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There is an exercise I struggle to solve. I hope you can give me a hint.

Let X be random variable taking values in $I\subset \mathbb{R}$. I have to show that the Value at Risk is invariant under any increasing and continuous function $f:I \rightarrow \mathbb{R}$ for each $\alpha \in (0,1)$, i.e., $VaR_{\alpha}(f(X))=f(VaR_{\alpha}(X))$.

I know that for $X_1\geq X_2,\mathbb{P}-a.s. \leftrightarrow VaR_{\alpha}(X_1)\geq VaR_{\alpha}(X_2)$. I probably have to show that $VaR_{\alpha}(f(X))\geq f(VaR_{\alpha}(X))$ and $VaR_{\alpha}(f(X))\leq f(VaR_{\alpha}(X))$.

I can write $f(X)=\tilde{X}$ and therefore probably $\tilde{X}\geq X, \mathbb{P}-a.s.$ which means $VaR_{\alpha}(f(X))\geq VaR_{\alpha}(X)$, but then I am stuck. Is this even the right way to start the proof?

I appreciate any tips and thoughts!

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Easy way would be to start with this definition of VaR:

$P\left[ X \le \mathrm{VaR}_\alpha\left(X\right)\right]=\alpha$

Now if f is increasing and continuous from the left, then easy to argue that:

$P\left[ f\left(X\right) \le \mathrm{VaR}_\alpha\left(f\left(X\right)\right)\right]=\alpha$

Next apply $f^{-1}$ to both sides of the inequality inside P:

$P\left[ X \le f^{-1}\left\{\mathrm{VaR}_\alpha\left(f\left(X\right)\right)\right\}\right]=\alpha$

Comparing this to the first equation, we can deduce that:

$\mathrm{VaR}_\alpha\left(X\right)=f^{-1}\left\{\mathrm{VaR}_\alpha\left(f\left(X\right)\right)\right\}$

$\Rightarrow f \left(\mathrm{VaR}_\alpha\left(X\right)\right)=\mathrm{VaR}_\alpha\left(f\left(X\right)\right)$

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  • $\begingroup$ Ah, I see. Thank you so much! :-) $\endgroup$ – Wombat Jun 21 at 12:54

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