- #1
Rasalhague
- 1,387
- 2
Var(X) = Cov(X,X) ??
[tex]Var(X)=\sum_{i=1}^N P(X_i)(X_i-EX)^2.[/tex]
[tex]Cov(X,Y) = \sum_{i=1}^N\sum_{j=1}^M P(X_i,Y_j)(X_i - EX)(Y_j - EY).[/tex]
If, for instance, [itex]P(X_i) = 1/N[/itex] and [itex]X = Y = (1,2,3)[/itex], then
[tex]Var(X) = \frac{1}{3} ((1-2)^2 + (2-2)^2 + (3-2)^2) = \frac{2}{3},[/tex]
but
[tex]Cov(X,X) = \sum_{i=1}^3 \sum_{j=1}^3 \frac{1}{9} (X_i - EX)(X_j - EX)[/tex]
[tex]=\frac{1}{9}((1-2)^2+(3-2)^2+2(1-2)(3-2)) = 2 - 2 = 0??[/tex]
There are 9 values of (X,Y); each occurs with equal probability. I've omitted the terms that contain (2-2) from the summation. Apparently I've misunderstood something about the definition of covariance, but what?
[tex]Var(X)=\sum_{i=1}^N P(X_i)(X_i-EX)^2.[/tex]
[tex]Cov(X,Y) = \sum_{i=1}^N\sum_{j=1}^M P(X_i,Y_j)(X_i - EX)(Y_j - EY).[/tex]
If, for instance, [itex]P(X_i) = 1/N[/itex] and [itex]X = Y = (1,2,3)[/itex], then
[tex]Var(X) = \frac{1}{3} ((1-2)^2 + (2-2)^2 + (3-2)^2) = \frac{2}{3},[/tex]
but
[tex]Cov(X,X) = \sum_{i=1}^3 \sum_{j=1}^3 \frac{1}{9} (X_i - EX)(X_j - EX)[/tex]
[tex]=\frac{1}{9}((1-2)^2+(3-2)^2+2(1-2)(3-2)) = 2 - 2 = 0??[/tex]
There are 9 values of (X,Y); each occurs with equal probability. I've omitted the terms that contain (2-2) from the summation. Apparently I've misunderstood something about the definition of covariance, but what?