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!


1 Answer 1


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:


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

  • $\begingroup$ Ah, I see. Thank you so much! :-) $\endgroup$
    – Wombat
    Commented Jun 21, 2020 at 12:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.