Help with Poisson process and conditional probability

Question

$N = \{N(t) | t \ge 0\}$ is a Poisson process with intensity $\lambda > 0$. Let T be the time of occurrence of the first event. Show that $T | N(t) = 1 \ \sim U(0,t)$.

Hint: It may be a good idea to start with

$P(T \le x | N(t) = 1), \ 0\le x\le t$

In the solution they say "It is clear by defintion of T that"

$P(T\le x | N(t) = 1) = 0 $ if $x< 0$ and $P(T\le x | N(t) = 1) = 1 $ if $x > t$ Why is this clear?

So take $0\le x \le t$ and observe that

$P(T \le x | N(t) = 1) = \frac{P(N(x) = 1, N(t) = 1)}{P(N(t) = 1}$

since $\{T \le x\} = \{N(x) \ge 1 \}$, but as $N(t)=1 $, we have $N(x) = 1$ for $x\le t$. How do we conclude this last part and what does it actually mean?

The rest of the calculation to conclude the sought distribution are quite straight forward.

Answer 1

On your first question, $T$ is defined to be the time of the fist event, which must occur not before the initial time, hence $T \ge 0$. Also you are assuming that $N(t)=1$ so by time $t$, the first event happened already, hence $T \le t$. Thus, $$ \mathbb{P}[T \le x | N(t) = 1] = \begin{cases} 0, & x < 0 \\ 1, & x > t \end{cases} $$

Answer 2

Let $N(t) \sim P(\lambda t)$. You are interested in the distribution of $\{T|N(t)=1\}$ on $0\le x\le t$. Hence, we can try to compute its cdf:

\begin{align} F_T(x) &= P\{T\le x|N(t)=1\} = \frac{P(\{T\le x\}\cap \{N(t)=1\})}{P(\{N(t)=1\})}, \end{align} in the nominator we want to compute the probability that the first event happen before time $x$ and that the total number of events in the interval $[0,t]$ was $1$, i.e., the first and the only arrival was in $[0,x]$, so \begin{align} P(\{T\le x\}\cap \{N(t)=1\}) &= P(\{N(x)=1\} \cap \{N(t-x)=0\})\\ &= \lambda xe^{-\lambda x} \lambda e^{-\lambda (t-x)}\\ &=\lambda x e^{-\lambda t}. \end{align} Plugging the result in the cdf we get,

\begin{align} F_T(x) &= \frac{\lambda x e^{-\lambda t}}{\lambda t e^{-\lambda t}}=\frac{x}{t}, & 0\le x\le t. \end{align} Hence, $\{T|N(t)=1\}\sim \mathrm{U}[0,t]$.