Stock&Watson question 202.5

[orit]

Hi David,
I was really struggling to find the formula for:
Cov(G,P)=Cov(G,0.5G+0.5S)=0.5Cov(G,G)+0.5Cov(G,S)
Based on the notes: Cov(G,P)=Var(G)+Var(P)+Cov(G,P)
Var(G) - is the Cov(G,G) but I don't understand the rest of the formula
Can you please explain the rational?

Thanks,
Orit

 

[David Harper CFA FRM]

Hi Orit,

For reference, the full line (http://forum.bionicturtle.com/threads/p1-t2-202-variance-of-sum-of-random-variables.4967/) is
cov(G,P) = cov(G,0.5G+0.5S) = 0.5cov(G,G) + 0.5cov(G,S) = 0.5var(G) + 0.5cov(G,S).

The first step elaborates on the portfolio (P) as consisting of two positions in equal weights P = 0.5G +0.5S
... this gets us into the partially-self-referential concept of a covariance between a position (G) and the portfolio (P) that contains the position G itself.

The second step is a property of covariance that I like to think of as a "distributive" property of covariance, where the full two variable form is given by
cov(ax + by, cw + dv) = ac*cov(x,w) + ad*cov(x,v) + bc*cov(y,w) + bd*cov(y,v) per http://en.wikipedia.org/wiki/Covariance#Properties
note that a, b, c, and d are constants: they "move out" of the covariance; but x,y,w, and v are random-variables

so the application, in this case, is:

• cov(G, aG+bS) = cov(G,aG) + cov(G,bS); see how it's sort of distributive. The proof is short, but the intuition is almost accessible if you just keep in mind that G and S are random variables
• since cov(G,aG) = a*cov(G,G) and cov(G,bS) = s*cov(GS), cov(G,aG) + cov(G,bS) = a*cov(G,G) + b*cov(G,S) = a*var(G) + b*cov(G,S); pulling out constants. Note that, unlike wikipedia, I am using capitals for random variables and small letters for constants.

So, to me, there are ~ three key properties that combine:

1. the random variables are "distributive" (which i sort of think of as: they can't escape fully participating in the covariance)
2. the constants always come out directly; e.g., cov(aG,aG) = a*a*cov(G,G) = a^2*var(G). This is commonly used: var(aG) = a^2*var(G)
3. cov(G,G) = var(G) as you already noted; where the var(G) can be thought of as the diagonal in a covariance matrix

I hope that helps,

 

[orit]

Thank you David, now its clear!!