If $X' \leq X$ almost surely, is it possible to prove that $P(X = s) \geq P(X' = s)$?

Question

With respect to my previous question, let us define $X$ as:

$$ X = \sum_j^r l^j Y^j, $$

where $l^j \geq 0$ and $Y^j$, $j = 1, \ldots, r$ is a Bernoulli random variable which takes on values in $\{-1, +1\}$, with $\Pr(Y^j = 1) = p^j \geq \frac{1}{2}$.

I would like to prove that $$Pr(X = s) \geq P(X = -s) = P(-X = s).$$ To this end, it is possible to show that, since $-Y^j$ is stochastically dominated by $Y^j$, there exists some $\tilde{Y}^j$ with the same distribution as $-Y^j$ so that $\tilde{Y}^j \leq Y^j$ almost surely. Hence, there exists a $\tilde{X} \leq X$ such that $\tilde{X}$ is distributed like $-X$.

Such a dominance should allow proving inequalities of the form $$ \Pr(X \geq s) \geq P(X' \geq s). $$

However, I'm strugling in harnessing the dominance for the case $$\Pr(X = s) \geq \Pr(X' = s) = \Pr(X = -s).$$

This is needed to prove that $\mathbb{E}\left[\left| X + l^{r+1}Y^{r+1}\right|\right] \geq \mathbb{E}\left[\left| X \right|\right]$.

Thanks.

Answer 1

Let us assume that the random variables $Y_j$ are independent and that one goal of the OP is to prove that, for every nonnegative $s$, $P(X=s)\geqslant P(X=-s)$.

Then an elementary counterexample is $r=2$, $(l_1,l_2)=(1,2)$ and $s=1$, then $$ P(X=1)=P(Y_1=-1)P(Y_2=+1)=(1-p_1)p_2, $$ and $$ P(X=-1)=P(Y_1=+1)P(Y_2=-1)=p_1(1-p_2), $$ hence $P(X=1)\lt P(X=-1)$ in the setting of the question for every $p_1\gt p_2\geqslant\frac12$.

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Turning to the OP's main goal, which is to prove that $$E[|X+l_{r+1}Y_{r+1}|]\geqslant E[|X|],$$ note that one can assume without loss of generality that $l_{r+1}=1$, then the result below implies the claim.

Consider two independent random variables $(U,V)$ such that $U$ dominates stochastically $-U$ and $V=\pm1$ with $P(V=1)=v\geqslant\frac12$. Then, $E[|U+V|]\geqslant E[|U|]$.

To prove this, note that $E[|U+V|]=A(v)$ is an affine function of $v$ hence $A(v)\geqslant E[|U|]$ for every $v$ in $[\frac12,1]$ if and only if $A(1)\geqslant E[|U|]$ and $A(\frac12)\geqslant E[|U|]$. By convexity of the function $|\cdot|$, $$A(\tfrac12)=\tfrac12E[|U+1|+|U-1|]\geqslant E[|U|],$$ for every integrable random variable $U$. On the other hand, $A(1)=E[|U+1|]$, hence it suffices to prove that $E[|U+1|]\geqslant E[|U|]$.

Consider the function $B$ defined by $$B(u)=\left\{\begin{array}{ccc}-1&\text{if}&u\lt-\tfrac12,\\ 2u&\text{if}&|u|\leqslant\tfrac12,\\1&\text{if}&u\gt\tfrac12.\end{array}\right. $$ Then $B$ is nondecreasing hence $E[B(U)]\geqslant E[B(-U)]$ by the hypothesis of stochastic domination, and $B(-U)=-B(U)$ because $B$ is odd, hence $E[B(U)]\geqslant0$. Finally, note that $$|u+1|-|u|=B(u+\tfrac12)\geqslant B(u),$$ hence $E[|U+1|-|U|]\geqslant E[B(U)]\geqslant0$, QED.

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To sum up, the result the OP is interested in holds, but not the intermediary step they imagined to prove it.