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.
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.
We want to calculate $\mathbb{E}(X_t)$,
$$\begin{align*} \mathbb{E}(X_t) & = \mathbb{E}(X_t1_{T \leq t}) + \mathbb{E}(X_t1_{T > t}) \\ & = \int_0^t \lambda e^{-\lambda s}\mathbb{E}(X_t | T = s)ds + e^{-\lambda t} \end{align*}$$
My question is: why $\mathbb{E}(X_t1_{T > t})=e^{-\lambda t}$?
I know $\mathbb{P}(T>t)=e^{-\lambda t}$ but I don't see how the expectation simplifies.
It $T>t$ then there is no birth till time $t$ so $X_t$ is same as $X_0$ which is $1$. Hence $EX_t 1_{T>t}=P(T>t)=e^{-\lambda t}$