I am currently studying the simple birth process and am stuck in trying to understand the following example.
EXAMPLE: Setup
Consider a population in which each individual gives birth after an exponential time of parameter $\lambda$, all independently.
If $i$ individuals are present then the first birth will occur after an exponential time of parameter $i\lambda$.
Then we have $i + 1$ individuals and, by the memoryless property, the process begins afresh.
Hence the population performs a birth-death-process with $q_i = i\lambda$.
We denote with $X_t$ the number of individuals at time $t$ and suppose $X_0 = 1$.
Let $T$ be the time of the first birth.
EXAMPLE: Solution
We want to calculate $\mathbb{E}(X_t)$,
$$\begin{align*} \mathbb{E}(X_t) & = \mathbb{E}(X_t1_{T \leq t}) + \mathbb{E}(X_t1_{T > t}) \\ & = \int_0^t \lambda e^{-\lambda s}\mathbb{E}(X_t | T = s)ds + e^{-\lambda t} \end{align*}$$
Set $\mu(t) = \mathbb{E}(X_t)$, then $\mathbb{E}(X_t | T = s) = 2\mu(t-s)$
Setting $r = t - s$ we get, $$\begin{align*} e^{\lambda t }\mu(t) = 2\lambda \int_0^t e^{\lambda r} \mu(r) dr \end{align*}$$
By differentiating we get,
$$\begin{align*} \mu^{'}(t) = \lambda \mu(t) \end{align*}$$
and hence we conclude that,
$$\begin{align*} \mathbb{E} (X_t) = e^{\lambda t} \end{align*}$$
QUESTION
I don't understand the following point in the argument above,
Why can we immediately conclude that $\mathbb{E}(X_t | T = s) = 2\mu(t-s)$ ?