Is Variance monotonious

[sapozan]

What is the best way to prove that Variance is monotonious as a risk measure?

 

[Detective]

Can you give us the definition of monotonicity you are using? If you can share your current proof, then I can comment on the validity and soundness.

I believe the proof would involve using VAR(X) = E(X^2) - E(X)^2.

 

[David Harper CFA FRM]

@sapozan I assume you refer to the coherence property of monotonicity? There is prior conversation here https://forum.bionicturtle.com/threads/coherent-risk-measure-monotonicity.8214/#post-33126

But do we think (or know) that variance (or standard deviation) satisfies this monotonicity? Admittedly, it's always vexing me, but intuitively, I'm just not sure ... Compare two portfolios, both with normally distributed returns:

1. Exp return = +15% with standard deviation of 3.0%; ie., N(0.15, 0.03^2), versus portfolio with less risk per variance but far less expected return:
2. Exp return = +4% with standard deviation of 2.0%; ie., N(0.04, 0.02^2)

The first portfolio almost exclusively dominates the second in terms of expected future value, but as a risk measure, variance says it is more risky. If variance is monotonic, then the higher variance of #1 suggests that more cash must be added to make its risk acceptable, but it appears that less cash must be added .... is #1 riskier? it doesn't seem so!

I have the view (admittedly not certain) that relative VaR is not monotonic, and variance is just a special case of relative VaR so I tend to think the same of variance, however absolute VaR (aka, VaR) might satisfy monotonicity; e.g.,

So this monotonicity (IMO) is saying: if the future value of portfolio (Y) is greater than the future value of portfolio (X), then ceteris paribus, a monotonic risk measure will be lower Y than X; i.e., it will say that X is riskier. Notice that absolute VaR does this: if absolute VaR = -μ + ασ, then for unchanged volatility (σ), higher return lowers the risk (absolute VaR).

 

[sapozan]

@Detective, Thank you very much.

I have tried to assume that rvs X>Y and therefore Z=X-Y>0

It follows that Var(Z)=\[ {sigma^2}_x-2sigma_{x,y}+{sigma^2}_y>0 ].

It looks OK to me, but I am not 100% sure

Could you share your thoughts please?

 

[sapozan]

@David Harper CFA FRM , thank you very much.

This is confirming my uncertainty.

There is plenty of papers assuming that SD and Variance are monotonous and some give some analytical proof but none of them actually give a good math proof.

Is it fair to say that it could be monotonous or not for different cases?

It is quite clear it is not coherent so not sure if monotonicity really matters in this case?

Thank you

 

[Detective]

@Detective, Thank you very much.

I have tried to assume that rvs X>Y and therefore Z=X-Y>0

It follows that Var(Z)=\[ {sigma^2}_x-2sigma_{x,y}+{sigma^2}_y>0 ].

It looks OK to me, but I am not 100% sure

Could you share your thoughts please?

OK so your definition is if X > Y everywhere then VAR(X) > VAR(Y)?

Few questions on your proof:

1) Suppose I had made Z = Y - X, then I get same Variance breakdown right?

2) VAR(anything) >= 0 by definition, right?

3) Even if I know VAR(Z) > 0, how do I know/show VAR(X) > VAR(Y)?

 

[sapozan]

Thank you @Detective, I thought the same about Z=Y-X, but unfortunately this was the best I could come up with mathematically.

Analytically one could say that in case the Variance is a measure of risk and X is 'bigger' (riskier) than Y, than Var(X)>Var(Y), But I am not too sure if this is defined enough to prove monotonicity.

Please share your thoughts.

Thank you

 

[Detective]

Thank you @Detective, I thought the same about Z=Y-X, but unfortunately this was the best I could come up with mathematically.

Analytically one could say that in case the Variance is a measure of risk and X is 'bigger' (riskier) than Y, than Var(X)>Var(Y), But I am not too sure if this is defined enough to prove monotonicity.

Please share your thoughts.

Thank you

Is this a math, risk or some hybrid class? Can you share the exact definition of showing risk measure is monotone you were provided with? For example, is it a strictly greater than or greater than or equal to is OK? Does the definition state anything about RV being perfectly correlated? Are constant random variables allowed?

If it is strictly greater than, then consider two random variables X and Y with Y = X-1. X>Y everywhere, but VAR(X) = VAR(X-1) = VAR(Y).

Another example, constant random variable:
X=2
Y=1

X>Y but VAR(X) = VAR(Y) = 0.

 

[sapozan]

Thank you @Detective. This is a Market Risk problems I am trying to tackle and the actual problem was to show that Variance is not a coherent Risk measure however I have tried to check every assumption of coherency for Variance so I can better understand it myseld.

 

[Detective]

Thank you @Detective. This is a Market Risk problems I am trying to tackle and the actual problem was to show that Variance is not a coherent Risk measure however I have tried to check every assumption of coherency for Variance so I can better understand it myseld.

I see. Yes, Variance is not a coherent risk measure. It’s easier to show the failure of the other properties rather than get bogged down with monotonicity in my opinion.

 

[sapozan]

Thank you @Detective, I just tried to prove it for myself, however this seems to be a bit tricky.

Appreciate your help

 

[ami44]

I’m not sure if this is still of interest, but the variance is not monotounius. That is also mentioned in Wikipedia:
https://en.m.wikipedia.org/wiki/Risk_measure#Variance
Davids intuition is correct.

Proof by counter example:
Assume X and Y are returns from two different Portfolios. Let X = c * B with B being a Bernoulli distributed random variable and c > 0 being a constant. In other words the return of portfolio X is +c or -c. The portfolio might for example consist of some kind of digital option.

The Y portfolio shall consist of 2 digital options of the same kind and some fixed income position, so that the return is given by Y = 2 * X + 2 * c. This means, if X is +c than Y is 4 * c and if X is -c than Y is 0.
Clearly X < Y for all possible outcomes. We also know that var(Y) = 4 * var(X) > var(X)
In other words Y dominates X, but the risk of Y measured by the variance would be greater than the risk of X.
this is a contradiction to the monotonicity property.

 

[David Harper CFA FRM]

Thank you @ami44 for the link, interesting.

@sapozan To me, starting with Z = X - Y > 0 does not lead to a proof because if Z = X - Y > 0, then Z' = Y - X < 0, but σ^2(Z) = σ^2(Z') as variance(X-Y) = variance(Y-Z) would would appear to allow for either Y < X or X > Y if ρ(X) < ρ(X). I think? ... but I see that @Detective already made this point above.

I like the simplicity of showing that variance is not translation invariant per the link https://en.m.wikipedia.org/wiki/Risk_measure#Variance
... i had not considered how easy it is: var(X + a) <> VaR(X) - a because VaR(X +a ) = VaR(X). Easy peasy, not coherent!