# Expectation maximum i.i.d rv´s

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.

• It would be helpful to make your problem readable. – Wadim Zudilin Jun 22 '10 at 13:07
• Can you point to wher one can read a proof of this "well known fact"? I understand the inequality if u is concave and non-decreasing, but don't see a proof if u is decreasing, say. – Jeff Schenker Jun 24 '10 at 12:12
• In fact, the inequality is false without further assumptions on u. E.g., take u(x)=Cx(1-x) with C sufficiently large and positive. Let Xi be variables taking only the values 0 and 1. Then the choice a1=1 and all others zero wins over the choice 1/N for all i. What else do you intend to assume about u? – Jeff Schenker Jun 24 '10 at 15:05

## 1 Answer

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 $i=N_1 + 1, 2, ...N_2$