Explaining $sum_{i=1}^nsum_{j=1}^nCov(Y_i,B_j)=sum_{i=1}^nsum_{j=1,jneq...
$begingroup$
Sorry for the non-specific title, I was unsure of how to word this.
I can't wrap my head around this calculation where they split the summation $$sum_{i=1}^n sum_{j=1}^n operatorname{Cov}(Y_i,B_j) = sum_{i=1}^nsum_{j=1, jneq i}^noperatorname{Cov}(Y_i,B_j)+ sum_{i=1}^n operatorname{Cov}(Y_i,B_i).$$
Can someone explain how this works?
Full calculation:
enter image description here
Nevermind, I think I worked it out. It's because the summation after the + sign is summing over the i-th terms of Y and B which was excluded from the summation the first summation after the = sign, due to j != i
probability statistics summation covariance
edited Dec 17 '18 at 22:44
Blue
48.6k870154
asked Dec 17 '18 at 19:59
Dilsaajan DuggalDilsaajan Duggal
113
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add a comment |
$begingroup$
Sorry for the non-specific title, I was unsure of how to word this.
I can't wrap my head around this calculation where they split the summation $$sum_{i=1}^n sum_{j=1}^n operatorname{Cov}(Y_i,B_j) = sum_{i=1}^nsum_{j=1, jneq i}^noperatorname{Cov}(Y_i,B_j)+ sum_{i=1}^n operatorname{Cov}(Y_i,B_i).$$
Can someone explain how this works?
Full calculation:
enter image description here
Nevermind, I think I worked it out. It's because the summation after the + sign is summing over the i-th terms of Y and B which was excluded from the summation the first summation after the = sign, due to j != i
probability statistics summation covariance
edited Dec 17 '18 at 22:44
Blue
48.6k870154
asked Dec 17 '18 at 19:59
Dilsaajan DuggalDilsaajan Duggal
113
$endgroup$
add a comment |
$begingroup$
Sorry for the non-specific title, I was unsure of how to word this.
I can't wrap my head around this calculation where they split the summation $$sum_{i=1}^n sum_{j=1}^n operatorname{Cov}(Y_i,B_j) = sum_{i=1}^nsum_{j=1, jneq i}^noperatorname{Cov}(Y_i,B_j)+ sum_{i=1}^n operatorname{Cov}(Y_i,B_i).$$
Can someone explain how this works?
Full calculation:
enter image description here
Nevermind, I think I worked it out. It's because the summation after the + sign is summing over the i-th terms of Y and B which was excluded from the summation the first summation after the = sign, due to j != i
probability statistics summation covariance
edited Dec 17 '18 at 22:44
Blue
48.6k870154
asked Dec 17 '18 at 19:59
Dilsaajan DuggalDilsaajan Duggal
113
$endgroup$
Sorry for the non-specific title, I was unsure of how to word this.
I can't wrap my head around this calculation where they split the summation $$sum_{i=1}^n sum_{j=1}^n operatorname{Cov}(Y_i,B_j) = sum_{i=1}^nsum_{j=1, jneq i}^noperatorname{Cov}(Y_i,B_j)+ sum_{i=1}^n operatorname{Cov}(Y_i,B_i).$$
Can someone explain how this works?
Full calculation:
enter image description here
Nevermind, I think I worked it out. It's because the summation after the + sign is summing over the i-th terms of Y and B which was excluded from the summation the first summation after the = sign, due to j != i
probability statistics summation covariance
probability statistics summation covariance
edited Dec 17 '18 at 22:44
Blue
48.6k870154
asked Dec 17 '18 at 19:59
Dilsaajan DuggalDilsaajan Duggal
113
edited Dec 17 '18 at 22:44
Blue
48.6k870154
asked Dec 17 '18 at 19:59
Dilsaajan DuggalDilsaajan Duggal
113
edited Dec 17 '18 at 22:44
Blue
48.6k870154
edited Dec 17 '18 at 22:44
Blue
48.6k870154
edited Dec 17 '18 at 22:44
Blue
48.6k870154
48.6k870154
asked Dec 17 '18 at 19:59
Dilsaajan DuggalDilsaajan Duggal
113
asked Dec 17 '18 at 19:59
Dilsaajan DuggalDilsaajan Duggal
113
asked Dec 17 '18 at 19:59
Dilsaajan DuggalDilsaajan Duggal
113
113
add a comment |
add a comment |
1 Answer
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$begingroup$
$$sum_{i=1}^n sum_{j=1}^n Cov(Yi,Bj) = sum_{i=1}^n (sum_{j=1, j!=i}^n Cov(Yi,Bj) + sum_{j=i}^nCov(Yi,Bj)space)$$
$$= sum_{i=1}^nsum_{j=1, j!=i}^n Cov(Yi,Bj) + sum_{i=1}^nsum_{j=i}^n Cov(Yi,Bj)$$
Second sum in the second term is only defined once at $space j=i$. Therefore, we can substitute $i$ for $j$ and eliminate the second summation.
