Time of first $N(t)$ arrivals in a Poisson process

Question

Let $\left \{ N(t), t>0 \right \}$ be a Poisson process of rate $\lambda$ and $X_i$ be interarrival times. Let $S_n = \sum_{i=1}^n X_i$. I want to find the distribution of $ S_{N(t)}$, which is the time of the last arrival before time $t$. $$ F_{S_{N(t)}}(s) = \mathbb{P}(S_{N(t)} \leq s) = \sum_{k=1}^{\infty} \mathbb{P}(S_{N(t)} \leq s \,|\, N(t) = k)\mathbb{P}(N(t) = k) =\sum_{k=1}^{\infty} \mathbb{P}(S_{k} \leq s \,|\, N(t) = k)\mathbb{P}(N(t) = k) $$ Now, given that there are $k$ arrivals in $(0,t]$, times of those arrivals have the same distribution as the order statistics of $k$ iid $\text{Unif}(0,t)$ random variables. Then $S_{N(t)}\,|\, N(t) = k$ has the same distribution as the $\max_{1 \leq i \leq k} U_i$, where $U_i \sim \text{Unif}(0,t)$. $$ \mathbb{P}(S_{k} \leq s \,|\, N(t) = k) = \mathbb{P}(\max_{1 \leq i \leq k} U_i\leq s) = \mathbb{P} (U_1\leq s, \ldots, U_k\leq s) = \left [\mathbb{P} (U_1\leq s)\right ]^k = \left (\frac{s}{t} \right )^k $$ Thus $$ \mathbb{P}(S_{N(t)} \leq s) = \sum_{k=1}^{\infty} \left (\frac{s}{t} \right )^k e^{-\lambda t} \frac{(\lambda t)^k}{k!} = e^{-\lambda t} \sum_{k=1}^{\infty} \frac{(\lambda s)^k}{k!} = e^{-\lambda t}(e^{\lambda s}-1) $$ And $$ f_{S_{N(t)}}(s) = \lambda e^{-\lambda t}e^{\lambda s}, \quad 0 < s \leq t $$ The problem is that $$ \int_0^t f_{S_{N(t)}}(s)\, \mathrm{d}s = 1 - e^{-\lambda t} \neq 1 $$ And I can't find where the mistake is.

Answer 1

The error is just that you forgot to consider the case when there is no arrival at all before time $t$…! ;-)