If I have a fixed positive integer $N$ and $N$ i.i.d rv´s. $X_1,X_2,...,X_N$, and parameters $a_i$ such that $\displaystyle\sum_{i=1}^N{a_i}=1$, it is well known that there is a global maximum of
$f(a_1,a_2,...,a_N)=E[u\Bigl(\displaystyle\sum_{i=1}^N{u(a_iX_i)}\Bigr)]$ give by $a_i=1/n$, for a concave function $u$.
How do I find the maximum if $N$ instead of being fixed is a discrete random variable that takes positive integers.
I do not think the answer changes much from the basic situation. Consider a situation where $N$ takes two possible values: $N_1$ and $N_2$ with probabilities $p$ and $1-p$ respectively. Then you want to maximize the following function:
$f(.) = p E\Bigl(\displaystyle\sum_{i=1}^{N_1} u(a_i X_i )\Bigr) + (1-p) E\Bigl(\sum_{i=1}^{N_2} u(a_i X_i )\Bigr) $
Without loss of generality assume that $N_2$ > $N_1$. Then we can simplify the above and re-write as:
$f(.) = E\Bigl(\displaystyle\sum_{i=1}^{N_1} u(a_i X_i )\Bigr) + (1-p) E\Bigl(\sum_{i=N_1 +1}^{N_2} u(a_i X_i )\Bigr) $
The maximum value of $f(.)$ can be computed by taking the maximum of the two terms independently as the set of 'parameters' ${a_i}$ are disjoint between the two terms. Thus, the maximum for the above case occurs at:
$a_i= \frac{1}{N_1}$ for $i=1, 2, ...N_1$
and
$a_i = \frac{1}{N_2-N_1}$ for $i=N_1 + 1, 2, ...N_2$