# Annualisation of Downside Deviation

Is it possible to annualise the downside deviation? If so, on the basis of what theory?

The downside deviation (DD) of a series of daily returns is computed according to the formula:

$$\text{DD} = \sqrt{\frac{1}{{T}}\sum_{t=1}^{T}(\text{min}(\text{ret}_t,\text{thr}))^2}$$

where T is the number of daily observations and thr is a threshold, for instance 0 or the average of the returns.

Other definitions are also used, according to what we want to be represented by the formula. E.g.:

$$\text{DD} = \sqrt{\frac{1}{{T}}\sum_{t=1}^{T}(\text{min}((\text{ret}_t - \text{thr}),0))^2}$$

DD is basically just the average of square returns conditional on returns being smaller than $$thr$$. If $$\rho(\xi)$$ is the distribution of returns, then the continous equivalent of your formula is:

$$E[\mathit{DD}]=\sqrt{\int_{-\infty}^{thr}\xi^2\rho(\xi)d\xi}$$

So basically as long as you assume $$\rho(\xi)$$ is stationary, $$E[\mathit{DD}]$$ will not be a function of observation time.

• There is a concern about the 'quality' of the estimator given sample data. Of course you are right, but your formula relies on knowledge of the pdf a priori which generally you won't have, so the question becomes if you use the estimator given by the OP does it have a bias dependent on T, what is its variance and/or is it a consistent estimator? I haven't tried but this might be quite difficult to calculate for this particular estimator...
– Attack68
Mar 15 '19 at 5:24
• @Attack68: not quite sure about that, but not having a proof..
– ZRH
Mar 15 '19 at 8:17
• @FrancescoVe: I found a definition of DD which actually differs from what you state above: $\mathit{DD}=\sqrt{\frac{1}{n}\sum_{r_i<c}^n (r_i-c)^2}$
– ZRH
Mar 15 '19 at 8:19
• yes, that is correct
– ZRH
May 23 '20 at 9:09
• thats exactly correct
– ZRH
May 23 '20 at 12:42

If you make the assumption that your returns are iid normally distributed $$R_i \sim \mathcal{N}(0, 1)$$, then with the second definition, and using @ZRH provided formula...

$$E[DD] = \sqrt{\int_{-\infty}^{c}x^2\frac{1}{\sqrt{2 \pi}}exp(-\frac{x^2}{2}) dx }$$

So expanding this out (integration by parts) you get;

$$E[DD] = \sqrt{ \left [ - x \frac{1}{\sqrt{2 \pi}} exp(-\frac{x^2}{2}) \right ]_{-\infty}^c + \int_{-\infty}^{c} \frac{1}{\sqrt{2 \pi}}exp(-\frac{x^2}{2})}$$

which is (where $$\Phi(c)$$ is the standard normal cumulative distribution function),

$$E[DD] = \sqrt{ -\frac{c}{\sqrt{2 \pi}} exp(-\frac{c^2}{2}) + \Phi(c) } \;.$$

Now you are proposing an estimator, $$\theta(T)$$, for this value based on the second definition. You should be concerned about the bias;

$$Bias(\theta) = E[\theta - E[DD]]$$ $$Bias(\theta) = E \left [\sqrt{\frac{1}{T}\sum_i^T \min(0, x_i-c)^2} \right ] - E[DD]$$

The reason I mention this is because presumably an annualisation will involve some multiplication of the result, and if the bias is not zero you will be amplifying this sample error, and if there is a large variance in this estimator then by annualising, again, you will be amplifying a potentially poor estimator.

I could not expand out the above due to lack to either lack of time or ability (it was not clear to me which it was) but even in this simple case of standard normal it is not clear to me what to do. This might not necessarily have been that bad: the bias might have been zero and the variance small.