I am struggling to understand the derivation of the expected prediction error per below (ESL), especially on the derivation of 2.11 and 2.12 (conditioning, the step towards point-wise minimum). Any pointers or links much appreciated.

Below I am reporting the excerpt from ESL pg. 18. The first two equations are, in order, equation 2.11 and 2.12.

Let $X \in \mathbb{R}^p$ denote a real valued random input vector, and $Y \in \mathbb{R}$ a real valued random output variable, with joint distribution $\text{Pr}(X,Y)$. We seek a function $f(X)$ for predicting $Y$ given values of the input $X$. This theory requires a loss function $L(Y,f(X))$ for penalizing errors in prediction, and by far the most common and convenient is squared error loss: $L(Y,f(X))=(Y-f(X))^2$. This leads us to a criterion for choosing $f$,

$$ \begin{split} \text{EPE}(f) &= \text{E}(Y - f(X))^2\\ & = \int [y - f(x)]^2 \text{Pr}(dx, dy) \end{split} $$

the expected (squared) prediction error. By conditioning on $X$, we can write EPE as

$$ \text{EPE}(f) = \text{E}_X \text{E}_{Y|X}([Y-f(X)]^2|X) $$

and we see that it suffices to minimize EPE point-wise:

$$ f(x) = \text{argmin}_c \text{E}_{Y|X}([Y-c]^2|X) $$

The solution is

$$ f(x) = \text{E}(Y|X=x) $$

the conditional expectation, also known as the regression function.

  • $\begingroup$ Swapping $X$ and $Y$ in the first equation in the Wikipedia article on Law of Total Expectation gives the equivalence of (2.9) and (2.11). Read that article for proofs. (2.12) is immediate, on the understanding that $f$ is to be chosen in order to minimize EPE. $\endgroup$
    – whuber
    Commented Jun 9, 2014 at 13:22
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    $\begingroup$ Side note: This is from Elements of Statistical Learning $\endgroup$
    – Zhubarb
    Commented Oct 1, 2016 at 13:53
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    $\begingroup$ For those also reading this book, check out these comprehensive notes by Weathermax and Epstein $\endgroup$
    – Dodgie
    Commented May 25, 2017 at 6:50
  • $\begingroup$ @Dodgie That link has died : ( $\endgroup$ Commented Jun 21, 2018 at 13:54
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    $\begingroup$ @MatthewDrury Fortunately a googling of "Weathermax and Epstein statistics" returned a link as the very first result ;) -- waxworksmath.com/Authors/G_M/Hastie/WriteUp/… $\endgroup$
    – Dodgie
    Commented Jun 21, 2018 at 17:09

5 Answers 5


\begin{align*} EPE(f) &= \int [y - f(x)]^2 Pr(dx, dy) \\ &= \int [y - f(x)]^2p(x,y)dxdy \\ &= \int_x \int_y [y - f(x)]^2p(x,y)dxdy \\ &= \int_x \int_y [y - f(x)]^2p(x)p(y|x)dxdy \\ &= \int_x\left( \int_y [y - f(x)]^2p(y|x)dy \right)p(x)dx \\ &= \int_x \left( E_{Y|X}([Y - f(X)]^2|X = x) \right) p(x)dx\\ &= E_{X}E_{Y|X}([Y - f(X)]^2| X = x) \end{align*}

