Dans la régression linéaire simple, d'où vient la formule de la variance des résidus?

L'intuition concernant les signes «plus» liés à la variance (du fait que même lorsque nous calculons la variance d'une différence de variables aléatoires indépendantes, nous ajoutons leurs variances) est correcte mais fatalement incomplète: si les variables aléatoires impliquées ne sont pas indépendantes , alors les covariances sont également impliquées - et les covariances peuvent être négatives. Il existe une expression qui est presque comme l'expression de la question a été pensé qu'il « devrait » être par l'OP (et moi), et il est la variance de la prédiction erreur , noterons $e^0 = y^0 - \hat y^0$ , où $y^0 = \beta_0+\beta_1x^0+u^0$ :

Var (e^{0}) = σ^{2} \cdot (1 + \frac{1}{n} + \frac{(x^{0} - \bar{x})^{2}}{S_{x x}})

$\text{Var}(e^0) = \sigma^2\cdot \left(1 + \frac 1n + \frac {(x^0-\bar x)^2}{S_{xx}}\right)$

La différence critique entre la variance de l'erreur de prédiction et la variance de l' erreur d' estimation (c'est-à-dire du résiduel) est que le terme d'erreur de l'observation prédite n'est pas corrélé avec l'estimateur , puisque la valeur $y^0$ n'a pas été utilisée dans la construction l'estimateur et le calcul des estimations, étant une valeur hors échantillon.

L'algèbre pour les deux se déroule exactement de la même manière jusqu'à un point (en utilisant au lieu de ), mais diverge ensuite. Plus précisément: $^0$ $_i$

Dans la régression linéaire simple , la variance de l'estimateur est toujours $y_i = \beta_0 + \beta_1x_i + u_i$ $\text{Var}(u_i)=\sigma^2$ $\hat \beta = (\hat \beta_0, \hat \beta_1)'$

Var (\hat{β}) = σ^{2} {(X^{'} X)}^{- 1}

$\text{Var}(\hat \beta) = \sigma^2 \left(\mathbf X' \mathbf X\right)^{-1}$

Nous avons

X^{'} X = [\begin{matrix} n & \sum x_{i} \\ \sum x_{i} & \sum x_{i}^{2} \end{matrix}]

$\mathbf X' \mathbf X= \left[ \begin{matrix} n & \sum x_i\\ \sum x_i & \sum x_i^2 \end{matrix}\right]$

et donc

{(X^{'} X)}^{- 1} = [\begin{matrix} \sum x_{i}^{2} & - \sum x_{i} \\ - \sum x_{i} & n \end{matrix}] \cdot {[n \sum x_{i}^{2} - {(\sum x_{i})}^{2}]}^{- 1}

$\left(\mathbf X' \mathbf X\right)^{-1}= \left[ \begin{matrix} \sum x_i^2 & -\sum x_i\\ -\sum x_i & n \end{matrix}\right]\cdot \left[n\sum x_i^2-\left(\sum x_i\right)^2\right]^{-1}$

Nous avons

[n \sum x_{i}^{2} - {(\sum x_{i})}^{2}] = [n \sum x_{i}^{2} - n^{2} {\bar{x}}^{2}] = n [\sum x_{i}^{2} - n {\bar{x}}^{2}] = n \sum (x_{i}^{2} - {\bar{x}}^{2}) \equiv n S_{x x}

$\left[n\sum x_i^2-\left(\sum x_i\right)^2\right] = \left[n\sum x_i^2-n^2\bar x^2\right] = n\left[\sum x_i^2-n\bar x^2\right] \\= n\sum (x_i^2-\bar x^2) \equiv nS_{xx}$

{(X^{'} X)}^{- 1} = [\begin{matrix} (1 / n) \sum x_{i}^{2} & - \bar{x} \\ - \bar{x} & 1 \end{matrix}] \cdot (1 / S_{x x})

