Find $\mathbb E((N(1,4]|N(3,10]=7))$ where $(N_t, t\geq 0)$ is a Poisson counting process with $\lambda=1$

Question

Find $\mathbb E((N(1,4]|N(3,10]=7))$ where $(N_t, t\geq 0)$ is a poisson counting process with rate $\lambda=1$

Attempt:

First determine conditional pmf: \begin{align} f_{N(1,4]|N(3,10]}(x|y) &= \frac{f_{N(1,4],N(3,10]}(x,y)}{f_{N(3,10]}(y)} \end{align} then \begin{align} \mathbb E((N(1,4]|N(3,10]=7)) &= \sum_x xf_{N(1,4]|N(3,10]}(x|y) \end{align} I'm having trouble determining the joint pmf since the intervals overlap, and does $x$ range from 0 to $\infty$ in the summation?

Answer 1

\begin{align} \mathbb{E} (N(1,4]|N(3,10]=7)&=\mathbb{E} (N(1,3)|N(3,10]=7) + \mathbb{E} (N[3,4]|N(3,10]=7)\\ &=\mathbb{E} (N(1,3)) + \mathbb{E} (N[3,4]|N(3,10]=7)\\ &= 2 + \frac{4-3}{10-3}\cdot 7 \end{align}

As for your question should $x$ range from $0$ to $\infty$, yes.