Expected Negative Return Under a Zero-Mean Normal Model
Summary
The document distinguishes the average return conditional on a loss from the unconditional contribution of losses to expected return. For a zero-mean normal return with volatility parameter sigma, it identifies the requested conditional quantity as the expectation given that the return is below zero. By symmetry, the negative side has the same magnitude as the mean absolute return, with a negative sign; the half-normal expression therefore gives the conditional average loss in this particular case.
The reply also points toward conditional tail expectation or tail value at risk for broader loss-tail questions. It cautions that a normal model may understate risk relative to empirical return distributions and suggests considering a Student t model, but supplies no comparative test for that recommendation. The calculation relies on zero mean and normality; it should not be generalized unchanged to nonzero means or treated as a complete tail-risk measure without specifying the loss threshold and conditioning event.
Key ideas
- Expected return conditional on a negative return differs from the unconditional expected loss contribution.
- For a zero-mean normal variable, the conditional mean below zero has magnitude sigma times the square root of two over pi.
- The half-normal relationship applies here because of symmetry and the zero-mean assumption.
- Tail conditional expectation is a related measure for losses beyond a specified threshold.
- The document cautions that normality may understate empirical tail risk.
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Full text
# Use Half-Normal to estimate Expected Loss
# Use Half-Normal to estimate Expected Loss
Say a stock return follows a normal distribution with 0% mean and 50% volatility. If I want to calculate the Expected Loss (ie. only the expected value of the negative returns), does it make sense to use Half-Normal Distribution's mean to approximate that value?
The Mean of a Half-Normal is sigma x sqrt( 2 / pi). Does it mean the Expected Loss should be -50% x 0.8 = -40%?
## Answer by ir7 (score 1)
https://quant.stackexchange.com/a/69543
You are looking for the distribution of $X$ conditional on $X<0$, in particular for the conditional expectation
$$E[X|X<0].$$
For $X$ normal with mean $0$ and variance $\sigma^2$, that turns out to be equal to
$$ E[X1_{X<0}]/P(X<0) = 2\sigma\int_{-\infty}^0 x \phi(x) dx $$ $$=2 \sigma \int_{-\infty}^0 -\phi’(x) dx = -\sigma\sqrt{2/\pi}$$
Note that when the mean is $0$, half-normal and one-sided truncated normal distributions coincide. In particular, we have:
$$ |X|= X1_{X>0}-X1_{X<0},$$
leading to
$$E[|X|]= \sigma/\sqrt{2\pi} + \sigma/\sqrt{2\pi} = \sigma\sqrt{2/\pi}. $$
## Answer by Mild_Thornberry (score 0)
https://quant.stackexchange.com/a/69541
You’re looking for a CTE or TVaR value. Check out Wikipedia or the associated references for formulas.
https://en.m.wikipedia.org/wiki/Tail_value_at_risk
http://uryasev.ams.stonybrook.edu/wp-content/uploads/2019/10/Norton2019_CVaR_bPOE.pdf
For the record though, you are probably underestimating your risk if you use a normal distribution. You should try using the Student T with ~4 degrees of freedom. This is more in line with empirical observation. See the introduction to the paper below:
https://www.uts.edu.au/sites/default/files/qfr-archive-02/QFR-rp194.pdfShown 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.