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    $\begingroup$ I understand what you wrote, but do you think if the OP was confused by the derivation shown in the question, that he/she will understand your answer? of course, I already understood the derivation shown in the question. $\endgroup$ Commented Jun 23, 2015 at 21:52
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    $\begingroup$ I got here from google with the same question and actually find this derivation to be exactly what I needed. $\endgroup$ Commented May 31, 2016 at 20:53
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    $\begingroup$ @MarkL.Stone - this might be a stupid question, but could you explain what is meant by $Pr(dx,dy)$ and how it becomes $p(x,y)dxdy$ ? Thanks a bunch $\endgroup$ Commented Jun 21, 2018 at 12:26
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    $\begingroup$ What is meant by the former is the latter. I think it is more common to instead use dP(x,y) or dF(x,y). In 1D, you will often see dF(x) to mean f(x)dx, where f(x) is the probability density function, but the notation can also allow for discrete probability mass function (in summation) or even a mixture of continuous density and discrete probability mass. $\endgroup$ Commented Jun 21, 2018 at 12:41
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    $\begingroup$ Wouldn't be more precise to say (last formula) $ E_{X}(E_{Y|X}([Y - f(X)]^2| X = x))$ ? $\endgroup$
    – D1X
    Commented Jan 31, 2019 at 10:07

The equation (2.11) is a consequence of the following little equality. For any two random variables $Z_1$ and $Z_2$, and any function $g$

$$ E_{Z_1, Z_2} (g(Z_1, Z_2)) = E_{Z_2}(E_{Z_1 \mid Z_2}(g(Z_1, Z_2) \mid Z_2)) $$

The notation $E_{Z_1, Z_2}$ is the expectation over the joint distribution. The notation $E_{Z_1 \mid Z_2}$ essentially says "integrate over the conditional distribution of $Z_1$ as if $Z_2$ was fixed".

It's easy to verify this in the case that $Z_1$ and $Z_2$ are discrete random variables by just unwinding the definitions involved

$$ \begin{align} E_{Z_2} & (E_{Z_1 \mid Z_2}(g(Z_1, Z_2) \mid Z_2)) \\ &= E_{Z_2} \left( \sum_{z_1} g(z_1, Z_2) Pr(Z_1 = z_1 \mid Z_2 ) \right) \\ &= \sum_{z_2} \left( \sum_{z_1} g(z_1, z_2) Pr(Z_1 = z_1 \mid Z_2 = z_2 ) \right) Pr(Z_2 = z_2) \\ &= \sum_{z_1, z_2} g(z_1, z_2) Pr(Z_1 = z_1 \mid Z_2 = z_2) Pr(Z_2 = z_2) \\ &= \sum_{z_1, z_2} g(z_1, z_2) Pr(Z_1 = z_1, Z_2 = z_2 ) \\ &= E_{Z_1, Z_2} (g(Z_1, Z_2)) \end{align} $$

The continuous case can either be viewed informally as a limit of this argument, or formally verified once all the measure theoretic do-dads are in place.

To unwind the application, take $Z_1 = Y$, $Z_2 = X$, and $g(x, y) = (y - f(x))^2$. Everything lines up exactly.

The assertion (2.12) asks us to consider minimizing

$$ E_X E_{Y \mid X} (Y - f(X))^2 $$

where we are free to choose $f$ as we wish. Again, focusing on the discrete case, and dropping halfway into the unwinding above, we see that we are minimizing

$$ \sum_{x} \left( \sum_{y} (y - f(x))^2 Pr(Y = y \mid X = x) \right) Pr(X = x) $$

Everything inside the big parenthesis is non-negative, and you can minimize a sum of non-negative quantities by minimizing the summands individually. In context, this means that we can choose $f$ to minimize

$$\sum_{y} (y - f(x))^2 Pr(Y = y \mid X = x)$$

individually for each discrete value of $x$. This is exactly the content of what ESL is claiming, only with fancier notation.

  • $\begingroup$ Why does choosing $f$ that minimizes the sum over $y$ also minimize the nested sum of over $x$ and over $y$? I still don't understand this because if you minimize only the stuff in the big parenthesis, isn't that minimizing for a single value of $x$ and not over all $x$? I understand that the summands are non-negative because we multiply a squared error (nonnegative) by a probability (nonnegative as well), but $\endgroup$
    – roulette01
    Commented Jul 15, 2020 at 20:40
  • $\begingroup$ I read this again, and I think I may have misunderstood you and ESL. You are NOT claiming that an $f(x)$ that minimizes the last equation, also minimizes the second to last equation right? $\endgroup$
    – roulette01
    Commented Jul 15, 2020 at 21:10

I find some parts in this book express in a way that is difficult to understand, especially for those who do not have a strong background in statistics.

