In simple linear regression, the error sum of squares is given by

$$ \text{SSE} = \sum_{i=1}^n(y_i - \hat{y_i})^2 \\ \hat{\sigma}^2 = s^2 = \dfrac{\text{SSE}}{n-2} $$ where $n-2$ is the degrees of freedom.

1. Why n-2?

Answers elsewhere:

  1. Most stop with telling us, n-2 because, we need to estimate $\beta_1,\beta_0$ before calculating $\hat{y}$ (source)
  2. An answer here, suggests, assuming errors are normally distribution ($\varepsilon \sim N(0,\sigma^2)$), the residual sum of squares will have a chi-squared distribution with n-2 df as below. $$\begin{aligned} \text{SSE} \sim \sigma^2 \text{Chi-Sq(df=n-2)} \end{aligned}$$ Here is the proof of above which again involves matrices and I was lost at orthogonal transformation. In another one here, in hat-matrix.

What did I do? 1. With hope of simpler proof, just like proving unbiased estimator of sample variance as shown here, I attempted as below, but stuck after few steps.

$$\begin{aligned} E(s^2) &= E\bigg(\dfrac{1}{n-2}\sum_{i=1}^n(y_i - \hat{y_i})^2\bigg) \\ &= E\bigg(\dfrac{1}{n-2}\sum_{i=1}^n(y_i^2 + \hat{y_i}^2 - 2y_i\hat{y_i})\bigg) \\ \end{aligned}$$

For any random variable X, $$ E\bigg( \sum_{i=1}^2 X_i \bigg) = E\bigg( X_1 + X_2 \bigg) = E(X_1) + E(X_2) = \sum_{i=1}^2 E(X_i) $$ That is, the expectation permeates in to the summation because $E(X+Y) = E(X) + E(Y)$.

Using same technique,

$$\begin{aligned} E(s^2) &= E\bigg(\dfrac{1}{n-2}\sum_{i=1}^n(y_i^2 + \hat{y_i}^2 - 2y_i\hat{y_i})\bigg) \\ &=\dfrac{1}{n-2} \sum_{i=1}^n \big( \ E(y_i^2) + E(\hat{y_i}^2) - 2E(y_i\hat{y_i}) \ \big) & \text{(1) stuck} \end{aligned}$$

I am stuck after this step. I wanted to show above ends up as $\sigma^2$., thus proving $s^2$ of SSE as unbiased.

Is it a duplicate Q?: I am learning these as part of "Intro to statistics" in Udacity, which is extremely limited in giving a mathematical background (its just basic intuition + formula => apply without understanding system) so I have been using few books 1, 2 as reference and during gaps, will use SE. Topics completed so far (Distributions, MLE, CI, Hypo.Testing) did not require matrices/vectors/quadratic forms yet because so far have been only dealing with single RVs (univariate?), (and chi-squared not yet covered). The books are "Introductory". However, many of the proofs I find here are using vectors/matrices which I find difficult to grasp, so with a hope of simpler answer for "introductory" student I am posting this Q, hopefully thus, also making it not a duplicate.

Previous Question

  • $\begingroup$ Suggestion: Based on your questions, you really want to study/work on statistics. (Real) Statistics is a branch of mathematics. Mathematics is differ from other discipline. In math, you need to go step by step, cannot jump. So for statistics, the required basic math background is calculus, linear algebra, probability, mathematical statistics. After you familiar with these math materials, the answers to your question can be found from textbook of general linear model. $\endgroup$
    – user158565
    Nov 23, 2018 at 17:29
  • $\begingroup$ I have been already familiar with them (about decade back when I finished grad), and also you can see, what I am studying is introductory statistics, not advanced. The books refer them as introductory and they start so as well from scratch, which I have been able to follow well. And any basic topic could involve advanced math, so I also define my borders (like here). At any topic there is depth and breadth. Breadth gives progress across topics, while deeper the depth, higher the understanding. Here, I just want to go to a depth of understanding df for this problem. $\endgroup$ Nov 24, 2018 at 3:14
  • $\begingroup$ I also see matrices/vector forms are needed only when you use multiple RVs. I seek for now, just proof for simple forms. I will definitely revise that to multiple RVs, when I finish relevant linear algebra topics, but that is for another day. To make progress, I wish I see proof for df for simple linear form, get convinced and move on. I just find difficult to digest, when book simply says its because we estimate 2 terms wo mathematically proving it, with the level intended for the reader (in introductory book, accordingly). $\endgroup$ Nov 24, 2018 at 3:19
  • $\begingroup$ $E(s^2) = E(\frac{1}{n-2}\sum_{i=1}^n(y_i - \hat{y_i})^2)$. Your first step is wrong. That is why I advised you to go back one step, and do not jump. $\endgroup$
    – user158565
    Nov 24, 2018 at 20:17
  • 1
    $\begingroup$ As I said, I have been already in basics multiple times, it does not help if you generically ask to go back, so can you please add clarity what is wrong in the step. that would be helpful for me to focus specifically. what exactly is going wrong? $\endgroup$ Nov 25, 2018 at 4:01

1 Answer 1


The key form you want to reach is expression (5) below. The derivation is all algebra.