$\left(\mathbf X' \mathbf X\right)^{-1}= \left[ \begin{matrix} (1/n)\sum x_i^2 & -\bar x\\ -\bar x & 1 \end{matrix}\right]\cdot (1/S_{xx})$

which means that

Var ({\hat{β}}_{0}) = σ^{2} (\frac{1}{n} \sum x_{i}^{2}) \cdot (1 / S_{x x}) = \frac{σ^{2}}{n} \frac{S_{x x} + n {\bar{x}}^{2}}{S_{x x}} = σ^{2} (\frac{1}{n} + \frac{{\bar{x}}^{2}}{S_{x x}})

$\text{Var}(\hat \beta_0) = \sigma^2\left(\frac 1n\sum x_i^2\right)\cdot \ (1/S_{xx}) = \frac {\sigma^2}{n}\frac{S_{xx}+n\bar x^2} {S_{xx}} = \sigma^2\left(\frac 1n + \frac{\bar x^2} {S_{xx}}\right)$

Var ({\hat{β}}_{1}) = σ^{2} (1 / S_{x x})

$\text{Var}(\hat \beta_1) = \sigma^2(1/S_{xx})$

Cov ({\hat{β}}_{0}, {\hat{β}}_{1}) = - σ^{2} (\bar{x} / S_{x x})

$\text{Cov}(\hat \beta_0,\hat \beta_1) = -\sigma^2(\bar x/S_{xx})$

The $i$ -th residual is defined as

{\hat{u}}_{i} = y_{i} - {\hat{y}}_{i} = (β_{0} - {\hat{β}}_{0}) + (β_{1} - {\hat{β}}_{1}) x_{i} + u_{i}

$\hat u_i = y_i - \hat y_i = (\beta_0 - \hat \beta_0) + (\beta_1 - \hat \beta_1)x_i +u_i$

The actual coefficients are treated as constants, the regressor is fixed (or conditional on it), and has zero covariance with the error term, but the estimators are correlated with the error term, because the estimators contain the dependent variable, and the dependent variable contains the error term. So we have

Var ({\hat{u}}_{i}) = [Var (u_{i}) + Var ({\hat{β}}_{0}) + x_{i}^{2} Var ({\hat{β}}_{1}) + 2 x_{i} Cov ({\hat{β}}_{0}, {\hat{β}}_{1})] + 2 Cov ([(β_{0} - {\hat{β}}_{0}) + (β_{1} - {\hat{β}}_{1}) x_{i}], u_{i})

$\text{Var}(\hat u_i) = \Big[\text{Var}(u_i)+\text{Var}(\hat \beta_0)+x_i^2\text{Var}(\hat \beta_1)+2x_i\text{Cov}(\hat \beta_0,\hat \beta_1)\Big] + 2\text{Cov}([(\beta_0 - \hat \beta_0) + (\beta_1 - \hat \beta_1)x_i],u_i)$

= [σ^{2} + σ^{2} (\frac{1}{n} + \frac{{\bar{x}}^{2}}{S_{x x}}) + x_{i}^{2} σ^{2} (1 / S_{x x}) + 2 Cov ([(β_{0} - {\hat{β}}_{0}) + (β_{1} - {\hat{β}}_{1}) x_{i}], u_{i})

$=\Big[\sigma^2 + \sigma^2\left(\frac 1n + \frac{\bar x^2} {S_{xx}}\right) + x_i^2\sigma^2(1/S_{xx}) +2\text{Cov}([(\beta_0 - \hat \beta_0) + (\beta_1 - \hat \beta_1)x_i],u_i)$

Pack it up a bit to obtain

Var ({\hat{u}}_{i}) = [σ^{2} \cdot (1 + \frac{1}{n} + \frac{(x_{i} - \bar{x})^{2}}{S_{x x}})] + 2 Cov ([(β_{0} - {\hat{β}}_{0}) + (β_{1} - {\hat{β}}_{1}) x_{i}], u_{i})

