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I am attempting to build a Walsh-Fourier spectral density and it appears that it is first required to compute the logical covariance which in turn involves a dyadic add.

I am not at all familiar with dyadics or their operations and introductory references are hard to come by. In Stoffer (1988) the logical covariance of a categorical series $X(0), X(1),\dots,X(N-1)$ is described as being: \begin{align} \tau(j)=N^{-1} \sum_{j=0}^{N-1} \gamma(j\oplus k-k) \end{align} where $j\oplus k$ is the dyadic addition. $\gamma$, is our usual autocovariance, $\gamma(h)=cov\{X(n), X(n+h)\}$.

The Walsh-Fourier spectral density is then:

\begin{align} f(\lambda)=\sum_{j=0}^{\infty}\tau(j)W(j, \lambda) \end{align}

where $W(j, \lambda)$ is the $j$th sequency (zero-crossings) with $0\leq\lambda < 1$.

I'm sure a HMM would be great for categorical times series but at the moment I am restricted to spectral analysis so I must continue with this approach. It's not exactly homework. It's a final project that has moved a little beyond the coursework. The professor is a little hard to get a hold of and thus the question is posed here.

Is it just addition? I haven't even been able to confirm that.

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After digging about and reading various papers I discovered that the definition is referenced as given in a paper by R.Kohn in 1980, On the Spectral Decomposition of Stationary Time Series Using Walsh Functions. I

In short there are two definitions depending upon whether or not the values being added are integers or reals. Luckily I am only concerned with the integer case which is a little simpler.

Kohn defines the $\oplus$ operation on integers as follows:

Let $m,n$ be integers and $m_j,n_j$ be restricted to 0 or 1, then

\begin{align} m=\sum_{j=0}^fm_j2^j \hspace{2mm} \text{and} \hspace{2mm} n=\sum_{j=0}^fn_j2^j \end{align}

the dyadic sum then is

\begin{align} m\oplus n=\sum_{j=0}^f|m_j-n_j|2^j \end{align}

He gives as an example $5\oplus 3=6$. While $f$ isn't defined it appears to mean the 'useful' domain of the function because

\begin{align} 5=2^0*1+2^1*0+2^2*1+\sum_{j=3}^{\infty}0*2^j\hspace{2mm} \text{and} \hspace{2mm} 3=2^0*1+2^1*1+\sum_{j=2}^{\infty}0*2^j \end{align}

Then

\begin{align} 5\oplus 3=2^0*|1-1|+2^1*|0-1|+2^2*|1-0|+\sum_{j=3}^{\infty}|0-0|*2^j=6 \end{align}

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so basically XOR –  MAB Nov 22 '12 at 3:22
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