Suppose we have two vectors $X = \{x_1, x_2, ....,x_n\}$ and $R = \{r_1, r_2, ...., r_n\}$ with $\sum_i^n{x_i} \leq 1$ and $0 \leq x_i \leq 1$. For $R$, we have $R \geq 0$, i.e. for all i, $r_i \geq 0$.
Then we have the operation called randomization here that let's pick up any element inside X with probability $P(x_k \mbox{ is picked}) = \frac{x_k}{\sum_i^n{x_i}}$, change the value of $x_i$ to 1 and the rest of elements in X to be $0$. Then the vector X is converted to $X^* = \{x^*_1, x^*_2, ....,x^*_n\}, x^*_i \in \{0,1\}$ Then we can know that $E[X^*R^T] = E[\sum_i^n{r_ix^*_i}] = \sum_i^n{r_iE[x^*_i]} \geq \sum_i^n{r_ix_i} = XR^T$
Then let's select an element $x_j$ inside X randomly with probability $P = \frac{x_j}{\sum_i^n{x_i}}$ and change its value to $0$, giving us $X_{new} = \{x_1, x_2, ..., x_{j-1}, 0, x_{j+1}, ...,x_n\}$. Let's repeat the same operation as described above to $X_{new}$, what can we say about the relationship between $E[X^*_{new}R^T]$ and $XR^T$? Ideally, I want to prove that $E[X^*_{new}R^T] \geq XR^T$.
Greatly appreciate any comments and hints!
Let $x_i^{new}$ be the $i^{th}$ index of $X_{new}$. \begin{align*} E[X^*_{new}R^T] = E[\sum x_i^{new}r_i] =\sum E[x_i^{new} r_i]\\ = \sum r_iE[x_i^{new}] = \sum r_i(0\cdot P(x_i^{new}=0)+x_i\cdot P(x_i^{new}=x_i))\\ = \sum r_ix_i(1-\frac{x_i}{\sum x_i})= \sum r_ix_i-\frac{\sum r_ix_i^2}{\sum x_i} \leq \sum r_ix_i=XR^T. \end{align*} All the sums are take on the index set $\{1, 2, \dots, n\}$.