# Value at Risk under increasing function

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!

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

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:

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

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

• Ah, I see. Thank you so much! :-) Jun 21 '20 at 12:54