Expression for Covariance- summation over duplicates
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?
covariance
asked Nov 24 at 21:19
ChinG
11411
|
show 1 more comment
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?
covariance
asked Nov 24 at 21:19
ChinG
11411
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 $$frac{1}{binom{n}{2} +n} sumlimits_{i=1}^Nsumlimits_{j=i}^N frac{mathrm{Cov}(A_i,A_j) }{mathrm{Cov}(Y_i,Y_j) }.$$
– Cain
Nov 26 at 22:18
Why does the duplication of entries make it incorrect?
– herb steinberg
Nov 26 at 22:46
@herb It doesn't make it incorrect, but I am simply not considering them.
– ChinG
Nov 27 at 4:42
@herbsteinberg . In your expression, the summation still is over all N for i and j. How will this ignore duplicates?
– ChinG
Nov 27 at 4:42
@ChinG Why ignore duplicates? The only thing to be careful of is the treatment of those terms where i=j.
– herb steinberg
Nov 27 at 4:54
|
show 1 more comment
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?
covariance
asked Nov 24 at 21:19
ChinG
11411
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?
covariance
covariance
asked Nov 24 at 21:19
ChinG
11411
asked Nov 24 at 21:19
ChinG
11411
asked Nov 24 at 21:19
ChinG
11411
asked Nov 24 at 21:19
ChinG
11411
asked Nov 24 at 21:19
ChinG
11411
11411
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 $$frac{1}{binom{n}{2} +n} sumlimits_{i=1}^Nsumlimits_{j=i}^N frac{mathrm{Cov}(A_i,A_j) }{mathrm{Cov}(Y_i,Y_j) }.$$
– Cain
Nov 26 at 22:18
Why does the duplication of entries make it incorrect?
– herb steinberg
Nov 26 at 22:46
@herb It doesn't make it incorrect, but I am simply not considering them.
– ChinG
Nov 27 at 4:42
@herbsteinberg . In your expression, the summation still is over all N for i and j. How will this ignore duplicates?
– ChinG
Nov 27 at 4:42
@ChinG Why ignore duplicates? The only thing to be careful of is the treatment of those terms where i=j.
– herb steinberg
Nov 27 at 4:54
|
show 1 more comment
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 $$frac{1}{binom{n}{2} +n} sumlimits_{i=1}^Nsumlimits_{j=i}^N frac{mathrm{Cov}(A_i,A_j) }{mathrm{Cov}(Y_i,Y_j) }.$$
– Cain
Nov 26 at 22:18
Why does the duplication of entries make it incorrect?
– herb steinberg
Nov 26 at 22:46
@herb It doesn't make it incorrect, but I am simply not considering them.
– ChinG
Nov 27 at 4:42
@herbsteinberg . In your expression, the summation still is over all N for i and j. How will this ignore duplicates?
– ChinG
Nov 27 at 4:42
@ChinG Why ignore duplicates? The only thing to be careful of is the treatment of those terms where i=j.
– herb steinberg
Nov 27 at 4:54
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 $$frac{1}{binom{n}{2} +n} sumlimits_{i=1}^Nsumlimits_{j=i}^N frac{mathrm{Cov}(A_i,A_j) }{mathrm{Cov}(Y_i,Y_j) }.$$
– Cain
Nov 26 at 22:18
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 $$frac{1}{binom{n}{2} +n} sumlimits_{i=1}^Nsumlimits_{j=i}^N frac{mathrm{Cov}(A_i,A_j) }{mathrm{Cov}(Y_i,Y_j) }.$$
– Cain
Nov 26 at 22:18
Why does the duplication of entries make it incorrect?
– herb steinberg
Nov 26 at 22:46
Why does the duplication of entries make it incorrect?
– herb steinberg
Nov 26 at 22:46
@herb It doesn't make it incorrect, but I am simply not considering them.
– ChinG
Nov 27 at 4:42
@herb It doesn't make it incorrect, but I am simply not considering them.
– ChinG
Nov 27 at 4:42
@herbsteinberg . In your expression, the summation still is over all N for i and j. How will this ignore duplicates?
– ChinG
Nov 27 at 4:42
@herbsteinberg . In your expression, the summation still is over all N for i and j. How will this ignore duplicates?
– ChinG
Nov 27 at 4:42
@ChinG Why ignore duplicates? The only thing to be careful of is the treatment of those terms where i=j.
– herb steinberg
Nov 27 at 4:54
@ChinG Why ignore duplicates? The only thing to be careful of is the treatment of those terms where i=j.
– herb steinberg
Nov 27 at 4:54
|
show 1 more comment
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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 $$frac{1}{binom{n}{2} +n} sumlimits_{i=1}^Nsumlimits_{j=i}^N frac{mathrm{Cov}(A_i,A_j) }{mathrm{Cov}(Y_i,Y_j) }.$$
– Cain
Nov 26 at 22:18
Why does the duplication of entries make it incorrect?
– herb steinberg
Nov 26 at 22:46
@herb It doesn't make it incorrect, but I am simply not considering them.
– ChinG
Nov 27 at 4:42
@herbsteinberg . In your expression, the summation still is over all N for i and j. How will this ignore duplicates?
– ChinG
Nov 27 at 4:42
@ChinG Why ignore duplicates? The only thing to be careful of is the treatment of those terms where i=j.
– herb steinberg
Nov 27 at 4:54