# Can Value at Risk and Expected Shortfall be equal?

I am struggling with the study of value at risk (VaR) and expected shortfall (ES). In particular, I am looking for some cases in which these two measures can be equal. I know that, given a loss cdf $$F_L(x)$$ VaR is defined as $$\inf\{x\in\mathbb{R}:F_L(x)\geq\alpha\}$$, for $$\alpha\in(0,1)$$. ES instead is $$(1-\alpha)^{-1}\int_{\alpha}^{1}\text{VaR}_L(u)du$$ for $$\alpha\in0$$.

Is there a distribution for which VaR and ES are equal? If yes, how can I prove it mathematically? I was wondering about a function with jumps but i'm not sure about that. Thanks for your help.

• Thanks for your answer. Can you give me an example for a particular quantile? I mean, for which distribution and quantile this should be possible? One case is sufficient, i just want to understand the reasoning behind that. Apr 30, 2021 at 9:48

Let us call the random variable of interest $$X$$. To make VaR and ES equal for a quantile level $$\alpha$$, we need $$q_\alpha=q_{\alpha'}$$ for all $$\alpha'<\alpha$$. (Otherwise, $$\text{ES}_\alpha<\text{VaR}_\alpha$$.) That implies the distribution of the random variable must have a point mass $$m\geq\alpha$$ at the minimum value of $$X$$: $$m:=P(\min(X))\geq\alpha$$. An example would be a Bernoulli distribution with parameter $$p$$ and any $$\alpha\leq1-p$$.
To make VaR and ES equal for all quantile levels $$\alpha$$, the point mass must be unity, turning the random variable into a constant.
• You have been very clear about discrete distributions. For continuous ones, I can only think of possible limiting cases (when $\alpha\rightarrow1^-$ for instance) but I am not sure if I am right, I will definitely look into this further. Thank you very much for the help. Apr 30, 2021 at 16:06