I will try to make it simple and hope that you can get rid of confusion.

Claim 1 (Smoothing) $E(X) = E(E(X|Y)),\forall X,Y$

Proof: Notice that E(Y) is a constant but E(Y|X) is a random variable depending on X. \begin{align} E(E(X|Y)) &= \displaystyle\int E(X|Y=y) f_Y(y) dy \\ &= \int \int x f_{X|Y} (x|y) dx f_Y(y) dy \\ &= \int \int x f_{X|Y} (x|y) f_Y(y) dx dy \\ &= \int \int x f_{XY} (x,y) dx dy \\ &= \int x \left(\int f_{XY} (x,y) dy \right) dx \\ &= \int x f_X(x) dx = E(X) \end{align}

Claim 2: $E(Y - f(X))^2 \geq E(Y - E(Y|X))^2, \forall f$

Proof: \begin{align} E((Y - f(X))^2 | X) &= E( ([Y - E(Y|X)] + [E(Y|X) - f(X)])^2|X) \\ &= E((Y-E(Y|X))^2 |X) + E((E(Y|X) - f(X))^2|X) + \\ &\qquad 2 E((Y - E(Y|X))(E(Y|X) - f(X))|X) \\ &=E((Y-E(Y|X))^2 |X) + E((E(Y|X) - f(X))^2|X) + \\ &\qquad 2 (E(Y|X) - f(X)) E(Y - E(Y|X))|X) \\[5pt] &( \text{ since } E(Y|X) - f(X) \text{ is constant given } X) \\[5pt] &= E((Y-E(Y|X))^2 |X) + E((E(Y|X) - f(X))^2|X) \text{ ( use Claim 1 }) \\ &\geq E((Y-E(Y|X))^2 |X) \end{align}

Taking expectation both sides of the above equation give Claim 2 (Q.E.D)

Therefore, the optimal f is $f(X) = E(Y|X)$

  • $\begingroup$ For claim 2, how would you solve this without knowing $E(Y|X)$ is the solution and adding $0$ in the first line of the equation? Could we find the argument of $f(X)$ that minimises the LHS by taking derivatives and setting to $0$? $\endgroup$ Commented May 22 at 16:08

I found the explanation in the textbook and the other answers here (and Confusion in regression function derivation and Confused by Derivation of Regression Function) inadequate, so I decided to add my own version. I know many readers of the book and this website will not have a background in measure theory but for me it makes everything clearer. Some of the difficulties the authors have - to my mind - comes from avoiding rigorous treatment of the probability parts of their arguments:

Let $X : \bigl(\Omega,\mathcal{F},\mathbb{P}\bigr) \to \bigl(\mathbf{R}^p,\mathcal{B}(\mathbf{R}^p)\bigr)$ and $Y : \bigl(\Omega,\mathcal{F},\mathbb{P}\bigr) \to \bigl(\mathbf{R},\mathcal{B}(\mathbf{R})\bigr)$ be random variables (there's no need for them not to be on the same underlying probability space, so we may as well do it like this). Define the expected prediction error as $$ EPE(f) = \mathbb{E}\Bigl[ \bigl(Y-f(X)\bigr)^2\Bigr]. $$

