
The solution: \begin{align} \mbox{S_1} & \mbox{: the time when the first man entered.}\\ \mbox{S_2} & \mbox{: the time when the second man entered.}\\ \mbox{N(t)} & \mbox{: the number of event occurred in (0, t]}\\ \end{align}
\begin{align} f_{S_1, S_2\mid N(1)=2} (s_1, s_2 \mid 2) = 2, && 0<s_1<s_2<1 \\ \end{align}
$$\therefore P\left.\left[S_2\le\frac12\right | N(1)=2\right] = \frac14$$
My trial: \begin{align} \mbox{S_1} & \mbox{: the time when the first man entered.}\\ \mbox{S_2} & \mbox{: the time when the second man entered.}\\ \mbox{N(t)} & \mbox{: the number of event occurred in (0, t]}\\ \end{align}
\begin{align} f_{S_1, S_2|N(1)=2} (s_1, s_2 \mid 2) = 2, && 0<s_1<s_2<1 \\ \end{align} \begin{align} P\left.\left[S_2\le\frac12\right|N(1)=2\right] &= P\left.\left[S_1\le\frac12, S_2\le\frac12\right|N(1)=2\right]\\ &=\int_{0}^{\frac12}\int_{0}^{\frac12} f_{S_1, S_2|N(1)=2} (s_1, s_2 | 2) {ds_1}{ds_2}\\ &=\int_{0}^{\frac12}\int_{0}^{\frac12} 2 {ds_1}{ds_2}\\ &=\frac12 \end{align}
My answer and the solution are different. I guess the solution used conditional Poisson process $\sim$ Uniform distribution. However, I want to solve the problem with conditional pdf. Can someone help me? Thank you for reading.
I'm not entirely sure what you are doing, but I think your bounds are wrong. It might be $0\leq s_1 \leq s_2$. \begin{align} P\left.\left[S_2\le\frac12\right|N(1)=2\right] &= P\left.\left[S_1\le\frac12, S_2\le\frac12\right|N(1)=2\right]\\ &=\int_{0}^{\frac12}\int_{0}^{s_2} f_{S_1, S_2|N(1)=2} (s_1, s_2 | 2) {ds_1}{ds_2}\\ &=\int_{0}^{\frac12}\int_{0}^{s_2} 2 {ds_1}{ds_2}\\ &=\int_0^\frac{1}{2}2[s]_0^{s_2}\,ds_2\\ &=\int_0^\frac{1}{2}2s_2\,ds_2\\ &=2\cdot\frac{1}{2}[s_2^2]_0^\frac{1}{2}\\ &=\frac{1}{4} \end{align}