# Formal proof that mean minimize squared error function

On an important book of Machine Learning, I've found this proof.

We want to minimize the cost function $J_0(X_0)$ defined by the formula $$J_0(x_0) = \sum_{k=1}^n \|x_0 - x_k \|^2.$$

The solution to this problem is given by $x_0=m$, where $m$ is the sample mean $m = \frac{1}{n}\sum_{k=1}^nx_k$.

Proof. $$\begin{array}{rcl} J_0(x_0) & = & \sum_{k=1}^n \|(x_0 - m)-(x_k - m) \|^2 \\ & = & \sum_{k=1}^n \|x_0 - m \|^2 -2(x_0-m)^T\sum_{k=1}^n(x_k-m) + \sum_{k=1}^n \|x_k - m \|^2 \\ & = & \sum_{k=1}^n \|x_0 - m \|^2 + \sum_{k=1}^n \|x_k - m \|^2. \end{array}$$

Since $\sum_{k=1}^n \|x_k - m \|^2$ is independent of $x_0$, this expression is obviously minimized by $x_0=m$.

I cannot understand this proof. What does assure that $\sum_{k=1}^n \|x_k - m \|^2$ is minimized?

• Maybe you can provide extra information about the background of the book: definitions, approaches, etc. Consider there are two interpretations of MSE, in probability theory and linear algebra. What is the title of the book? $x_0$ and $m$ are vectors? Oct 10, 2014 at 22:04
• If you want a formal proof (wondering why on earth you would want such a thing), you need to specify what formal system the proof should be in.
– bof
Oct 11, 2014 at 0:07

## 5 Answers

Assume $X_0$ is a space of "values" to optimize, and so the values are constant. Note that $\sum_{k=1}^n \|x_k− m\|^2$ is constant because it does not depend of $x_0$ ($x_k$ and $m$ are calculated from $X_0$). So in the last line of the proof you have $J_0(x_0) = \sum_{k=1}^n \|x_0 - m \|^2 + \sum_{k=1}^n \|x_k - m \|^2$, i.e., $$J_0(x_0) = \sum_{k=1}^n \|x_0 - m \|^2 + C,$$ for some non-negative integer $C \ge 0$. Thus, you need to minimize $\sum_{k=1}^n \|x_0 - m \|^2$. Clearly, it is minimized when $x_0 = m$.

Edit 1. The term $2(x_0-m)^T\sum_{k=1}^n(x_k-m)$ vanishes because $$\sum_{k=1}^n (x_k - m) = \sum_{k=1}^n x_k - \sum_{k=1}^n m = \sum_{k=1}^n x_k - m \times n.$$ Since $m = \frac{1}{n} \sum_{k=1}^n x_k$ we have $$\sum_{k=1}^n x_k = m \times n.$$ Thus $$\begin{array}{rcl} 2(x_0-m)^T\sum_{k=1}^n(x_k-m) & = & 2(x_0-m)^T(m \times n - m \times n) \\ & = & 2(x_0-m)^T \times 0 \\ & = & 0. \end{array}$$

Edit 2. If $x_0 = v$ and $v \ne m$ we obtain the same. $$\begin{array}{rcl} J_0(x_0) & = & \sum_{k=1}^n \|(x_0 - v)-(x_k - v) \|^2 \\ & = & \sum_{k=1}^n \|x_0 - v \|^2 -2(x_0-v)^T\sum_{k=1}^n(x_k-v) + \sum_{k=1}^n \|x_k - v \|^2. \end{array}$$ The cross term vanishes when $x_0 = v$, so \begin{array}{rcl} J_0(x_0) & = & \sum_{k=1}^n \|x_0 - x_0 \|^2 -2(x_0-x_0)^T\sum_{k=1}^n(x_k-v) + \sum_{k=1}^n \|x_k - x_0 \|^2 \\ & = & 0 + 0 + \sum_{k=1}^n \|x_k - x_0 \|^2 \\ & = & \sum_{k=1}^n \|x_k - x_0 \|^2 \\ & = & \sum_{k=1}^n \|x_0 - x_k \|^2 . \end{array}

Your proof is a proof of existence. This type of proofs can be done picking some value $m$ and proving that it satisfies the claim, but it does not prove the uniqueness, so one can imagine that there is another value which satisfies the claim. Your proof does not prove the uniqueness, maybe because this is "clearly".

Aditional. Simplifying your problem, I will assume that $X_0$ is a scalar space (a collection of numbers), to give a alternative proof.

If we have $J_0(x_0) = \sum_{k=1}^n (x_0 - x_k)^2$, to minimize we derive respect to $x_0$, thus

$$\begin{array}{rcl} \frac{\partial}{\partial x_0} J_0(x_0) & = & \frac{\partial}{\partial x_0} \sum_{k=1}^n (x_0 - x_k)^2 \\ & = & 2 \times \sum_{k=1}^n (x_0 - x_k) \\ & = & 2 \times \left(\sum_{k=1}^n x_0 - \sum_{k=1}^n x_k\right) \\ & = & 2 \times \left(n x_0 - \sum_{k=1}^n x_k\right). \end{array}$$

Now to minize $J_0(x_0)$, we need that $\partial J_0(x_0) /\partial x_0 = 0$. Thus \begin{array}{rcl} 2 \times \left(n x_0 - \sum_{k=1}^n x_k\right) & = & 0 \\ n x_0 & = & \sum_{k=1}^n x_k \\ x_0 & = & \frac{1}{n} \sum_{k=1}^n x_k \\ x_0 & = & m. \qquad\qquad \Box \end{array}

To obtain tha same result for vectors you need to use the gradient and proceed similarly. Surely, you will learn later about normal equations.

