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Suppose I have a random variable: $$ Y_{it}=A_{it}+B_{it} $$ which is made up of two random variables $A$ and $B.$ $i$ is the unit of observation that runs through $i=1....N$ for each time period $t=1...T$. Now taking the covariance of $Y$ for any two units $i$ and $j$ over time, we have: $$ cov(Y_{i},Y_{j})=cov(A_{i},A_{j})+cov(B_{i},B_{j})+cov(A_{i},B_{j})+cov(A_{j},B_{i}) $$ Now, I want to figure out the average contribution of the term $cov(A_{i},A_{j})$ to the term $cov(Y_{i},Y_{J}).$ What I have is $$ C=\frac{1}{N^2}\sum_{i=1}^{N}\sum_{j=1}^{N}\frac{cov(A_{i},A_{j})}{cov(Y_{i},Y_{j})} $$

Notice that the variance-covariance matrix is symmetric. As such, there are ionly $N/(N-1)/2$ unique entries ($cov(A_{1},A_{_2}$)=$cov(A_{2},A_{1})$. I know that the expression above is incorrect (as it generates duplicates). How can I modify the above expression to correct notation such that it contains only a unique occurence of symmetric covariances?

ChinG
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  • Assuming that the only duplicates are of the form $\mathrm{Cov}(A_i,A_j) = \mathrm{Cov}(A_j,A_2)$ you could simply rewrite the summation as $$\frac{1}{\binom{n}{2} +n} \sum\limits_{i=1}^N\sum\limits_{j=i}^N \frac{\mathrm{Cov}(A_i,A_j) }{\mathrm{Cov}(Y_i,Y_j) }.$$ – Cain Nov 26 '18 at 22:18
  • Why does the duplication of entries make it incorrect? – herb steinberg Nov 26 '18 at 22:46
  • @herb It doesn't make it incorrect, but I am simply not considering them. – ChinG Nov 27 '18 at 04:42
  • @herbsteinberg . In your expression, the summation still is over all N for i and j. How will this ignore duplicates? – ChinG Nov 27 '18 at 04:42
  • @ChinG Why ignore duplicates? The only thing to be careful of is the treatment of those terms where i=j. – herb steinberg Nov 27 '18 at 04:54
  • $Cov(A_i,A_i)$ has to be treated differently from $Cov(A_i,A_j)$ where $i\ne j$. To get the average, you need to divide by $N$ for the terms where $i=j$ and divide by $N(N-1)$ for the others. Combining these averages is dependent on what you mean by average. – herb steinberg Nov 27 '18 at 17:12

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