Probability in Poisson Process

Question

Let $ {N(t)}$ be a Poisson process with $\lambda$ parameter. Let $t \in [0,T]$.

What is the probability that number of jumps in Poisson process at first half of $[0,T]$ interval will be greater than the number of jumps at second half of this interval provided that there were $n$ jumps at $[0,T]$ interval.

My attempt:

I came up with this idea: But $N(0) = 0$ , then we have $N(\frac{1}{2} T) > N(\frac{1}{2} T)$ which is clear non-sense. Where is the catch?

Answer 1

For $0\leqslant k\leqslant n$ we have \begin{align} \mathbb P(N(T/2) = k\mid N(T)=n) &= \frac{\mathbb P(N(T/2)=k, N(T)=n)}{\mathbb P(N(T)=n)}\\ &= \frac{\mathbb P(N(T/2)=n, N(T)-N(T/2)=n-k}{\mathbb P(N(T)=n)}\\ &= \frac{(e^{-\lambda T/2}(\lambda T/2)^k/k!)(e^{-\lambda T/2}(\lambda T/2)^{n-k}/(n-k)!)}{e^{-\lambda T}(\lambda T)^n/n!)}\\ &= \binom nk \left(\frac12\right)^n. \end{align} Hence $$ \mathbb P(N(T/2) > N(T) - N(T/2) \mid N(T)= n) = \left(\frac12\right)^n\sum_{k=\lceil n/2\rceil}^n \binom nk. $$

Answer 2

Attempt number 2:

It looks better to me, but i am still not sure about my way of thinking.