I am trying to understand if the following statement is true. Let $M,N$ and $X$ be random variables. If the following inequality holds for any concave non-decreasing function $u$ \begin{equation} \mathbb{E}[u(N)]\leq\mathbb{E}[u(M)] \end{equation} Then \begin{equation} \mathbb{E}[u(N+X)]\leq\mathbb{E}[u(M+X)] \end{equation} Note that there is no an assumed relation between $N$ and $M$.
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1$\begingroup$ While this isn't true, I suspect it might be true if $X$ is independent of $(M,N)$. $\endgroup$– Thomas LumleyCommented Jun 12 at 23:01
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$\begingroup$ Do you think that it could be true only if I restrict $X$ to be a constant, or at least a nonnegative r.v.? $\endgroup$– Don P.Commented Jun 13 at 0:01
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$\begingroup$ There shouldn't be any need to make $X$ non-negative, because you don't know if $M$ and $N$ are positive or negative. It reminds me of what Pollard calls Anderson's Lemma, which is specialised to $N=0$ and symmetric $M$ $\endgroup$– Thomas LumleyCommented Jun 13 at 0:14
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$\begingroup$ I get it. Thank you very much for your answers. I will learn more about Anderson's Lemma. $\endgroup$– Don P.Commented Jun 13 at 0:30
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The implication does not hold.
For example, let $M \equiv 0$, $N$ be the Radamacher random variable (i.e., $P(N = \pm 1) = \frac{1}{2}$). Then for any concave function $u$, it follows by Jensen's inequality that \begin{align*} E[u(N)] \leq u(E(N)) = u(0) = u(M) = E[u(M)]. \end{align*}
On the other hand, for $X = -2N$, we have \begin{align*} & E[u(N + X)] = E[u(-N)] = \frac{1}{2}u(1) + \frac{1}{2}u(-1), \\ & E[u(M + X)] = E[u(-2N)] = \frac{1}{2}u(2) + \frac{1}{2}u(-2). \end{align*}
Obviously, $u(1) + u(-1) \leq u(2) + u(-2)$ cannot hold for any concave and non-decreasing function $u$. For example, it cannot hold for \begin{align*} u(x) = (x + 1)I_{(-\infty, 0)}(x) + I_{[0, \infty)}(x), \end{align*} for which $u(1) + u(-1) = 1 + 0 = 1 > u(2) + u(-2) = 1 + (-1) = 0$.
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$\begingroup$ Thank you very much for your answer. I have some comments, $u(x)=x$ is a concave non-decreasing function for which the inequality holds. Do you think that the statement could be true if $X$ is restricted to be nonnegative r.v. or maybe for a constant?. $\endgroup$– Don P.Commented Jun 13 at 0:11
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$\begingroup$ Or maybe conditions for $M$ and $N$? Actually I am looking for conditions under the statement is true. Thank you very much again. $\endgroup$– Don P.Commented Jun 13 at 0:26