Explication intuitive du

Si $X$ est de rang plein, l'inverse de $X^TX$ existe et nous obtenons les moindres carrés

$$\hat\beta = (X^TX)^{-1}XY$$ et $$\operatorname{Var}(\hat\beta) = \sigma^2(X^TX)^{-1}$$

Comment expliquer intuitivement $(X^TX)^{-1}$ dans la formule de variance? La technique de dérivation est claire pour moi.

Considérons une régression simple sans terme constant et où le régresseur unique est centré sur sa moyenne d'échantillon. Alors $X'X$ est ( $n$ fois) sa variance d'échantillon, et $(X'X)^{-1}$ son recirpocal. Ainsi, plus la variance = variabilité dans le régresseur est élevée, plus la variance de l'estimateur de coefficient est faible: plus nous avons de variabilité dans la variable explicative, plus nous pouvons estimer avec précision le coefficient inconnu.

Pourquoi? Parce que plus un régresseur est varié, plus il contient d'informations. Lorsque les régresseurs sont nombreux, cela se généralise à l'inverse de leur matrice variance-covariance, qui prend également en compte la co-variabilité des régresseurs. Dans le cas extrême où $X'X$ est diagonal, la précision de chaque coefficient estimé ne dépend que de la variance / variabilité du régresseur associé (compte tenu de la variance du terme d'erreur).

Une façon simple de voir $\sigma^2 \left(\mathbf{X}^{T} \mathbf{X} \right)^{-1}$ est comme l'analogue matriciel (multivarié) de $\frac{\sigma^2}{\sum_{i=1}^n \left(X_i-\bar{X}\right)^2}$ , qui est la variance du coefficient de pente dans la régression OLS simple. On peut même obtenir$\frac{\sigma^2}{\sum_{i=1}^n X_i^2}$ pour cette variance en omettant l'ordonnée à l'origine dans le modèle, c'est-à-dire en effectuant une régression à travers l'origine.

De l'une ou l'autre de ces formules, on peut voir qu'une variabilité plus grande de la variable prédictive conduira généralement à une estimation plus précise de son coefficient. C'est l'idée souvent exploitée dans la conception des expériences, où en choisissant des valeurs pour les prédicteurs (non aléatoires), on essaie de rendre le déterminant de ( X T X ) aussi grand que possible, le déterminant étant une mesure de la variabilité.$\left(\mathbf{X}^{T} \mathbf{X} \right)$

La transformation linéaire de la variable aléatoire gaussienne est-elle utile? En utilisant la règle que si, x ∼ N ( μ , Σ ) , alors A x + b ∼ N ( A μ + b , A T Σ A ) .$x \sim \mathcal{N}(\mu,\Sigma)$$Ax + b ~ \sim \mathcal{N}(A\mu + b,A^T\Sigma A)$

En supposant que Y = X β + ϵ est le modèle sous-jacent et ϵ ∼ N ( 0 , σ 2 ) .$Y = X\beta + \epsilon$$\epsilon \sim \mathcal{N}(0, \sigma^2)$

$$\therefore Y \sim \mathcal{N}(X\beta,\sigma^2)\\ X^TY \sim \mathcal{N}(X^TX\beta, X\sigma^2 X^T)\\ (X^TX)^{-1}X^TY \sim \mathcal{N}[\beta,(X^TX)^{-1} \sigma^2]$$

So $(X^TX)^{-1}X^T$ is just a complicated scaling matrix that transforms the distribution of $Y$.

Hope that was helpful.

I'll take a different approach towards developing the intuition that underlies the formula $\text{Var}\,\hat{\beta}=\sigma^2 (X'X)^{-1}$. When developing intuition for the multiple regression model, it's helpful to consider the bivariate linear regression model, viz.,

$$y_i=\alpha+\beta x_i + \varepsilon_i, \quad i=1,\ldots,n.$$ $\alpha+\beta x_i$ is frequently called the deterministic contribution to $y_i$, and $\varepsilon_i$ is called the stochastic contribution. Expressed in terms of deviations from the sample means $(\bar{x},\bar{y})$, this model may also be written as $$(y_i-\bar{y}) = \beta(x_i-\bar{x})+(\varepsilon_i-\bar{\varepsilon}), \quad i=1,\ldots,n.$$

To help develop the intuition, we will assume that the simplest Gauss-Markov assumptions are satisfied: $x_i$ nonstochastic, $\sum_{i=1}^n(x_i-\bar{x})^2>0$ for all $n$, and $\varepsilon_i \sim \text{iid}(0,\sigma^2)$ for all $i=1,\ldots,n$. As you already know very well, these conditions guarantee that

$$\text{Var}\,\hat{\beta}=\tfrac{1}{n}\sigma^2(\text{Var}\,x)^{-1}\text{,}$$ where $\text{Var}\,x$ is the sample variance of $x$. In words, this formula makes three claims: "The variance of $\hat{\beta}$ is inversely proportional to the sample size $n$, it is directly proportional to the variance of $\varepsilon$, and it is inversely proportional to the variance of $x$."

