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ArixII · #3 January 12th, 2009, 03:06 AM

Quote:

Originally Posted by Laurent

It seems to be simply a consequence of ${\rm Var}(\lambda X)=\lambda^2{\rm Var}(X)$ , which is easily seen from the definition of the variance.

So the variance of $X=0.005 A$ is ${\rm Var}(X)=0.005^2 {\rm Var}(A)=0.005^2\cdot 0.2^2$ .

Again, this is in fact quite simple, and very specific to your setting: you have $X=\lambda A$ from above, and $V=\mu A$ where $\lambda=0.005$ and $\mu=0.1$ (with appropriate units), so that ${\rm Cov}(X,V)=E[(X-E[X])(V-E[V])]$ $=E[\lambda(A-E[A])\mu(A-E[A])]=\lambda\mu E[(A-E[A])^2]$ $=\lambda\mu{\rm Var}(A)=\sqrt{\lambda^2{\rm Var}(A)}\sqrt{\mu^2{\rm Var}(A)}$ $=\sqrt{{\rm Var}(X)}\sqrt{{\rm Var}(V)}=\sigma(X)\sigma(V)$ .

I hope I'm right, I didn't try to understand the physics beneath. The whole thing relies on the way proportional quantities behave for the variance/covariance.

Thank you!! It helped me to understand )