Question on effective days of an exponentially weighted moving average model

Question

I have been reading the book "RiskMetrics —Technical Document" by Longerstaey (J.P.Morgan) and Spencer (Reuters) (4th Edition, 1996). I am wondering what the effective days of the exponentially weighted moving average model (EWMA) mentioned in the book would be if I use a 2-day percentage change as also explained below.

On pages 93-95 of the book, it is stated that $K$, the effective days of the EWMA, can be calculated by: $$ K = \frac{\log(\alpha)}{\log(\lambda)}.$$

For example, if one is interested in a confidence level of $\alpha=1\%$ with $\lambda=0.99$, $K$ comes out as 458 days, as also shown in the table on page 94.

On page 93, it is also stated that the formula above for $K$ is what one gets if the following equation is solved for $K$:

$$\lambda^{K}(1-\lambda)(1+\lambda+\lambda^2+\dots)=\alpha.$$

I have proven this as follows:

\begin{align*} \lambda^{K}(1-\lambda)(1+\lambda+\lambda^2+\dots)&=\alpha \\ \lambda^{K}&=\frac{ \alpha}{(1-\lambda)(1+\lambda+\lambda^2+\dots)} \\ \log( \lambda^{K})&=\log\left(\frac{ \alpha}{(1-\lambda)(1+\lambda+\lambda^2+\dots)} \right) \end{align*}

Because $1+\lambda+\lambda^2+\dots$ is a geometric series and $|\lambda|<1$, it converges to $\frac{1}{1-\lambda}$. Then,

\begin{align*} K\log( \lambda) &= \log\left(\frac{ \alpha}{(1-\lambda)(\frac{1}{1-\lambda})} \right)\\ K\log( \lambda) &= \log(\alpha)\\ K &= \frac{\log(\alpha)}{\log( \lambda)}. \\ \end{align*}

My questions are:

1. What happens if I use a 2-day percentage change such as $r_{t} = \frac{y_{t-2}}{y_t}-1$?
2. With $r_{t}$ defined just as above, how would the effective number of days change when, for example, $\lambda=0.99$ and $\alpha=1\%$? Or would it remain the same at 458, and why?

Answer 1

The EWMA is a weighting scheme. Calculating the EWMA $Z$ of some input signal $X$, the EWMA is defined as:

$$ Z_t\equiv (1-\lambda)\sum_{t=1}^T \lambda^{t-1}X_t $$

At closer inspection, we find that, usually, $T<<\infty$, and hence the weights do not sum to 1:

$$ (1-\lambda)\sum_{t=1}^T\lambda^{t-1}=(1-\lambda)\frac{1-\lambda^T}{1-\lambda}=1-\lambda^T<1 $$

If you'd simply replace the leading $1-\lambda$ with $\frac{1-\lambda}{1-\lambda^T}$, there would be no need for a 'tolerance level'. It has nothing to do with a confidence level (in the statistical sense).

Regarding your question: It does not matter whether you use daily or 2-day returns. Instead of $N$ daily returns, you now use $N$ returns, looking back for a longer time period.

HTH?

Answer 2

Define the continuous-time analogue of the discrete time EWMA: $ \quad e^{-H \tau} := \lambda^M$

Therefore $H = -252 ln (\lambda)$, assuming $M=252$ (number of daily observations in $\tau = 1yr$)

We wish to calculate the effective window length, define as $L = \frac{ \int_0^T \tau . e^{-H \tau} d \tau } { \int_0^T e^{-H \tau} d \tau }$

Integrating top and bottom we find $L = \frac{ \frac{1}{H^2} \Big [ 1 - e^{-H T}(1 + H T) \Big ] } { \frac{1}{H} \Big [ 1 - e^{-H T} \Big] }$

As $T \rightarrow \infty$, we have $L \rightarrow \frac{ \frac{1}{H^2} \Big [ 1 - 0 \Big ] } { \frac{1}{H} \Big [ 1 - 0 \Big] } = \frac{1}{H}$

So given $\lambda$ (which is typically quoted as a daily-discrete observation, you find that the effective window length is $\frac{1}{-252 ln(\lambda)}$.

Obviously if your $\lambda$ is monthly, then this becomes $L= \frac{1}{-12 ln(\lambda)}$, etc

Aside: the nice thing about the continuous-time EWMA is it can be applied $\textit{without adjustment}$ to observations which are asynchronous. i.e. intraday measures where the return between adjacent spot observations include close-to-open as well as intraday ones.