The title is not actually correct, but I chose appeal over correctness ;)
I'd like to model a flower blossoming cycle, and these are the assumptions:
1) The instant $T$ in which each flower starts to blossom is modeled by a probability distribution $D_1$, which I assume to be $\sim N(\mu,\sigma)$;
2) Each flower has a lifespan $L$ which follows a probability distribution $D_2$, assumed for the moment to be another normal distribution $N(\lambda,\nu)$.
(The real model supposes that $D_2$ depends on the time of initial blossoming of the flower, but let's forget it for the time being.)
Question: how to find the instant(s) in time when the most flowers are blossoming together?
What I've tried so far: Let, as stated, $T\sim N(\mu,\sigma)$ and $L\sim N(\lambda,\nu)$. Let also be $X(t)$ be the number of flowers still blossoming (i.e. alive) at time $t$.
I think it's safe to say that \begin{equation} P(\text{flower $F$ is alive at time $t$})=P(\text{$F$ blossomed before $t$}\ \cap \text{$F$ is not dead at time $t$}) \end{equation} so \begin{equation} P(\text{flower $F$ is alive at time $t$})=P(T\leq t\ \cap\ L\geq t-T) \end{equation} I assume also that the two events are independent, so that probability is just the product of the two, but I'm stuck here, since I cannot imagine a way to count those flowers. Any hint is greatly appreciated!
Suppose there are $n$ flowers. The number of flowers still blossoming at time $t$ is a random variable $X(t)$, equal to $$X(t)=\sum_{f=1}^n I(\text{flower $f$ is alive at time $t$})\tag1$$ where the indicator function $I(A)$ equals one if event $A$ is true, zero otherwise. The expectation of $X(t)$ is then $$E(X(t))=n P(\text{flower $1$ is alive at time $t$})\tag2$$ since all flowers have the same blossoming behavior. If you're seeking the place where (2) is maximized as a function of $t$, you don't need to know the number of flowers; you get the same answer if you consider just one flower. Go ahead and calculate $$P(\text{flower $1$ is alive at time $t$}) = P(T\le t\cap T+L\ge t)\tag3$$ and maximize it as a function of $t$.
Note: the assumption that the random variables $T$ and $L$ are independent does not imply that the two events on the RHS of (3) are independent. Evaluate (3) by conditioning on $T$: $$ \begin{align} P(T\le t\cap T+L\ge t)&=\int_{u=-\infty}^\infty P(T\le t\cap T+L\ge t\mid T=u)\,f_T(u)\,du\\ &=\int_{u=-\infty}^t P(L\ge t-u)\,f_T(u)\,du \end{align} $$ To compute the derivative of (3) you'll need to differentiate under the integral sign.