This problem comes from Sheldon Ross's introduction to probability models and I'm trying to determine that actual mathematics involved instead of a verbal argument.
Let $\{N(t), t \geq 0 \}$ be a Poisson process with rate $\lambda$. For $s < t$ calculate $E[N(s) \mid N(t) = 4]$.
I can argue that because we know that these events will be uniformly distributed across $[0,t]$, it makes sense that if we were to want to find the expected number of events that have occurred up to a point would be the proportion of time $s/t$ times the number of events. This results in the solution given by the text which is $\dfrac{4s}{t}$.
My attempts at showing this mathematically begin by acknowledging that the events are uniformly distributed with pdf $\dfrac{1}{t}$. Because there are 4 events, the distribution is better defined as $\dfrac{4}{t}$. If we want to find the expected number of events from $[0,s]$, we would integrate $$\int_0^s \frac{4}{t} dx = \frac{4s}{t}.$$
Is this a reasonable argument or is there something more elaborate behind this?
Assuming that "conditioned on the event $N(t)=4$, the four arrival times are the order statistics $U_{(1)}, \ldots, U_{(4)}$ of four i.i.d. $\text{Unif}(0,t)$ random variables $U_1, \ldots, U_4$," then, conditioned on $N(t)=4$, we have $N(s)=I_1 + I_2 + I_3 + I_4$ where $I_i$ is an indicator for the event $U_i \le s$. Thus $E[N(s) \mid N(t)=4] = \sum_{i=1}^4 E[I_i \mid N(t)=4] = 4 \frac{s}{t}$.
It remains to verify the assumption.