Thread: Exercise 4.7
View Single Post
  #1  
Old 10-23-2013, 12:15 PM
Sweater Monkey Sweater Monkey is offline
Junior Member
 
Join Date: Sep 2013
Posts: 6
Default Exercise 4.7

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 g^- 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!
Reply With Quote