[proofplan] We decompose the squared prediction error into an irreducible conditional-noise part and an approximation part involving only the conditional mean. The key point is that the residual $Y-m$ is orthogonal in $L^2$ to every $\sigma(X)$-measurable square-integrable random variable, hence in particular to every affine function of $X$ and to $m$ itself. This makes minimizing $\mathbb E[(Y-Z)^2]$ over affine predictors exactly the same problem as projecting $m$ onto $\operatorname{span}\{1,X\}$. The final equivalence follows from the elementary fact that a point is its own projection onto a subspace exactly when it already belongs to that subspace. [/proofplan]

[step:Define the affine subspace and the conditional residual] Let \begin{align*} \mathcal A := \operatorname{span}\{1,X\} = \{a+bX : a,b \in \mathbb R\} \subset L^2(\Omega,\mathcal F,\mathbb P). \end{align*} Since $X \in L^2(\Omega,\mathcal F,\mathbb P)$ and $1 \in L^2(\Omega,\mathcal F,\mathbb P)$, every $Z \in \mathcal A$ belongs to $L^2(\Omega,\mathcal F,\mathbb P)$. Also $m \in L^2(\Omega,\mathcal F,\mathbb P)$, because conditional expectation is an $L^2$ contraction: \begin{align*} \mathbb E[m^2] \leq \mathbb E[Y^2] < \infty. \end{align*} Define the residual random variable \begin{align*} r:\Omega &\to \mathbb R \\ \omega &\mapsto Y(\omega)-m(\omega). \end{align*} Then $r \in L^2(\Omega,\mathcal F,\mathbb P)$. [/step]

[step:Show the conditional residual is orthogonal to every square-integrable function of $X$]Let $W \in L^2(\Omega,\mathcal F,\mathbb P)$ be $\sigma(X)$-measurable. We prove \begin{align*} \mathbb E[rW]=0. \end{align*} For each $n \in \mathbb N$, define the bounded $\sigma(X)$-measurable truncation \begin{align*} W_n:\Omega &\to \mathbb R \\ \omega &\mapsto \max\{-n,\min\{W(\omega),n\}\}. \end{align*} By the defining property of conditional expectation, since $W_n$ is bounded and $\sigma(X)$-measurable, \begin{align*} \mathbb E[YW_n] = \mathbb E[mW_n], \end{align*} and hence \begin{align*} \mathbb E[rW_n]=0. \end{align*} Moreover $W_n \to W$ pointwise and $|W_n-W| \leq 2|W|$. Since $r,W \in L^2(\Omega,\mathcal F,\mathbb P)$, the [Cauchy-Schwarz inequality](/theorems/432) gives $rW \in L^1(\Omega,\mathcal F,\mathbb P)$, and \begin{align*} |r(W_n-W)| \leq 2|r||W| \in L^1(\Omega,\mathcal F,\mathbb P). \end{align*} The [dominated convergence theorem](/theorems/4) gives \begin{align*} \lim_{n\to\infty}\mathbb E[rW_n] = \mathbb E[rW]. \end{align*} Therefore $\mathbb E[rW]=0$.[/step]

[guided]The conditional expectation identity is initially guaranteed against bounded $\sigma(X)$-measurable test random variables. Since we need to test against square-integrable random variables such as $m-Z$, we first reduce to the bounded case by truncation. Let $W \in L^2(\Omega,\mathcal F,\mathbb P)$ be $\sigma(X)$-measurable. For each $n \in \mathbb N$, define \begin{align*} W_n:\Omega &\to \mathbb R \\ \omega &\mapsto \max\{-n,\min\{W(\omega),n\}\}. \end{align*} Each $W_n$ is bounded and $\sigma(X)$-measurable because it is obtained from $W$ by composing with the continuous truncation map $t \mapsto \max\{-n,\min\{t,n\}\}$. By the defining property of $m=\mathbb E[Y\mid \sigma(X)]$, \begin{align*} \mathbb E[YW_n]=\mathbb E[mW_n]. \end{align*} Subtracting the right-hand side from the left-hand side gives \begin{align*} \mathbb E[(Y-m)W_n]=\mathbb E[rW_n]=0. \end{align*} It remains to pass from $W_n$ to $W$. We have $W_n \to W$ pointwise and $|W_n-W|\leq 2|W|$. Since $r,W\in L^2(\Omega,\mathcal F,\mathbb P)$, the Cauchy-Schwarz inequality gives \begin{align*} \mathbb E[|rW|] \leq \mathbb E[r^2]^{1/2}\mathbb E[W^2]^{1/2}<\infty. \end{align*} Thus $2|r||W|$ is an integrable dominating random variable, and \begin{align*} |r(W_n-W)|\leq 2|r||W|. \end{align*} By dominated convergence, \begin{align*} \mathbb E[rW]=\lim_{n\to\infty}\mathbb E[rW_n]=0. \end{align*} This proves that the conditional residual $r=Y-m$ is orthogonal in $L^2$ to every square-integrable $\sigma(X)$-measurable random variable.[/guided]

