Consider a nonhomogeneous Poisson process on $[0, T]$ with mean value function $m(t)$ for $t \in [0, T]$.
Let $X_1$ denote the time of the first arrival. Show that $(X_1\,|\,N(T) = 1)$ has the following cdf:
$$F(x) = \frac{m(x)}{m(T)},\quad x \in [0, T]$$
Can someone help me out here? In this case does x refer to the event $(X_1\,|\,N(T) = 1)$? Why is it not just $F(x)= m(T)$?
A cdf refers to a random variable, not an event.
By definition,
$$F(x)=\mathbb{P}(X_1\le x\,|\,N(T)=1).$$
Now note that $\{X_1\le x,N(T)=1\}=\{N(x)=1,N(T)-N(x)=0\}$. Therefore,
$$F(x)=\frac{\mathbb{P}(N(x)=1,N(T)-N(x)=0)}{\mathbb{P}(N(T)=1)}$$
and since $N(x)$ and $N(T)-N(x)$ are independent for a Poisson process,
$$F(x)=\frac{\mathbb{P}(N(x)=1)\mathbb{P}(N(T)-N(x)=0)}{\mathbb{P}(N(T)=1)}=\frac{m(x)e^{-m(x)}e^{-(m(T)-m(x))}}{m(T)e^{-m(T)}}=\frac{m(x)}{m(T)}.$$