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This question already has an answer here:

While reading a book on statistics, I encountered the following:

$$f(x) = \mathop{\text{argmin}}\limits_c E_{Y|X}([Y-c^2]|X=x)\tag{1}$$ which somehow equates to $$f(x) = E(Y|X=x)\tag{2}$$ How do we get from result $(1)$ to $(2)$?

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marked as duplicate by Dilip Sarwate, user158565, kjetil b halvorsen, jpmuc, gung Aug 2 at 14:08

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Simply differentiate $f$ with respect to $c$ (your square operation must be outside the differencing): $$\begin{align}\frac{\partial f}{\partial c}&=\frac{\partial}{\partial c}\left(E[(Y-c)^2|X=x]\right)\\&=\frac{\partial}{\partial c}\left(E[Y^2|X=x]-2cE[Y|X=x]+c^2\right)\\&=-2E[Y|X=x]+2c=0\end{align}$$ which yields $c=E[Y|X=x]$. This $c$ minimizes $f$, because the second derivative is greater than $0$. Also, it's obvious that $f$ is a parabola with respect to $c$, and $E[Y|X=x]$ is the bottom of it.

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The answer follows exactly the same path than non conditional probability.

Say your proba $\mathbb{P}(.|X=x)$ is just a another proba $\mathbb{Q}$, then you know

$argmin_c \mathbb{E}_{\mathbb{Q}}[Y-c]^2$ is $\mathbb{E}_{\mathbb{Q}}[Y]$. It comes from the fact that you can find the minimum of a polynomial equation. Since $\partial_c\mathbb{E}_{\mathbb{Q}}[Y-c]^2=-2\mathbb{E}_{\mathbb{Q}}[Y-c]$. Or by any other way.

Now since your conditional proba is a normal proba the result follow if you replace $\mathbb{Q}$.

Hope it helps.

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