Assume the model $Y_i=\beta_0+\beta_1X_i+\varepsilon_i$. Then $\bar Y=\beta_0 +\beta_1\bar X+\bar\varepsilon$ so that $$Y_i-\bar Y = \beta_1(X_i-\bar X) + (\varepsilon_i-\bar\varepsilon).\tag1$$ The least squares estimators of $\beta_0$ and $\beta_1$ are, respectively, $$ \hat{\beta_0}:=\bar Y -\hat{\beta_1}\bar X \qquad{\text {and}}\qquad \hat{\beta_1}:=\frac{\sum(X_i-\bar X)(Y_i-\bar Y)}{\operatorname{SSX}},\tag2 $$ where $\operatorname{SSX}:=\sum(X_i-\bar X)^2$. Substitute (1) into the expression for $\hat\beta_1$: $$ \hat\beta_1=\frac{\sum(X_i-\bar X)\left(\beta_1(X_i-\bar X) + (\varepsilon_i-\bar\varepsilon)\right)}{\operatorname{SSX}}=\beta_1+\frac{\sum(X_i-\bar X)(\varepsilon_i-\bar\varepsilon)}{\operatorname{SSX}} $$ to obtain an alternative expression for $\hat\beta_1$ for later use: $$\hat\beta_1-\beta_1=\frac{\sum(X_i-\bar X)(\varepsilon_i-\bar\varepsilon)}{\operatorname{SSX}}.\tag{3} $$ Now derive an expression for $\operatorname{SSE}$. Plug $\hat{\beta_0}$ into $\hat Y_i:=\hat{\beta_0}+\hat{\beta_1}X_i$ to get $$ Y_i-\hat {Y_i} = (\varepsilon_i-\bar\varepsilon) - (\hat{\beta_1}-\beta_1)(X_i-\bar X).\tag4 $$ Square both sides of (4) and sum over $i$. This yields $$ \begin{aligned} \operatorname{SSE}&:=\sum(Y_i-\hat {Y_i})^2\\ &=\sum(\varepsilon_i-\bar\varepsilon)^2-2(\hat\beta_1-\beta_1)\sum(\varepsilon_i-\bar\varepsilon)(X_i-\bar X)+(\hat\beta_1-\beta_1)^2\sum(X_i-\bar X)^2 \\ &\stackrel{(3)}=\sum(\varepsilon_i-\bar\varepsilon)^2 - (\hat{\beta_1}-\beta_1)^2\operatorname{SSX}. \end{aligned} $$ Writing $\sum(\varepsilon_i-\bar\varepsilon)^2=\sum\varepsilon_i^2-n\bar\varepsilon^2$, divide through by $\sigma^2$ in this last expression for SSE and rearrange to the form $$ \boxed{\sum_{i=1}^n\left[\frac{\varepsilon_i}\sigma\right]^2= \left[\frac{\bar\varepsilon}{\sigma/\sqrt n}\right]^2 + \left[\frac{\hat{\beta_1}-\beta_1}{\sigma/\sqrt{\operatorname{SSX}}}\right]^2+\frac{\operatorname{SSE}}{\sigma^2}.}\tag5 $$ Now for some distribution theory. You can check$^\color{red}a$ that each of the bracketed items in (5) has a standard normal distribution. The expectation of the square of a standard normal equals its variance, which is 1. Conclude from (5) the expectation of $\operatorname{SSE}/\sigma^2$ is $n-2$, so $\operatorname{SSE}/(n-2)$ is an unbiased estimator of $\sigma^2$.

We can go further and derive the chi-square($n-2$) distribution for $\operatorname{SSE}/\sigma^2$. What is not obvious, and this is the step that requires matrix algebra and/or multivariable calculus to prove, is that the three terms on the RHS of (5) are mutually independent$^\color{red}b$. Using this, and the fact that the LHS of (5) is the sum of squares of $n$ independent standard normal variables, it follows that $\operatorname{SSE}/\sigma^2$ must have$^\color{red}c$ the distribution of the sum of squares of $n-2$ independent standard normal variables. This is the chi-square($n-2$) distribution.

$\color{red}{a}$: Since $\sum(X_i-\bar X)(\varepsilon_i-\bar\varepsilon)=\sum(X_i-\bar X)\varepsilon_i$, we deduce from (3) that $$E(\hat\beta_1)=\beta_1\qquad\text{and}\qquad\operatorname{Var}(\beta_1)=\frac{\sigma^2}{\operatorname{SSX}}.$$ $\color{red}b$: Define a change of variables from $(\varepsilon_1,\ldots,\varepsilon_n)$ to $(Z_1,\ldots,Z_n)$ by $$ \begin{aligned} Z_1&:=\bar\varepsilon\\ Z_2&:=\hat\beta_1\\ Z_i&:=Y_i-\hat Y_i,\qquad i=3,\ldots,n. \end{aligned} $$ From (4) we see that $\sum(Y_i-\hat Y_i)=0$ and $\sum(X_i-\bar X)(Y_i-\hat Y)=0$, implying that we can solve for $Y_1-\hat Y_1$ and $Y_2-\hat Y_2$ in terms of $Z_3,\ldots,Z_n$, and therefore that $\operatorname{SSE}$ is a function of $Z_3,\ldots,Z_n$. Given (5), we see the joint density of $Z_1,\ldots,Z_n$ has the form $$ g(z_1)h(z_2) k(z_3,\ldots,z_n); $$ note the Jacobian of the transformation is free of $z$ since the map from $\varepsilon$ to $z$ is a multiplication by a constant matrix. This factorization means that $Z_1$, $Z_2$, and $(Z_3,\ldots,Z_n)$ are mutually independent, therefore so are the three terms on the RHS of (5).

$\color{red}c$: One way to prove: Use moment generating functions, independence, and the fact that the moment generating function determines the distribution uniquely.

  • $\begingroup$ Is this just the $n-p$ denominator for an unbiased estimate of the variance in the special case of simple linear regression with $p=2$? $\endgroup$
    – Dave
    May 3, 2020 at 0:46
  • 1
    $\begingroup$ @Dave Yes, this is the $p=2$ case. $\endgroup$
    – grand_chat
    May 3, 2020 at 1:34

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