$$ = sum_{i=1}^nsum_{j=1, j!=i}^n Cov(Yi,Bj)+ sum_{i=1}^n Cov(Yi,Bi)$$
answered Dec 17 '18 at 20:17
EDZEDZ
46418
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1 Answer
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1 Answer
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$begingroup$
$$sum_{i=1}^n sum_{j=1}^n Cov(Yi,Bj) = sum_{i=1}^n (sum_{j=1, j!=i}^n Cov(Yi,Bj) + sum_{j=i}^nCov(Yi,Bj)space)$$
$$= sum_{i=1}^nsum_{j=1, j!=i}^n Cov(Yi,Bj) + sum_{i=1}^nsum_{j=i}^n Cov(Yi,Bj)$$
Second sum in the second term is only defined once at $space j=i$. Therefore, we can substitute $i$ for $j$ and eliminate the second summation.
$$ = sum_{i=1}^nsum_{j=1, j!=i}^n Cov(Yi,Bj)+ sum_{i=1}^n Cov(Yi,Bi)$$
answered Dec 17 '18 at 20:17
EDZEDZ
46418
$endgroup$
add a comment |
$begingroup$
$$sum_{i=1}^n sum_{j=1}^n Cov(Yi,Bj) = sum_{i=1}^n (sum_{j=1, j!=i}^n Cov(Yi,Bj) + sum_{j=i}^nCov(Yi,Bj)space)$$
$$= sum_{i=1}^nsum_{j=1, j!=i}^n Cov(Yi,Bj) + sum_{i=1}^nsum_{j=i}^n Cov(Yi,Bj)$$
Second sum in the second term is only defined once at $space j=i$. Therefore, we can substitute $i$ for $j$ and eliminate the second summation.
$$ = sum_{i=1}^nsum_{j=1, j!=i}^n Cov(Yi,Bj)+ sum_{i=1}^n Cov(Yi,Bi)$$
answered Dec 17 '18 at 20:17
EDZEDZ
46418
$endgroup$
add a comment |
$begingroup$
$$sum_{i=1}^n sum_{j=1}^n Cov(Yi,Bj) = sum_{i=1}^n (sum_{j=1, j!=i}^n Cov(Yi,Bj) + sum_{j=i}^nCov(Yi,Bj)space)$$
$$= sum_{i=1}^nsum_{j=1, j!=i}^n Cov(Yi,Bj) + sum_{i=1}^nsum_{j=i}^n Cov(Yi,Bj)$$
Second sum in the second term is only defined once at $space j=i$. Therefore, we can substitute $i$ for $j$ and eliminate the second summation.
$$ = sum_{i=1}^nsum_{j=1, j!=i}^n Cov(Yi,Bj)+ sum_{i=1}^n Cov(Yi,Bi)$$
answered Dec 17 '18 at 20:17
EDZEDZ
46418
$endgroup$
$$sum_{i=1}^n sum_{j=1}^n Cov(Yi,Bj) = sum_{i=1}^n (sum_{j=1, j!=i}^n Cov(Yi,Bj) + sum_{j=i}^nCov(Yi,Bj)space)$$
$$= sum_{i=1}^nsum_{j=1, j!=i}^n Cov(Yi,Bj) + sum_{i=1}^nsum_{j=i}^n Cov(Yi,Bj)$$
Second sum in the second term is only defined once at $space j=i$. Therefore, we can substitute $i$ for $j$ and eliminate the second summation.
$$ = sum_{i=1}^nsum_{j=1, j!=i}^n Cov(Yi,Bj)+ sum_{i=1}^n Cov(Yi,Bi)$$
answered Dec 17 '18 at 20:17
EDZEDZ
46418
answered Dec 17 '18 at 20:17
EDZEDZ
46418
answered Dec 17 '18 at 20:17
EDZEDZ
46418
answered Dec 17 '18 at 20:17
EDZEDZ
46418
46418
add a comment |
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