Consider a particle $X_t^1$ which moves in space and assume that we have a countable family of exponential times ($exp(\nu)$ distributed) $T_1,....,T_N$ $N \in \mathbb{N}$, where $T_1$ is the time when particle 1 dies and splits into two new particles which get numbers 2 and 3. Then at $T_2$ particle $2$ dies and splits into two new particles and so on. This obviously leads to a tree structure. I now want to know the expectation for the number of particles at a given time $ \tilde{t}$.
My idea is as follows: The exponential distribution is memoryless so the probability for particle 1 to proliferate 3 times ( just hypothetically) is the same as if three particles proliferate each once. So I define $ \mathfrak{N}_t $ as the random variable which gives the probability for particle $1$ to split n times. As the occurence of these events is exponetial distributed $ \mathfrak{N}_t $ is a poisson process. If finally $\mathfrak{A}_t$ denotes the number of particles at time $t$ we have obviously $ \mathfrak{A}_t = \mathfrak{N}_t +1 $ and so the expecation is:
$ \mathbb{E}(\mathfrak{A}_t) = \mathbb{E}(\mathfrak{N}_t) +1 = \sum_{i=0}^{\infty}\frac{(\nu t)^{i}}{i!}e^{- \nu t}i + 1 = \nu t +1 $
So after time $t$ one expects $ (\nu t +1)$ particles. Is this correct?