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Var(X) = Cov(X,X) ? 
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#1
Jan2413, 12:52 AM

P: 1,402

[tex]Var(X)=\sum_{i=1}^N P(X_i)(X_iEX)^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} ((12)^2 + (22)^2 + (32)^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}((12)^2+(32)^2+2(12)(32)) = 2  2 = 0??[/tex] There are 9 values of (X,Y); each occurs with equal probability. I've omitted the terms that contain (22) from the summation. Apparently I've misunderstood something about the definition of covariance, but what? 


#2
Jan2413, 03:56 AM

Mentor
P: 18,279

Here is how you calculate it. By definition, the covariance is [tex]Cov(X,X)=E[ (XEX)(XEX) ][/tex] So define the random variable [itex]Z=(XEX)(XEX)=(X2)^2[/itex]. The covariance is EZ. Now, if X takes on the values 1,2 and 3. Then Z takes on the values 0,1. Furthermore [itex]P\{Z=0\}=P\{X=2\}=1/3[/itex] and [itex]P\{Z=1\}=P(\{X=1\}\cup \{X=3\}) = 2/3[/itex]. Thus [tex]Cov(X,X)=EZ = \sum_{k=0}^1 k P\{Z=k\} = 2/3[/tex] 


#3
Jan2413, 04:07 AM

P: 4,573

Hey Rasalhague.
I don't know what you did, but Ill use the expanded form of covariance in your definition. Cov(X,X) = E[(X  E[X])(X  E[X])] = E[X^2]  E[X]^2. You are not applying the expectation operator correctly since you are need to apply the definition of the expectation to the whole definition (i.e (XE[X])(XE[X) and this means taking into account shifts by the mean. If you expand the Covariance operator you get: Cov(X,Y) = E[XY]  E[X]E[Y] and this is done using some simple algebra which leaves us with Cov(X,X) = E[X^2]  E[X]^2 which is the same as the variance. You are not calculating the variance or covariance but something that I have absolutely no idea with. 


#4
Jan2413, 08:40 AM

P: 1,402

Var(X) = Cov(X,X) ?
The formula defines covariance for discrete variables in Simon & Blume (1994): Mathematics for Economists, end of section A5.4, and in Robert J. Serfling's online intro 'Covariance and Correlation', formula (1) which he identifies with E[(XEX)(YEY)P(X,Y)] in the formula which follows that. Serfling also states that P(X,Y) means
[tex]P_X(X)\cdot P_Y(Y)[/tex] which in my example makes P(X,X) = (1/3)(1/3) = 1/9. Perhaps you could explain how you would calculate an example where [itex]X \neq Y[/itex], e.g. X = (1, 2, 3) and Y = (1, 4, 9). 


#5
Jan2413, 08:55 AM

P: 1,402

I'm not sure how to reconcile Serfling's formula (1) with the way Wolfram Mathworld writes it out explicitly for the case where N = M:
http://mathworld.wolfram.com/Covariance.html Are there two somewhat different concepts each called covariance, each corresponding to its own way of defining the mean of the product of two random variables? 


#6
Jan2413, 11:32 AM

P: 1,402

Ah, reading further on that Mathworld article, it seems one definition concerns realvalued random variables from a finite sample space, another definition concerns tuples of such random variables. But still, there appear to be a variety of concepts here to which the name covariance is attached, with disagreement over certain points, and Mathworld doesn't give an explicit version of the more general definition.



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