Setup:
Assume that we have an interval coded random variable $T$ which takes values in the set of half-open intervals $\{[L_1, U_1), \ldots, [L_K, U_K]\}$, where $U_k$ and $L_k$ are the upper and lower limits of the $k$-th interval; $k = 1, \ldots, K$. Denote the intervals as $I_k$.
Denote the distribution of the variable as $p_k \equiv \mathbb{P}[T \in I_k]$, written compactly in vector form as $\boldsymbol{p}$.
Let one way of computing the mean of the random variable be: $$ \mu \equiv \mathbb{E}(T) = \sum_k p_k \underbrace{\left(\dfrac{U_k + L_k}{2}\right)}_{M_k} $$
Problem statement:
I want to compute the change in the distribution ($\boldsymbol{p}_1$ to $\boldsymbol{p}_2$) of the random variable that would imply a change in the mean from $\mu_1$ (for which the $\boldsymbol{p}_1$ is known) to $\mu_2$ for which the $\boldsymbol{p}_2$ is unknown.
I understand that this is an ill-posed inverse problem, but one idea might be to find the vector $\boldsymbol{p}_2$ which is nearest in $L_1$ norm and also implies that the mean is equal to $\mu_2$.
$$ \begin{align} \hat{\boldsymbol{p}}_2 &= \arg\max_{\boldsymbol{p}_2}\|\boldsymbol{p}_2 - \boldsymbol{p}_1\|_1 \\ \text{subject to } &\sum_k p_{2k} M_k = \mu_2 \\ & p_{2k} \geq 0 \forall k \\ & \sum_k p_{2k} = 1 \end{align} $$
Questions:
- Is this a known class of problems? Is there a known way to solve this problem?
- Is this a reasonable way to try and solve this problem? Is there a better way?