Aside: Recall that given some other random variable $Z : \bigl(\Omega,\mathcal{F},\mathbb{P}\bigr) \to \bigl(\mathbf{R},\mathcal{B}(\mathbf{R})\bigr)$, we have that $\mathbb{E}(Z |X ) : \bigl(\Omega,\mathcal{F},\mathbb{P}\bigr) \to \bigl(\mathbf{R},\mathcal{B}(\mathbf{R})\bigr)$ is a random variable of the form $\omega \mapsto G(X(\omega))$ f. When we write $\mathbb{E}(Z|X=x)$, we are evaluating the random variable $\mathbb{E}(Z |X )$ at some $\omega \in \Omega$ for which $X(\omega) = x$. And recall that the definition of the conditional expectation means that $$ \mathbb{E}(Z) = \mathbb{E}\biggl( \mathbb{E}(Z |X ) \biggr). $$ Also: From a rigorous perspective, there isn't really such a thing as $\mathbb{E}_X$ or $\mathbb{E}_{Y|X}$; this is just confusing: If everything is on the same probability space then "$\mathbb{E}$" means "integrate over the probability space'' and nothing else, every time.

So \begin{align} EPE(f) &= \mathbb{E}\Bigl[ \bigl(Y-f(X)\bigr)^2\Bigr]\\ & = \mathbb{E}\biggl(\mathbb{E}\Bigl[ \bigl(Y-f(X)\bigr)^2\big| X \Bigr]\biggr)\\ &= \int_{\Omega} \mathbb{E}\Bigl[ \bigl(Y-f(X)\bigr)^2\big| X\Bigr] d\mathbb{P}(\omega) \end{align} By taking $Z = \bigl(Y-f(X)\bigr)^2$ in our aside above, we can see that the integrand here is of the form $G(X(\omega))$, i.e. we can write this as $\int_{\Omega} G(X(\omega)) d\mathbb{P}(\omega)$ and - using the definition of the the law $\mu_X$ of $X$ - this is equal to $\int_{\mathbf{R}^p} G(x) d\mu_X(x)$. This certainly makes it clear that to choose an $f$ which minimizes the EPE you should minimize the integrand pointwise as a function of $x$. So let's try to do that: By linearity and 'taking out what is known' (i.e. the fact that $f(X)$ is $\sigma(X)$-measurable): \begin{align*} \mathbb{E}\Bigl[ \bigl(Y-f(X)\bigr)^2\big| X\Bigr] &= \mathbb{E}\bigl[ Y^2 | X\bigr] - 2\mathbb{E}\bigl[ Y f(X) | X\bigr] + \mathbb{E}\bigl[ f(X)^2 | X\bigr] \\ &= \mathbb{E}\bigl[ Y^2 | X\bigr] - 2f(X)\mathbb{E}\bigl[ Y | X\bigr] + f(X)^2 \\ &= \mathbb{E}\bigl[ Y^2 | X\bigr] - \mathbb{E}\bigl[ Y | X\bigr]^2 + \biggl(f(X) - \mathbb{E}\bigl[ Y | X\bigr] \biggr)^2 \end{align*} Then when we evaluate both sides at some $\omega \in \Omega$ for which $X(\omega) = x$, we get $$ \mathbb{E}\Bigl[ \bigl(Y-f(X)\bigr)^2\big| X=x\Bigr] = \mathbb{E}\bigl[ Y^2 | X=x\bigr] - \mathbb{E}\bigl[ Y | X=x\bigr]^2 + \biggl(f(x) - \mathbb{E}\bigl[ Y | X=x\bigr] \biggr)^2, $$ and we can see that the choice of $f(x)$ which minimizes this expression is $f(x) = \mathbb{E}\bigl[ Y | X=x\bigr]$, the so-called 'regression function'


I will try to explain this as per my understanding. The main idea of the section 2.4 Statistical Decision Theory is to provide a framework for developing models(e.g. least-squares regression, k-NN).

As a first step(it is what author of this topic is asking about) in that section we consider regression function.

Idea of the step: to show that we can use conditional expectation as a linear regression function. So

$$f(x) = mx+b = \text{E}(Y|X=x)$$

where $m$ - slope, $b$ - intercept ($X$ is vector in the example of the book but I hope it doesn't confuse.)

