Deriving CVaR as the Average Loss Beyond VaR
Summary
The note clarifies the integral form of Conditional Value at Risk by expressing it as a conditional expectation over outcomes in the tail. Integrating the outcomes below the relevant return threshold gives their probability-weighted total; dividing by the tail probability, represented by the confidence level under the stated convention, produces the average tail outcome. This explains why the integral includes a probability normalization factor.
A key caveat is sign convention. If VaR is defined as a positive loss threshold while the variable represents returns, the cutoff and the sign of the conditional expectation must be adjusted consistently. The responses illustrate the normalization using an indicator-function expression and offer a discrete weighted-average interpretation, where observations outside the tail receive zero weight. These explanations address the formula conceptually but do not cover complications such as distributions with probability mass at the threshold or alternative CVaR definitions.
Key ideas
- CVaR can be written as the conditional average of outcomes in the selected tail.
- The tail integral must be divided by the probability of entering that tail.
- VaR and CVaR signs depend on whether the variable represents returns or losses.
- A weighted-sum view assigns zero weight to observations outside the tail.
Tags
Full text
# CVaR reformulation correct?
# CVaR reformulation correct?
Conditional Value at Risk (CVaR) is given as: $$CVaR_\alpha(X)=\frac{1}{\alpha}\int_{0}^{\alpha}VaR_\beta(X)d\beta=-E(X|X\leq-VaR_\alpha(X))=-\frac{1}{\alpha}\int_{-\infty}^{-VaR_\alpha(X)}x \cdot f(x)\,dx$$
I am not sure if the last term is correct regarding multiplication with $1/\alpha$?
The average is already only up to $VaR_\alpha$.
## Answer by vanguard2k (score 3, accepted)
https://quant.stackexchange.com/a/15455
It is correct!
You can also see it this way:
$$ \text{CVaR}_\alpha(X)=\mathbb{E}(X|X\leq \text{VaR}_\alpha(X)) = \frac{\int_{\mathbb{R}} x\cdot 1_{X\leq \text{VaR}_\alpha(X)}dF(x)}{\int_\mathbb{R}1_{X\leq \text{VaR}_\alpha(X)}dF(x)} = \frac{1}{\alpha} \int_{-\infty}^{\text{VaR}_\alpha(X)}xdF(x) $$
The sign problem still remains (in both versions). If you define $\text{VaR}_\alpha (X) = - F_X^{-1}(\alpha)$ then you probably want to define $\text{CVaR}_\alpha(X) = - \mathbb{E}(X|X\leq -\text{VaR}_\alpha(X))$ you will get your result.
## Answer by John (score 1)
https://quant.stackexchange.com/a/15457
I also have been puzzled by the intuition of this formula in the past.
What made sense to me was converting the integral to a summation. You can then do the calculation quite simply in an excel document with some simulated data and see where everything is coming from.
CVaR is really just calculating the average return given that it is less than a certain amount. The calculation is like a weighted average where every observation above VaR has a zero weight. You do the sum and then you divide by the weighted divisor. But since we know that the VaR is at a particular confidence, the divisor should equal that confidence every time.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.