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| − | So then... <math>Var[z] = E[z^2] - (E[z])^2</math> and since <math>E[z]=0</math> <math>Var[z] = E[z^2] = (x\cos(\theta)+sin(\theta))^2</math> | + | So then... <math>Var[z] = E[z^2] - (E[z])^2</math> and since <math>E[z]=0</math> <math>Var[z] = E[z^2] = (x\cos(\theta)+y\sin(\theta))^2</math> |
Latest revision as of 09:14, 9 December 2008
Well from observation we know $ E[z] = E[w] = 0 $ due to them being periodic.
We also know that $ E[x^2]=\sigma_x $ and $ E[Y^2] =\sigma_y $.
So then... $ Var[z] = E[z^2] - (E[z])^2 $ and since $ E[z]=0 $ $ Var[z] = E[z^2] = (x\cos(\theta)+y\sin(\theta))^2 $
