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I want to show that if $X$ and $Y$ are the two loss variables such that $X\leq Y$, then $\text{VaR}_\delta(X)\leq\text{VaR}_\delta(Y)$.

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A link would be in place here. (Note: VaR=Variance) – AD. Nov 22 '10 at 15:58
@A.D. "VaR" likely means "Value at Risk": . The real question here is what is meant by "$X \le Y$", because the usual definition (of "Loss") is unrelated to probabilities (which is what VaR is defined in terms of) and therefore the assertion is false. But if $\le$ refers to stochastic dominance, there is no difficulty. – whuber Nov 22 '10 at 16:31
up vote 2 down vote accepted

From the link given by whuber, we have $$ {\rm VaR}_\alpha (L) = \inf \lbrace l \in \mathbb{R}:{\rm P}(L > l) \le 1 - \alpha \rbrace. $$ If $X$ and $Y$ are random variables such that $X \leq Y$, then obviously ${\rm P}(X > l) \leq {\rm P}(Y > l)$ for any $l \in \mathbb{R}$. So, if $l$ belongs to the set $\lbrace l \in \mathbb{R}:{\rm P}(Y > l) \le 1 - \alpha \rbrace$, then $l$ belongs also to the set $\lbrace l \in \mathbb{R}:{\rm P}(X > l) \le 1 - \alpha \rbrace$. Hence, in particular, $$ \inf \lbrace l \in \mathbb{R}:{\rm P}(X > l) \le 1 - \alpha \rbrace \leq \inf \lbrace l \in \mathbb{R}:{\rm P}(Y > l) \le 1 - \alpha \rbrace, $$ that is ${\rm VaR}_\alpha (X) \leq {\rm VaR}_\alpha (Y)$. Of course, nothing changes if the $\leq$ in $X \leq Y$ refers to stochastic dominance.

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thanks i am answered – Vaolter Nov 22 '10 at 18:45

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