$\text{Var}(\hat u_i)=\left[\sigma^2\cdot \left(1 + \frac 1n + \frac {(x_i-\bar x)^2}{S_{xx}}\right)\right]+ 2\text{Cov}([(\beta_0 - \hat \beta_0) + (\beta_1 - \hat \beta_1)x_i],u_i)$

The term in the big parenthesis has exactly the same structure with the variance of the prediction error, with the only change being that instead of $x_i$ we will have $x^0$ (and the variance will be that of $e^0$ and not of $\hat u_i$ ). The last covariance term is zero for the prediction error because $y^0$ and hence $u^0$ is not included in the estimators, but not zero for the estimation error because $y_i$ and hence $u_i$ is part of the sample and so it is included in the estimator. We have

2 Cov ([(β_{0} - {\hat{β}}_{0}) + (β_{1} - {\hat{β}}_{1}) x_{i}], u_{i}) = 2 E ([(β_{0} - {\hat{β}}_{0}) + (β_{1} - {\hat{β}}_{1}) x_{i}] u_{i})

$2\text{Cov}([(\beta_0 - \hat \beta_0) + (\beta_1 - \hat \beta_1)x_i],u_i) = 2E\left([(\beta_0 - \hat \beta_0) + (\beta_1 - \hat \beta_1)x_i]u_i\right)$

= - 2 E ({\hat{β}}_{0} u_{i}) - 2 x_{i} E ({\hat{β}}_{1} u_{i}) = - 2 E ([\bar{y} - {\hat{β}}_{1} \bar{x}] u_{i}) - 2 x_{i} E ({\hat{β}}_{1} u_{i})

$=-2E\left(\hat \beta_0u_i\right)-2x_iE\left(\hat \beta_1u_i\right) = -2E\left([\bar y -\hat \beta_1 \bar x]u_i\right)-2x_iE\left(\hat \beta_1u_i\right)$

the last substitution from how $\hat \beta_0$ is calculated. Continuing,

. . . = - 2 E (\bar{y} u_{i}) - 2 (x_{i} - \bar{x}) E ({\hat{β}}_{1} u_{i}) = - 2 \frac{σ^{2}}{n} - 2 (x_{i} - \bar{x}) E [\frac{\sum (x_{i} - \bar{x}) (y_{i} - \bar{y})}{S_{x x}} u_{i}]

$...=-2E(\bar yu_i) -2(x_i-\bar x)E\left(\hat \beta_1u_i\right) = -2\frac {\sigma^2}{n} -2(x_i-\bar x)E\left[\frac {\sum(x_i-\bar x)(y_i-\bar y)}{S_{xx}}u_i\right]$

= - 2 \frac{σ^{2}}{n} - 2 \frac{(x_{i} - \bar{x})}{S_{x x}} [\sum (x_{i} - \bar{x}) E (y_{i} u_{i} - \bar{y} u_{i})]

$=-2\frac {\sigma^2}{n} -2\frac {(x_i-\bar x)}{S_{xx}}\left[ \sum(x_i-\bar x)E(y_iu_i-\bar yu_i)\right]$

= - 2 \frac{σ^{2}}{n} - 2 \frac{(x_{i} - \bar{x})}{S_{x x}} [- \frac{σ^{2}}{n} \sum_{j \neq i} (x_{j} - \bar{x}) + (x_{i} - \bar{x}) σ^{2} (1 - \frac{1}{n})]

$=-2\frac {\sigma^2}{n} -2\frac {(x_i-\bar x)}{S_{xx}}\left[ -\frac {\sigma^2}{n}\sum_{j\neq i}(x_j-\bar x) + (x_i-\bar x)\sigma^2(1-\frac 1n)\right]$

= - 2 \frac{σ^{2}}{n} - 2 \frac{(x_{i} - \bar{x})}{S_{x x}} [- \frac{σ^{2}}{n} \sum (x_{i} - \bar{x}) + (x_{i} - \bar{x}) σ^{2}]

