Let {X(t) : t ≥ 0} be a Poisson process with rate λ
Suppose it is known that X(1) = 2. Determine the mean of S1S2, the product of the first two arrival times.
I am not sure what to do with the first bit of information, that X(1)=2. I figure that the arrival time would be ~ Exp(λ) hence I am thinking that
E[S1*S2] = E[(S2-S1)*S1 + S1^2] = E[S2-S1] * E[S1] + E[S1^2] = λ(2λ+1)
But this does not make use of that bit of information. Any help would be appreciated
The fact that $X(1)=2$ means that there are two arrivals within the first unit time, so that $S_1, S_2 \in [0,1]$.
The problem with your answer is that you are taking expectation over all possibly values of $S_1$ and $S_2$ instead of just in the interval $[0,1]$.
What I am trying to say is that, for example, $\Bbb P(S_1S_2>1)=0$, yet you have (implicitly) included it as a positive value in your expectation.
So what we should do is to figure out the probability (density) that $S_1$ take a particular value GIVEN that $X(1)=2$.
$$f_{S_1|X(1)=2}(s)=f_{S_1|S_1 \in [0,1]}(s)=\frac{f_{S_1}(s)}{\Bbb P(S_1 \in [0,1])}=\frac{\lambda e^{-\lambda s}}{\int_0^1 \lambda e^{-\lambda s}ds}=\frac{\lambda e^{-\lambda s}}{1-e^{-\lambda}}$$
And we know that the inter-arrival times are iid, so we also have
$$f_{S_2-S_1|X(1)=2}(s)=\frac{\lambda e^{-\lambda s}}{1-e^{-\lambda}}$$
Hence
\begin{align} \Bbb E [S_1S_2] & = \Bbb E[S_2-S_1] \cdot \Bbb E[S_1]+\Bbb E[S_1^2] \\ & = \biggl(\int_0^1 sf_{S_2-S_1|X(1)=2}(s)ds \bigg)\bigg(\int_0^1 sf_{S_1|X(1)=2}(s)ds \bigg) + \bigg(\int_0^1 s^2f_{S_1|X(1)=2}(s)ds\bigg) \\ & = \biggl(\int_0^1 s\frac{\lambda e^{-\lambda s}}{1-e^{-\lambda}}ds \bigg)\bigg(\int_0^1 s\frac{\lambda e^{-\lambda s}}{1-e^{-\lambda}}ds \bigg) + \bigg(\int_0^1 s^2\frac{\lambda e^{-\lambda s}}{1-e^{-\lambda}}ds\bigg) \\ & = \cdots \end{align}