Half life of Exponetial Weighted Moving Average

I am trying to apply a volatility strategy. I am reading a paper where the authors defined the volatility as: "Exponential Weighted Volatility of returns with a 1-year window and 3-month half-life"

I am having a hard time understanding the mathematical formula underlying it. The 1-year window part is easily understood as a summation of weighted square return deviation up to 12 months back. I think that the 3-month half-life is used for the weights but cannot figure out the exact mathematical representation. Any help on this is appreciated.

The Exponentially Weighted Moving Average (EWMA for short) is characterized my the size of the lookback window $$N$$ and the decay parameter $$\lambda$$.

The corresponding volatility forecast is then given by: $$\sigma_t^2 = \sum_{k = 0}^N \lambda^k x_{t-k}^2$$

Sometimes the above expression is normed such that the sum of the weights is equal to one. However, for large $$N$$ this makes no difference.

Coming to your question, instead of providing $$\lambda$$ the half-life $$\tau$$ can be provided as well. The half-life is the time lag at which the exponential weights decay by one half, i.e. $$\lambda^\tau = \frac 12 \iff \tau = - \frac{\ln2}{\ln \lambda} \iff \lambda = \biggl(\frac 12\biggr)^{\frac 1\tau}.$$

In your case $$\tau = \frac 14$$ which means that after 3 months the weights in the EWMA are less or equal than $$\frac 12$$. The corresponding value for $$\lambda$$ is then given by $$\lambda = \bigl(\frac 12\bigr)^{\frac 1\tau} = \frac {1}{16}$$.

• Thank you, i have a question regarding that 2 in the half life formula, is it in 'years' unit? What i mean to say is, do i use half life = 0.25 or 3? – Dhruv Mahajan Jun 19 '19 at 12:12
• Usually it is measured in years, I updated my answer. – Cettt Jun 19 '19 at 12:14
• @Cettt: Nice answer but two things, neither of which are a big deal. 1) the EMA is usually defined to go all the way back to the beginning of the data so that $N = \infty$. 2) If the OP wants to estimate the volatility of returns, and the mean is not assumed to be zero, then some mean estimate should be subtracted off of $x_{t}^2$. This adds complications ( what estimate should one use ) so hopefully it can be assumed to be zero. – mark leeds Jun 21 '19 at 19:31