[step:Decompose the risk for every affine predictor]Fix $Z \in \mathcal A$. Since $Z$ is a real linear combination of $1$ and $X$, it is $\sigma(X)$-measurable. Also $m$ is $\sigma(X)$-measurable by definition of conditional expectation, so $m-Z$ is $\sigma(X)$-measurable and belongs to $L^2(\Omega,\mathcal F,\mathbb P)$. Applying the orthogonality just proved with $W=m-Z$ gives \begin{align*} \mathbb E[r(m-Z)] = 0. \end{align*} Using $Y-Z=r+(m-Z)$, expanding the square, and using this orthogonality, \begin{align*} \mathbb E[(Y-Z)^2] &= \mathbb E[(r+(m-Z))^2] \\ &= \mathbb E[r^2] + 2\mathbb E[r(m-Z)] + \mathbb E[(m-Z)^2] \\ &= \mathbb E[r^2] + \mathbb E[(m-Z)^2]. \end{align*} The first term $\mathbb E[r^2]$ is independent of $Z$.[/step]

[guided]We now compare the least-squares risk for predicting $Y$ with the least-squares risk for approximating $m$. Fix an arbitrary affine predictor $Z \in \mathcal A$. By definition of $\mathcal A$, there exist $a,b\in\mathbb R$ such that $Z=a+bX$. Therefore $Z$ is $\sigma(X)$-measurable. The conditional mean $m$ is also $\sigma(X)$-measurable, so the difference $m-Z$ is $\sigma(X)$-measurable. Since $m,Z\in L^2(\Omega,\mathcal F,\mathbb P)$, also $m-Z\in L^2(\Omega,\mathcal F,\mathbb P)$. The previous step applies with $W=m-Z$, giving \begin{align*} \mathbb E[r(m-Z)] = 0. \end{align*} Now use the identity \begin{align*} Y-Z = (Y-m)+(m-Z)=r+(m-Z). \end{align*} Expanding the square in $L^2$ gives \begin{align*} \mathbb E[(Y-Z)^2] &= \mathbb E[(r+(m-Z))^2] \\ &= \mathbb E[r^2] + 2\mathbb E[r(m-Z)] + \mathbb E[(m-Z)^2] \\ &= \mathbb E[r^2] + \mathbb E[(m-Z)^2]. \end{align*} The term $\mathbb E[r^2]$ does not depend on the affine predictor $Z$. Thus the only part of the risk that can be changed by choosing $Z$ is the squared distance from $Z$ to the conditional mean $m$.[/guided]

[step:Identify affine regression with projection of the conditional mean] Let $\widehat Z \in \mathcal A$. From the risk decomposition, for every $Z \in \mathcal A$, \begin{align*} \mathbb E[(Y-\widehat Z)^2] \leq \mathbb E[(Y-Z)^2] \end{align*} holds if and only if \begin{align*} \mathbb E[(m-\widehat Z)^2] \leq \mathbb E[(m-Z)^2]. \end{align*} Therefore the minimizers of $Z \mapsto \mathbb E[(Y-Z)^2]$ over $\mathcal A$ are exactly the minimizers of $Z \mapsto \mathbb E[(m-Z)^2]$ over $\mathcal A$. This is precisely the assertion that the population affine regression predictor is the $L^2$-projection of $m$ onto $\mathcal A$. [/step]

[step:Characterize when the affine predictor equals the conditional mean] If $m \in \mathcal A$, then choosing $Z=m$ gives \begin{align*} \inf_{Z\in\mathcal A}\mathbb E[(m-Z)^2]=0, \end{align*} so every $L^2$-projection of $m$ onto $\mathcal A$ equals $m$ almost surely. Conversely, if an affine regression predictor $\widehat Z$ equals $m$ almost surely, then $\widehat Z\in\mathcal A$ by definition of affine regression, and hence $m\in\mathcal A$ as an element of $L^2(\Omega,\mathcal F,\mathbb P)$. Thus the affine regression predictor agrees with the conditional mean almost surely if and only if $m\in\operatorname{span}\{1,X\}$. [/step]