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Sweater Monkey · #1 10-23-2013, 12:15 PM

I feel like I'm overthinking Exercise 4.7 (b) and I am hoping for a little bit of insight.

My gut instinct says that $Var[E_{\text{val}}(g^-)] = \frac{1}{K} (P[g^-(x),y])^2$

I arrived at this idea by considering that the probability is similar to the standard deviation which is the square root of the variance so since:
$Var[E_{\text{val}}(g^-)] = \frac{1}{K}Var_{x}[e(g^-(x),y)]$ and $P[{g^-(x)}\neq{y}] = P[e(g^-(x),y)]$ does $Var_{x}[e(g^-(x),y)] = P[{g^-(x)}\neq{y}]^2$ ???

Then for part (c) on the exercise, assuming that the above is true, I used the notion that $P[{g^-(x)}\neq{y}] \le 0.5$ because if the probability of error were greater than 0.5 then the learned g would just flip its classification. Therefore this shows that for any

in a classification problem,
$Var[E_{\text{val}}(g^-)] \le \frac{1}{K}0.5^2$ and therefore:
$Var[E_{\text{val}}(g^-)] \le \frac{1}{4K}$

Any indication as to whether I'm working along the correct lines would be appreciated!