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Old 10-20-2012, 10:32 AM
axelrv axelrv is offline
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Default Variance of Eval

I'm confused about how to simplify expressions involving Var[Eval(g-)].

I know that Var[Eval(g-)] = E [ ( Eval(g-) - E[Eval(g-)] )^2] = E [ ( Eval(g-) - Eout(g-) )^2] and that for classification P[g-(x) != y] = Eout(g-). I'm not sure how to bring K into any of these expressions.

Any help would be greatly appreciated.
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Old 10-21-2012, 07:51 AM
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magdon magdon is offline
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Default Re: Variance of Eval

Here are two useful facts from probability:

The variance of a sum of independent terms is the sum of the variances:
Var\left(\sum_{k=1}^K X_k\right)=\sum_{k=1}^KVar(X_k)

When you scale a random quantity its variance scales quadratically:
Var(aX)=a^2Var(X)

[Hint: so, if you scale something by \frac{1}{K} its variance scales by \frac{1}{K^2}; the validation error is the average of K independent things (What things? Why are they independent?)]

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Originally Posted by axelrv View Post
I'm confused about how to simplify expressions involving Var[Eval(g-)].

I know that Var[Eval(g-)] = E [ ( Eval(g-) - E[Eval(g-)] )^2] = E [ ( Eval(g-) - Eout(g-) )^2] and that for classification P[g-(x) != y] = Eout(g-). I'm not sure how to bring K into any of these expressions.

Any help would be greatly appreciated.
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