Poisson process question

Anyway I'm pretty sure I interpreted your question correctly: say, for definiteness, there are two sources of meteors happening continuously at rates $\lambda_1, \lambda_2$. This question is asking, in plain english, for the expected value of the last time $Last_{1, 2}(t) =: S(t)$ you've seen a Meteor at time $t$ (arguably a more interesting statistic in real life is the expected value of $t - Last_{1, 2}(t)$, but never mind). 

It is annoying to compute anything about a statistic like $T^{[1]}_{N_1(t)}$, so let's do everything conditionally: $$ E[S(t)] = \sum_{n = 0}^\infty\sum_{m = 0}^\infty E[S(t)|N_1(t) = n, N_2(t) = m] P(N_1(t) = n, N_2(t) = m)$$

Now we have an ugly infinite sum but both factors of each term are very computable. Let's simplify: $$E[S(t)|N_1(t) = n, N_2(t) = m] = E[Max(T_n^{[1]}, T_m^{[2]})].$$ Let's think about this in a clearer way, so let's focus on just one Poisson process $N_1$. Then, conditioned on $N_t(t) = n$ we know that there are $n$ arrival times $T_1, \dots, T_n$ in the interval $[0, t]$. Let's just look at the distribution of the first one to start, we'll try to compute the conditional cdf: $$P(T_1 \leq s|N_1(t) = n) = 1 - P(N_1(s) = 0 | N_1(t) = n) $$ and this latter term gives by the memoryless property of the Poisson distribution: $$1 - P(N_1(s) = 0 | N_1(t) = n) = 1 - \frac{P(N_1(s) = 0 \cap N_1(t) = n)}{P(N_t(t) = n)}$$recallying that the pmf of the number of events in a Poisson process is, by definition $P(N(t) = n) = \frac{(\lambda \cdot t)^ne^{-\lambda t}}{n!}$ and is the same when any interval of length $t$ replaces $[0, t]$ thus: $$= 1 - \frac{P(N_1(s) = 0 \cap N_1([s, t]) = n)}{P(N_t(t) = n)} = 1 - \frac{e^{-\lambda \cdot s} \cdot (t - s)^n e^{-\lambda(t-s)}}{e^{-\lambda \cdot t }t^n}$$$$ = 1 - \frac{(t - s)^n}{t^n} $$ so we see that, as the gaps between arrival times are independent, the $n$ arrival times are uniformly distributed. Thus we see that $P(\text{Max}(T_1, \dots, T_n) < s) = \frac{s^n}{t^n}$, completely independent of $\lambda$ (when tacitly conditioned on $n$). This tells us that $$E[S(t) | N_1(t) = n, N_2(t) = m] = \int_{0}^t P(S(t) \geq s)ds$$ which gives us the very beautiful result $$ = \int_{0}^t (1 - P(T^{[1]}_n < s, T^{[2]}_m < s))dt = \int_{0}^t \left(1 - \frac{s^{n + m}}{t^{n + m}}\right) ds $$$$ = \frac{t(n + m)}{n + m + 1}.$$ On the other hand, we know by definition of a Poisson process and independence that: $$P(N_1(t) = n, N_2(t) = m) = \frac{(\lambda_1 \cdot t)^ne^{-\lambda_1 t}}{n!} \frac{(\lambda_2 \cdot t)^me^{-\lambda_2 t}}{m!}$$ more legibly: $$ = \frac{(\lambda_1^n\lambda_2^m)t^{n + m}e^{-(\lambda_1 + \lambda_2) t}}{n!m!}$$

Putting this all together, this gives:

$$\left(\sum_{n = 0}^\infty \sum_{m = 0}^\infty \frac{(n +m)(\lambda_1^n \lambda_2^m)}{(n + m + 1)\cdot n!\cdot m!} t^{m + n + 1} \right)e^{-(\lambda_1 + \lambda_2)t}$$

Letting $N = n + m$ we get the more pleasing: $$\left(\sum_{N = 0}^\infty \sum_{n = 0}^N \frac{N(\lambda_1^n \lambda_2^{N - n})}{(N + 1)\cdot n!\cdot (N - n)!} t^{N + 1} \right)e^{-(\lambda_1 + \lambda_2)t}$$

Finally as $\lambda_1 = 1, \lambda_2 = 3$ we get:

$$= \left(\sum_{N = 0}^\infty \sum_{n = 0}^N \frac{N\cdot 3^{N - n}}{(N + 1)\cdot n!\cdot (N - n)!} t^{N + 1} \right)e^{-4t}$$