Annualizing Downside Deviation and Estimator Limits

# Annualisation of Downside Deviation


# 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.:

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

## Answer by ZRH (score 1)

https://quant.stackexchange.com/a/44574

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.

## Answer by Attack68 (score 1)

https://quant.stackexchange.com/a/44586

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.