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BojanVujatovic · #4 07-10-2014, 04:31 PM

Quote:

Originally Posted by tatung2112

(c) If p = 0.9, what is the probability that S will produce a better hypothesis than C?
Answer: P[P(Sy = f) > P(Cy = f)] where Sy is the output hypothesis of S, Cy is the output hypothesis of C
+ Since yn = +1, Sy = +1. Moreover, P[f(x) = +1] = 0.9 --> P(Sy = f) = 0.9
+ We have, P(Cy = +1) = 0.5, P(Cy = -1) = 0.5, P[f(x) = +1] = 0.9, P[f(x) = -1] = 0.1
--> P[Cy = f] = 0.5*0.9 + 0.5*0.1 = 0.5
Since 0.9 > 0.5, P[P(Sy = f) > P(Cy = f)] = 1

Can you please elaborate more on why P(Cy = -1) = 0.5, I cannot understand that part?
Here is my reasoning for the (c) part: the event S produces a better hypothesis than C means that $E_{out}\left(\text(S)\right)$ is smaller than $E_{out}\left(\text(C)\right)$ , so

$P\left[E_{out}(S(\mathcal{D})) < E_{out}(C(\mathcal{D})) \right] = \\ P\left[E_{out}(h_1) < E_{out}(h_2) \right] =$

$= P\left[P\left[f(x) \neq h1\right] < P\left[f(x) \neq h2\right] \right] = P\left[P\left[f(x) = -1\right] < P\left[f(x) =+1\right] \right] =$

$= P\left[1-p< p \right] = P\left[0.1 < 0.9 \right] = 1$