I understand this mapping next way. Linear regression is a process to find a line closest to every point on scatter-plot. So to predict $y$ we can use expected value(or mean) of $y$ for given $x$ instead of $mx+b$.

How to prove that we are right with above assumption?
1. We need a loss function(squared error): $$L(Y, f(X))=(Y-f(X))^2$$    Therefore the expected squared prediction error for our regression function will be: $$EPE(f) = E(L[Y, f(x)]) = E([Y-f(X)]^2) \text{ - this is how we're getting 2.9}$$

2. Then how to derive 2.11 from 2.10 and 2.10 from 2.9. Generally we need to follow one of properties for conditional expectation

$$E(E[X|Y]) = E[X|Y = y] P(Y = y) \text{ - by law of unconscious statistician}$$    and $$E[X|Y = y] P(Y = y) = E[X] \text{ - by partition theorem from above we get this.}$$

   We can do the next steps:

   2.9 to 2.10:

   $EPE(f)= E([Y-f(X)]^2) = \int[y−f(x)]^2Pr(dx,dy)$ - this is by definition of expectation($E(X)=∫xf(x)dx$ for continuous case) probably except for $Pr(dx, dy)$
   There are 3 parts:
   $\int$ - because we're using continuous random variables
   $[y−f(x)]^2$ - this is our x from definition
   $Pr(dx, dy)$ - just notation for $p(x,y)dxdy$ where $p(x,y)$ is probability

   2.10 to 2.11: $$\int [y−f(x)]^2Pr(dx,dy) \text{ - 2.10 formula}$$ $$=\int[y−f(x)]^2\mathbf{p(x,y)dxdy} \text{ - from the above}$$ $$=\mathbf{\int_{x}\int_{y}}[y−f(x)]^2p(x,y)dxdy \text{ - just more precise integrals}$$ $$=\int_{x}\int_{y}[y−f(x)]^2\mathbf{p(x)p(y|x)}dxdy \text{ - by multiplication rule we got this}$$ $$=\int_{x}\mathbf{(\int_{y}[y−f(x)]^2p(y|x)dy)}p(x)dx \text{ - just regrouped members}$$ $$=\int_{x}\mathbf{(E_{Y|X}([Y−f(X)]^2|X=x))}p(x)dx \text{ - by definition of conditional expectation}$$

$$=E_{X}[E_{Y|X}([Y−f(X)]^2|X=x)] \text{ - by law of unconscious statistician we get this}$$

   So $E_{X}E_{Y|X}([Y−f(X)]^2|X=x)$ is generally $E(E[Y|X])$.

3. So far we've worked on $EPE(f)$ and proved that $E([Y-f(X)]^2)$ can be represented like this $E_{X}E_{Y|X}([Y−f(X)]2|X=x)$

   Then authors say that it suffices to minimize $EPE$ pointwise for $f(x)$. $$f(x) = argmin_{c} E_{Y|X}([Y − c]^2|X) = x$$    I thought of simple notations for regression to realize what authors mean. Specifically we can minimize squared error of regression line with partial derivatives.
   a. we can represent this $$SE_{line} = (y_{0}-(mx_{0}+b))^2+(y_{1}-(mx_{1}+b))^2 +...+(y_{n}-(mx_{n}+b))^2$$    like this $$SE_{line} = n\overline{y^2}-2mn\overline{yx}-2bn\overline{y}+m^2n\overline{x^2}+2mbn\overline{x}+nb^2$$    It is the same actually.
   b. Then we can find partial derivatives of the above with respect of $m$(slope) and $b$(intersect) to find minima for those variables.

   c. So we can use $m$ and $b$ in $mx+b$ to get predicted $y$ with minimum error.

   The same idea is in the book. We want to find some $c$ to get minimum for $$E_{Y|X}([Y − c]^2|X = x)\text{ (2.12)}$$

   So the best prediction of $Y$ at any point $X$ is the conditional mean(mean of $Y$'s for $X$) when the best is measured by average squared error.

   Hope it helps.


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