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$\newcommand\ep\varepsilon$Your stochastic domination condition $X\le_1-X+c$ can be written as $$P(X>x)\le P(-X+c>x)\tag{1}\label{1}$$ for all real $x$.

If $c\ge2M$, then $X\le M\le2M-X\le-X+c$, so that $X\le-X+c$ and hence $X\le_1-X+c$.

On the other hand, if $c<2M$ and $P(X=M)=1$, then $$P(X>c/2)=1\not\le0=P(X<c/2)=P(-X+c>c/2)$$ so that \eqref{1} does not hold for $x=c/2$ and hence it is not true that $X\le_1-X+c$.

(If you insist that the support of $X$ be the entire interval $[-M,M]$, just use an approximation. For instance, you can assume that $P(X\in B)=(1-\ep)\,1(M\in B)+\ep\dfrac{|B\cap[-M,M]|}{2M}$ for any $\ep\in(0,1/2]$ and all Borel subsets of $\Bbb R$. Then the support of the distribution of $X$ will be the entire interval $[-M,M]$, whereas, for $c<2M$,
$$P(X>c/2)>P(X=M)=1-\ep\ge\frac12$$ and hence $$P(-X+c>c/2)=P(X<c/2)=1-P(X>c/2)<\frac12,$$ so that $P(X>c/2)>P(-X+c>c/2)$ and hence \eqref{1} does not hold for $x=c/2$. So, it is not true that $X\le_1-X+c$ if $c<2M$.)

So, the best lower bound on $c$ under the given conditions is $2M$. (There is no finite upper bound on $c$.

$\newcommand\ep\varepsilon$Your stochastic domination condition $X\le_1-X+c$ can be written as $$P(X>x)\le P(-X+c>x)\tag{1}\label{1}$$ for all real $x$.

If $c\ge2M$, then $X\le M\le2M-X\le-X+c$, so that $X\le-X+c$ and hence $X\le_1-X+c$.

On the other hand, if $c<2M$ and $P(X=M)=1$, then $$P(X>c/2)=1\not\le0=P(X<c/2)=P(-X+c>c/2)$$ so that \eqref{1} does not hold for $x=c/2$ and hence it is not true that $X\le_1-X+c$.

(If you insist that the support of $X$ be the entire interval $[-M,M]$, just use an approximation. For instance, you can assume that $P(X\in B)=(1-\ep)\,1(M\in B)+\ep\dfrac{|B\cap[-M,M]|}{2M}$ for any $\ep\in(0,1/2]$ and all Borel subsets of $\Bbb R$, where $|\cdot|$ denotes the Lebesgue measure. Then the support of the distribution of $X$ will be the entire interval $[-M,M]$, whereas, for $c<2M$,
$$P(X>c/2)>P(X=M)=1-\ep\ge\frac12$$ and hence $$P(-X+c>c/2)=P(X<c/2)=1-P(X>c/2)<\frac12,$$ so that $P(X>c/2)>P(-X+c>c/2)$ and hence \eqref{1} does not hold for $x=c/2$. So, it is not true that $X\le_1-X+c$ if $c<2M$.)

So, the best lower bound on $c$ under the given conditions is $2M$. (There is no finite upper bound on $c$.

$\newcommand\ep\varepsilon$Your stochastic domination condition can be rewritten as $$g(x):=P(X\le x)+P(X\le c-x)\ge1 \tag{1}\label{1}$$ for all real $x$.

If $c\ge2M$, then for all real $x$ $$g(x)\ge\max(P(X\le x),P(X\le c-x)) =P(X\le\max(x,c-x))\ge P(X\le c/2)\ge P(X\le M)=1,$$ so that \eqref{1} holds.

On the other hand, if $c<2M$ and $P(X=M)=1$, then $$g(c/2)=2P(X\le c/2)=0,$$ so that \eqref{1} does not hold.

(If you insist that the support of $X$ be the entire interval $[-M,M]$, just use an approximation. For instance, you can assume that $P(X\in B)=(1-\ep)\,1(M\in B)+\ep\dfrac{|B\cap[-M,M]|}{2M}$ for any $\ep\in(0,1/2]$ and all Borel subsets of $\Bbb R$. Then the support of the distribution of $X$ will be the entire interval $[-M,M]$, whereas $g(c/2)=2P(X\le c/2)=2\ep\dfrac{M+c/2}{2M}<1$ if $c<2M$.)