• Thank you, you proof is much clear than the one reported in the book, nevertheless I wanted to understand the second step, why is it crucial that the cross term vanishes. Having the cross term and a generic vector $v$, instead of $m$, don't we minimize also for $x_0=v$? Oct 10, 2014 at 22:14
• Ah! Ok. Let see. You have $\sum_{k=1}^n (x_k - m)$, i.e., $\sum_{k=1}^n x_k - \sum_{k=1}^n m$, so $\sum_{k=1}^n x_k - n m$. Since $m = (\sum_{k=1}^n x_k) / n$, we have $\sum_{k=1}^n x_k = m n$. Thus, $mn - nm = 0$. Oct 10, 2014 at 22:54
• I've understood that. But my point of view is, if I choose a random vector $v$ it also minimize the function in the same manner (when $x_0 = v$), in fact $-2\sum(x_0-v)^t(x_k-v)$ sums to $0$ when $x_0-v = 0$, and thus first and second term of the equation vanish. i.e.$\sum\|x_0-v\|^2-2\sum(x_0-v)^t(x_k-v)+\sum\|x_k-v\|^2$ becomes $\sum\|x_k-v\|^2$ when $x_0 = v$. So the key point is why $\sum\|x_k-m\|^2$ is minimized respect to $\sum\|x_k-v\|^2$? Oct 10, 2014 at 23:21
• See the definition of your cost function, if you use $v$ you obtain the same expansion of the formula, but you cannot vanish the cross term unless you define $v = x_0$, and so you obtain the initial formula. Oct 10, 2014 at 23:24
• Your proof is a proof of existence. This type of proofs can be done picking some value $m$ and proving that it satisfies the claim, but it does not prove the uniqueness, so one can imagine that there is another value which satisfies the claim. Your proof does not prove the uniqueness (maybe because this is "clearly"). Another way is a direct proof, for instance, using the gradient to obtain the desired value. Oct 10, 2014 at 23:56

I believe that this expression is constant. The only variable anywhere in sight is $x_0$. You are supposed to pick $x_0$ that minimizes $J_0(x_0)$. The only expression involving $x_0$ is nonnegative, so it is minimized if it is zero. Choosing $x_0=m$ does this because $$J_0(x_0) = \underbrace{\sum_{k=1}^n \|x_0 - m \|^2}_{\textrm{zero when }x_0=m} + \underbrace{\sum_{k=1}^n \|x_k - m \|^2}_{\textrm{constant w.r.t. }x_0}$$

Once you understand the computation (the key relation is $\sum (x_k - m) = 0$) you are left with $$J(x_0) = \sum_{k=1}^n |x_0 - m|^2 + \sum_{k=1}^n |x_k - m|^2 \ge \sum_{k=1}^n |x_k - m|^2 = J(m)$$

This proves that the minimum is for $x_0 = m$.

• I missed the relation $\sum_{k_1}^n(x_k - m) = 0$, but if instead of $m$ I choose a generic vector $v$, isn't it also minimized by $x_0 = v$ since $(x_0 -v)^T(x_k - v)$ become a scalar multiplication for a null vector? Oct 10, 2014 at 21:57
• on it is not. The key is that the cross term vanishes! Oct 10, 2014 at 21:59

The squared error function is convex and differentiable. Hence it has a unique minimizer $\mu$ and its gradient exists. To obtain that minimum, we take the gradient of $J$ at $x$:

$\nabla J(x) = 2\sum_k (x-x_k)$

From the necessary conditions of optimality follows that the gradient at the unique minimizer $\mu$ is zero. Thus,

$\nabla J(\mu) = 2\sum_k (\mu-x_k) = 0$

Solving the above equation for $\mu$ gives the arithmetic mean:

$\quad 2\sum_k (\mu-x_k) = 0$

$\Rightarrow \sum_k \mu- \sum_k x_k = 0$

$\Rightarrow n \cdot \mu = \sum_k x_k$

$\Rightarrow \mu = \frac{1}{n}\sum_k x_k$

Let be $c\in\mathbb{R}$ a constant and $$f(x) = g(x)+ c\text{.}$$ Define

$$m_f = \min \{f(x)\mid x\in \mathcal{D}\}\quad\text{and}\quad g(x_0) = m_g:= \{g(x)\mid x\in \mathcal{D}\}\text{,}$$ where $\mathcal{D}$ is the domain of $f$ and $g$.

We claim that $$f(x_0) = m_f\Leftrightarrow g(x_0) = m_g \text{.}$$

Proof:

1. $(\Rightarrow)$ if there is a $x_0\in\mathcal{D}$ for which $f(x_0) = m_f$ and $g(x_0)\neq m_g$ then there is a $x_1\in\mathcal{D}$ such that $m_g = g(x_1) < g(x_0)$. It follows that $$f(x_1) = g(x_1) + c < g(x_0) + c = f(x_0) = m_f\text{,}$$ hence $f(x_1)<m_f$, which leads us to contradiction.
2. $(\Leftarrow)$ Analogically.

In your case, you have $$f(x) = J_0(x),\;\; g(x) =\sum_{k=1}^n \|x - m \|^2\quad\text{and}\quad c= \sum_{k=1}^n \|x_k - m \|^2$$ since the latest sum is a constant indeed, because $x_1,\dots , x_n$ are given (hence $m = \sum_{i\geq 1} x_i$ is given also). Obviously $g(x)\geq 0$ and it attains minimum for those $x_0$'s for which $g(x_0) = 0$. There is only one such $x_0$ and it equals $m$.