$=-2\frac {\sigma^2}{n}-2\frac {(x_i-\bar x)}{S_{xx}}\left[ -\frac {\sigma^2}{n}\sum(x_i-\bar x) + (x_i-\bar x)\sigma^2\right]$

= - 2 \frac{σ^{2}}{n} - 2 \frac{(x_{i} - \bar{x})}{S_{x x}} [0 + (x_{i} - \bar{x}) σ^{2}] = - 2 \frac{σ^{2}}{n} - 2 σ^{2} \frac{(x_{i} - \bar{x})^{2}}{S_{x x}}

$=-2\frac {\sigma^2}{n}-2\frac {(x_i-\bar x)}{S_{xx}}\left[ 0 + (x_i-\bar x)\sigma^2\right] = -2\frac {\sigma^2}{n}-2\sigma^2\frac {(x_i-\bar x)^2}{S_{xx}}$

Inserting this into the expression for the variance of the residual, we obtain

Var ({\hat{u}}_{i}) = σ^{2} \cdot (1 - \frac{1}{n} - \frac{(x_{i} - \bar{x})^{2}}{S_{x x}})

$\text{Var}(\hat u_i)=\sigma^2\cdot \left(1 - \frac 1n - \frac {(x_i-\bar x)^2}{S_{xx}}\right)$

So hats off to the text the OP is using.

(I have skipped some algebraic manipulations, no wonder OLS algebra is taught less and less these days...)

SOME INTUITION

So it appears that what works "against" us (larger variance) when predicting, works "for us" (lower variance) when estimating. This is a good starting point for one to ponder why an excellent fit may be a bad sign for the prediction abilities of the model (however counter-intuitive this may sound...).
The fact that we are estimating the expected value of the regressor, decreases the variance by $1/n$ . Why? because by estimating, we "close our eyes" to some error-variability existing in the sample,since we essentially estimating an expected value. Moreover, the larger the deviation of an observation of a regressor from the regressor's sample mean, the smaller the variance of the residual associated with this observation will be... the more deviant the observation, the less deviant its residual... It is variability of the regressors that works for us, by "taking the place" of the unknown error-variability.

But that's good for estimation. For prediction, the same things turn against us: now, by not taking into account, however imperfectly, the variability in $y^0$ (since we want to predict it), our imperfect estimators obtained from the sample show their weaknesses: we estimated the sample mean, we don't know the true expected value -the variance increases. We have an $x^0$ that is far away from the sample mean as calculated from the other observations -too bad, our prediction error variance gets another boost, because the predicted $\hat y^0$ will tend to go astray... in more scientific language "optimal predictors in the sense of reduced prediction error variance, represent a shrinkage towards the mean of the variable under prediction". We do not try to replicate the dependent variable's variability -we just try to stay "close to the average".

Alecos Papadopoulos
la source

Thank you for a very clear answer! I'm glad that my "intuition" was correct.

Eric

Alecos, I really don't think this is right.

Glen_b -Reinstate Monica

@Alecos the mistake is in taking the parameter estimates to be uncorrelated with the error term. This part:

Var ({\hat{u}}_{i}) = Var (u_{i}) + Var ({\hat{β}}_{0}) + x_{i}^{2} Var ({\hat{β}}_{1}) + 2 x_{i} Cov ({\hat{β}}_{0}, {\hat{β}}_{1})

$\text{Var}(\hat u_i) = \text{Var}(u_i)+\text{Var}(\hat \beta_0)+x_i^2\text{Var}(\hat \beta_1)+2x_i\text{Cov}(\hat \beta_0,\hat \beta_1)$ isn't right.

Glen_b -Reinstate Monica

@Eric I apologize for misleading you earlier. I have tried to provide some intuition for both formulas.

Alecos Papadopoulos

+1 You can see why I did the multiple regression case for this... thanks for going to the extra effort of doing the simple-regression case.

Glen_b -Reinstate Monica

Dans la régression linéaire simple, d'où vient la formule de la variance des résidus?

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