So, the best lower bound on $c$ under the given conditions is $2M$. (There is no finite upper bound on $c$.

$\newcommand\ep\varepsilon$Your stochastic domination condition $X\le_1-X+c$ can be written as $$P(X>x)\le P(-X+c>x)\tag{1}\label{1}$$ for all real $x$.

If $c\ge2M$, then $X\le M\le2M-X\le-X+c$, so that $X\le-X+c$ and hence $X\le_1-X+c$.

On the other hand, if $c<2M$ and $P(X=M)=1$, then $$P(X>c/2)=1\not\le0=P(X<c/2)=P(-X+c>c/2)$$ so that \eqref{1} does not hold for $x=c/2$ and hence it is not true that $X\le_1-X+c$.

(If you insist that the support of $X$ be the entire interval $[-M,M]$, just use an approximation. For instance, you can assume that $P(X\in B)=(1-\ep)\,1(M\in B)+\ep\dfrac{|B\cap[-M,M]|}{2M}$ for any $\ep\in(0,1/2]$ and all Borel subsets of $\Bbb R$. Then the support of the distribution of $X$ will be the entire interval $[-M,M]$, whereas, for $c<2M$,
$$P(X>c/2)>P(X=M)=1-\ep\ge\frac12$$ and hence $$P(-X+c>c/2)=P(X<c/2)=1-P(X>c/2)<\frac12,$$ so that $P(X>c/2)>P(-X+c>c/2)$ and hence \eqref{1} does not hold for $x=c/2$. So, it is not true that $X\le_1-X+c$ if $c<2M$.)

So, the best lower bound on $c$ under the given conditions is $2M$. (There is no finite upper bound on $c$.

$\newcommand\ep\varepsilon$Your stochastic domination condition can be rewritten as $$g(x):=P(X\le x)+P(X\le c-x)\ge1 \tag{1}\label{1}$$ for all real $x$.

If $c\ge2M$, then for all real $x$ $$g(x)\ge\max(P(X\le x),P(X\le c-x)) =P(X\le\max(x,c-x))\ge P(X\le c/2)\ge P(X\le M)=1,$$ so that \eqref{1} holds.

On the other hand, if $c<2M$ and $P(X=M)=1$, then $$g(c/2)=2P(X\le c/2)=0,$$ so that \eqref{1} does not hold.

(If you insist that the support of $X$ be the entire interval $[-M,M]$, just use an approximation. For instance, you can assume that $P(X\in B)=(1-\ep)\,1(M\in B)+\ep\dfrac{|B\cap[-M,M]|}{2M}$ for a any $\ep\in(0,1/2]$ and all Borel subsets of $\Bbb R$. Then the support of the distribution of $X$ will be the entire interval $[-M,M]$, whereas $g(c/2)=2P(X\le c/2)=2\ep\dfrac{M+c/2}{2M}<1$ if $c<2M$.)

So, the best lower bound on $c$ under the given conditions is $2M$. (There is no finite upper bound on $c$.

$\newcommand\ep\varepsilon$Your stochastic domination condition can be rewritten as $$g(x):=P(X\le x)+P(X\le c-x)\ge1 \tag{1}\label{1}$$ for all real $x$.

If $c\ge2M$, then for all real $x$ $$g(x)\ge\max(P(X\le x),P(X\le c-x)) =P(X\le\max(x,c-x))\ge P(X\le c/2)\ge P(X\le M)=1,$$ so that \eqref{1} holds.

On the other hand, if $c<2M$ and $P(X=M)=1$, then $$g(c/2)=2P(X\le c/2)=0,$$ so that \eqref{1} does not hold.

(If you insist that the support of $X$ be the entire interval $[-M,M]$, just use an approximation. For instance, you can assume that $P(X\in B)=(1-\ep)\,1(M\in B)+\ep\dfrac{|B\cap[-M,M]|}{2M}$ for any $\ep\in(0,1/2]$ and all Borel subsets of $\Bbb R$. Then the support of the distribution of $X$ will be the entire interval $[-M,M]$, whereas $g(c/2)=2P(X\le c/2)=2\ep\dfrac{M+c/2}{2M}<1$ if $c<2M$.)

So, the best lower bound on $c$ under the given conditions is $2M$. (There is no finite upper bound on $c$.