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Value at Risk Under Increasing Transformations

Article Quant Q&A · Author: Wombat
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Summary

The document asks whether Value at Risk commutes with an increasing continuous transformation of a random variable. The proposed identity is that VaR of the transformed variable equals the transformation applied to the original VaR. The response sketches a proof using a quantile definition of VaR: express the quantile through a probability threshold, apply the inverse transformation to the event, and compare the resulting threshold with the original quantile.

The argument is concise and relies on conditions that deserve care. The response invokes a left-continuity condition and uses an inverse, so strict monotonicity and the existence of an appropriate inverse matter; increasing alone can include flat regions. Quantile definitions also vary at distributions with atoms. The document therefore offers an intuitive proof route, but its assumptions should be checked against the precise VaR convention and transformation in a given risk analysis.

Key ideas

  • The proposed identity maps a quantile through an increasing transformation.
  • The proof strategy rewrites the transformed VaR event using the inverse function.
  • The argument depends on suitable continuity and inverse conditions for the transformation.
  • Quantile conventions and distributions with atoms can affect whether the identity applies as stated.

Tags

Risk managementStatistics
Full text
# Value at Risk under increasing function


# 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!

## Answer by Magic is in the chain (score 1, accepted)

https://quant.stackexchange.com/a/55100

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:

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

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

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.