# Var(X) = Cov(X,X) ?

1. Jan 24, 2013

### Rasalhague

Var(X) = Cov(X,X) ??

$$Var(X)=\sum_{i=1}^N P(X_i)(X_i-EX)^2.$$

$$Cov(X,Y) = \sum_{i=1}^N\sum_{j=1}^M P(X_i,Y_j)(X_i - EX)(Y_j - EY).$$

If, for instance, $P(X_i) = 1/N$ and $X = Y = (1,2,3)$, then

$$Var(X) = \frac{1}{3} ((1-2)^2 + (2-2)^2 + (3-2)^2) = \frac{2}{3},$$

but

$$Cov(X,X) = \sum_{i=1}^3 \sum_{j=1}^3 \frac{1}{9} (X_i - EX)(X_j - EX)$$

$$=\frac{1}{9}((1-2)^2+(3-2)^2+2(1-2)(3-2)) = 2 - 2 = 0??$$

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?

2. Jan 24, 2013

### micromass

Staff Emeritus
Re: Var(X) = Cov(X,X) ??

This formula is wrong.

Here is how you calculate it. By definition, the covariance is

$$Cov(X,X)=E[ (X-EX)(X-EX) ]$$

So define the random variable $Z=(X-EX)(X-EX)=(X-2)^2$. The covariance is EZ. Now, if X takes on the values 1,2 and 3. Then Z takes on the values 0,1. Furthermore $P\{Z=0\}=P\{X=2\}=1/3$ and $P\{Z=1\}=P(\{X=1\}\cup \{X=3\}) = 2/3$.

Thus

$$Cov(X,X)=EZ = \sum_{k=0}^1 k P\{Z=k\} = 2/3$$

3. Jan 24, 2013

### chiro

Re: Var(X) = Cov(X,X) ??

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 (X-E[X])(X-E[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. Jan 24, 2013

### Rasalhague

Re: 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[(X-EX)(Y-EY)P(X,Y)] in the formula which follows that. Serfling also states that P(X,Y) means

$$P_X(X)\cdot P_Y(Y)$$

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 $X \neq Y$, e.g. X = (1, 2, 3) and Y = (1, 4, 9).

5. Jan 24, 2013

### Rasalhague

Re: Var(X) = Cov(X,X) ??

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. Jan 24, 2013

### Rasalhague

Re: Var(X) = Cov(X,X) ??

Ah, reading further on that Mathworld article, it seems one definition concerns real-valued 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.

Last edited: Jan 24, 2013