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I have a statement that reads:

If $Z_1,Z_2$ are random variables such that $Z_1 \geq Z_2$, then $\rho(Z_1) \geq \rho(Z_2)$.

where $\rho$ is a function.

What is the meaning of $Z_1 \geq Z_2$? I am particularly interested in the monotonicity of coherent risk measures. For discrete random variables, say:

$\Pr[Z_1=x_i] = p_i$ and $\Pr[Z_2=y_j] = q_j$

how can I write the condition $Z_1 \geq Z_2$ using $x_i,p_i,y_j,q_j$?

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Inequality $Z_1 \geqslant Z_2$ holds surely, i.e for all $\omega \in \Omega$, $Z_1(\omega) \geqslant Z_2(\omega)$. Examples could be $Z_1 = X -1$, $Z_2 = X$ or, for independent uniform random variables $U_1 \sim \mathcal{U}(2,3)$ and $U_2 \sim \mathcal{U}(0,1)$, $Z_1 = U_1$ and $Z_2 = U_2$. – Sasha Aug 27 '12 at 3:36
up vote 2 down vote accepted

First, a comment states that $Z_1\ge Z_2$ means the inequality holds surely, i.e., that it holds for every $\omega$ in the probability space $\Omega$. (There must be such a space, or else the notions of probability makes no sense, though it is quite common not to specify or even mention it explicitly.) But in practice, one uses the weaker statement $Z_1\ge Z_2$ almost surely (abbreviated a.s.), which means $\Pr[Z_1<Z_2]=0$. Quite frequently, I suspect this is the intended meaning even when the author omits the “a.s.” part.

You cannot write the condition $Z_1\ge Z_2$ using the quantities you list. For that, you need the combined probability distribution, i.e. you need to know the probabilities $P_{ij}=\Pr[Z_1=x_i \mathbin\& Z_2=y_j]$. In this case, $Z_1\ge Z_2$ a.s. if and only if $P_{ij}=0$ whenever $x_i<y_j$. If you don't include the “a.s.”, then even the full probability distribution won't provide the answer. You need the underlying probability space in order to answer the question in that case.

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+1. And one knows exactly what are the marginals (the distributions of $Z_1$ and $Z_2$) such that there exists a joint distribution (of $(Z_1,Z_2)$) with the property that $Z_1\leqslant Z_2$ almost surely. Keyword: coupling. – Did Aug 27 '12 at 7:07

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