Could you please explain me how the standard risk is obtained in this slide ?
I guess that we integrate over the interval [0 ; +inf[ because it is the probability of making a mistake when y = -1 but why don't we also integrate over ]-inf ; 0] which is the probability of making a mistake when y = 1 ?
It seems like you have already figured it out, but just in case the same question concerns more people: the calculations go as follows: \(Pr(mistake) = Pr(\hat{y} \cdot y = -1) = Pr(\hat{y}=1, y=-1\ OR\ \hat{y}=-1, y=1) = \) \(Pr(y=-1) \cdot Pr(\hat{y}=1 | y=-1) + Pr(y=1) \cdot Pr(\hat{y}=-1 | y=1) = \) \(0.5 \cdot Pr(\hat{y}=1 | y=-1) + 0.5 \cdot Pr(\hat{y}=-1 | y=1) = \) \(0.5 \cdot Pr(x_1>0 | y=-1) + 0.5 \cdot Pr(x_1<0 | y=1) = \) \(0.5 \cdot \int + 0.5 \cdot \int = \int \),
where \(\int\) is the integral expression from the slide shown above. The key part here is that to realize that \(Pr(x_1>0 | y=-1)\) = \(Pr(x_1<0 | y=1) = \int\) by symmetry around 0. Then the same \(\int\) applies to both Gaussians and since \(Pr(y=-1) = Pr(y=1)\), we essentially just need to compute the \(\int\) once which will be the final answer.
Lecture 9c risk
Hi,
Could you please explain me how the standard risk is obtained in this slide ?
I guess that we integrate over the interval [0 ; +inf[ because it is the probability of making a mistake when y = -1 but why don't we also integrate over ]-inf ; 0] which is the probability of making a mistake when y = 1 ?
Best regards,
Ali
edit : Nevermind I figured it out
1
Hi Ali,
It seems like you have already figured it out, but just in case the same question concerns more people: the calculations go as follows:
\(Pr(mistake) = Pr(\hat{y} \cdot y = -1) = Pr(\hat{y}=1, y=-1\ OR\ \hat{y}=-1, y=1) = \)
\(Pr(y=-1) \cdot Pr(\hat{y}=1 | y=-1) + Pr(y=1) \cdot Pr(\hat{y}=-1 | y=1) = \)
\(0.5 \cdot Pr(\hat{y}=1 | y=-1) + 0.5 \cdot Pr(\hat{y}=-1 | y=1) = \)
\(0.5 \cdot Pr(x_1>0 | y=-1) + 0.5 \cdot Pr(x_1<0 | y=1) = \)
\(0.5 \cdot \int + 0.5 \cdot \int = \int \),
where \(\int\) is the integral expression from the slide shown above. The key part here is that to realize that \(Pr(x_1>0 | y=-1)\) = \(Pr(x_1<0 | y=1) = \int\) by symmetry around 0. Then the same \(\int\) applies to both Gaussians and since \(Pr(y=-1) = Pr(y=1)\), we essentially just need to compute the \(\int\) once which will be the final answer.
I hope that helps.
Best,
Maksym
2
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