In cs229, the problem set1 of 2019 summer, question 2(e):
why $p(y^{(i)}=1)$ equal to $E[p(y^{(i)}=1|x^{(i)})]$ ?
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1See https://en.wikipedia.org/wiki/Law_of_total_expectation – geetha290krm Nov 15 '22 at 07:19
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2Please do not use pictures for critical portions of your post. Pictures may not be legible, cannot be searched and are not view-able to some, such as those who use screen readers. – Christian E. Ramirez Nov 15 '22 at 07:37
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A probability mass is an expectation of the indicator variable for the event. The Law of Total Expectation becomes applicable. Similarly, the conditional expectation is also a conditional probability.
$$\begin{align}p(y^{(i)}=1) &=\mathsf E(\mathbf 1\{y^{(i)}=1\})\\ &=\mathsf E(\mathsf E(\mathbf 1\{y^{(i)}=1\}\mid x^{(i)}))\\ &=\mathsf E(p(y^{(i)}=1\mid x^{(i)}))\end{align}$$
Graham Kemp
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Thank you! And I want to ask for one more question: is $h(x^{(i)})$ and ${1}(y^{(i)}=1}$ independent? – doraemon Nov 17 '22 at 07:50
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I am sorry that I don't know how to type the indicator function of $y^{(i)}$ properly... – doraemon Nov 17 '22 at 07:54
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1@gczday (1) You haven't provided any information on the random variables. This holds whether they are or are not. Still, if they are independent you don't need conditioning. (2)
\mathbf 1\{y^{(i)}=1\}produces $\mathbf 1{y^{(i)}=1}$ . – Graham Kemp Nov 17 '22 at 13:15