Optimizing over probability distributions

Question

For simplicity, let's consider the simple problem of maximizing the mean of a random variable $X$ with a density $u$ supported on $[0,1],$ $$ \max \int_0^1 xu(x)dx \\ \text{subject to } \int_0^1 u = 1, u\geq 0. $$ Lagrange multiplier method does not work, because the problem with multiplier $\lambda$ $$ \int_0^1 (x-\lambda)u(x) dx + \lambda $$ has a trivial Euler-Lagrange equation $x-\lambda =0,$ which makes sense, because the optimal solution does not have a density function. It is the constant RV $X =1.$

How do I systematically treat this type of problem? It would be very helpful if anyone can provide a reference for the techniques used for optimization problems of this type.

Answer 1

The maximum value is not attained. The supremum is $1$:

$\int_0^{1}xu(x)dx\leq \int_0^{1}u(x)dx=1$ so $1$ is an upper bound. Now let $\epsilon >0$ and $u(x)=\frac 1 {\epsilon}$ if $1-\epsilon <x<1$ and $0$ otherwise. Then $\int_0^{1}xu(x)dx\geq \frac 1 {\epsilon} \int_{1-\epsilon}^{1}xdx=\frac {2\epsilon-\epsilon ^{2}} {2\epsilon} \to 1$ as $\epsilon \to 0$.

Note that we cannot have $\int_0^{1}xu(x)dx =1$. This is because we would then have $\int_0^{1}(1-x)u(x)dx =0$ and you can use the non-negativity of the integrand to show that this cannot happen. [I am leaving the details to you].

Answer 2

You just need to center the weight of the distribution at 1.

Actuall this should be written as

\begin{equation}\textrm{sup} \int_{0}^{1}xu(x)dx=1 \end{equation} because \begin{equation}\int_{0}^{1}xu(x)dx=1 \leq \int_{0}^{1}u(x)dx=1\end{equation} use the following functions \begin{equation} u_n(x)= \begin{cases} 2n^2(x-(1-\frac{1}{n})) && , 1-\frac{1}{n}\leq x \leq 1 \\ 0 &&, otherwise \end{cases} \end{equation} where $n \rightarrow \infty$.