Why should doubling the sample size, ceteris paribus, cause the variance of $\hat{\beta}$ to be cut in half? This result is intimately linked to the iid assumption applied to $\varepsilon$: Since the individual errors are assumed to be iid, each observation should be treated ex ante as being equally informative. And, doubling the number of observations doubles the amount of information about the parameters that describe the (assumed linear) relationship between $x$ and $y$. Having twice as much information cuts the uncertainty about the parameters in half. Similarly, it should be straightforward to develop one's intuition as to why doubling $\sigma^2$ also doubles the variance of $\hat{\beta}$.

Let's turn, then, to your main question, which is about developing intuition for the claim that the variance of $\hat{\beta}$ is inversely proportional to the variance of $x$. To formalize notions, let us consider two separate bivariate linear regression models, called Model $(1)$ and Model $(2)$ from now on. We will assume that both models satisfy the assumptions of the simplest form of the Gauss-Markov theorem and that the models share the exact same values of $\alpha$, $\beta$, $n$, and $\sigma^2$. Under these assumptions, it is easy to show that $\text{E}\,\hat{\beta}{}^{(1)}=\text{E}\,\hat{\beta}{}^{(2)}=\beta$; in words, both estimators are unbiased. Crucially, we will also assume that whereas $\bar{x}^{(1)}=\bar{x}^{(2)}=\bar{x}$, $\text{Var}\,x^{(1)}\ne \text{Var}\,x^{(2)}$. Without loss of generality, let us assume that $\text{Var}\,x^{(1)}>\text{Var}\,x^{(2)}$. Which estimator of $\hat{\beta}$ will have the smaller variance? Put differently, will $\hat{\beta}{}^{(1)}$ or $\hat{\beta}{}^{(2)}$ be closer, on average, to $\beta$? From the earlier discussion, we have $\text{Var}\,\hat{\beta} {}^{(k)} =\tfrac{1}{n}\sigma^2/\text{Var}\,x{}^{(k)})$ for $k=1,2$. Because $\text{Var}\,x^{(1)}>\text{Var}\,x^{(2)}$ by assumption, it follows that $\text{Var}\,\hat{\beta}{}^{(1)} <\text{Var}\,\hat{\beta}{}^{(2)}$. What, then, is the intuition behind this result?

Because by assumption $\text{Var}\,x^{(1)}>\text{Var}\,x^{(2)}$, on average each $x_i^{(1)}$ will be farther away from $\bar{x}$ than is the case, on average, for $x_i^{(2)}$. Let us denote the expected average absolute difference between $x_i$ and $\bar{x}$ by $d_x$. The assumption that $\text{Var}\,x^{(1)}>\text{Var}\,x^{(2)}$ implies that $d_x^{(1)} >d_x^{(2)}$. The bivariate linear regression model, expressed in deviations from means, states that $d_y = \beta d_x^{(1)}$ for Model $(1)$ and $d_y = \beta d_x^{(2)}$ for Model $(2)$. If $\beta\ne0$, this means that the deterministic component of Model $(1)$, $\beta d_x^{(1)}$, has a greater influence on $d_y$ than does the deterministic component of Model $(2)$, $\beta d_x^{(2)}$. Recall that the both models are assumed to satisfy the Gauss-Markov assumptions, that the error variances are the same in both models, and that $\beta^{(1)}=\beta^{(2)}=\beta$. Since Model $(1)$ imparts more information about the contribution of the deterministic component of $y$ than does Model $(2)$, it follows that the precision with which the deterministic contribution can be estimated is greater for Model $(1)$ than is the case for Model $(2)$. The converse of greater precision is a lower variance of the point estimate of $\beta$.

It is reasonably straightforward to generalize the intuition obtained from studying the simple regression model to the general multiple linear regression model. The main complication is that instead of comparing scalar variances, it is necessary to compare the "size" of variance-covariance matrices. Having a good working knowledge of determinants, traces and eigenvalues of real symmetric matrices comes in very handy at this point :-)

Say we have $n$ observations (or sample size) and $p$ parameters.

The covariance matrix $\operatorname{Var}(\hat{\beta})$ of the estimated parameters $\hat{\beta}_1,\hat{\beta}_2$ etc. is a representation of the accuracy of the estimated parameters.

If in an ideal world the data could be perfectly described by the model, then the noise will be $\sigma^2= 0$. Now, the diagonal entries of $\operatorname{Var}(\hat{\beta})$ correspond to $\operatorname{Var}(\hat{\beta_1}),\operatorname{Var}(\hat{\beta_2})$ etc. The derived formula for the variance agrees with the intuition that if the noise is lower, the estimates will be more accurate.

In addition, as the number of measurements gets larger, the variance of the estimated parameters will decrease. So, overall the absolute value of the entries of $X^TX$ will be higher, as the number of columns of $X^T$ is $n$ and the number of rows of $X$ is $n$, and each entry of $X^TX$ is a sum of $n$ product pairs. The absolute value of the entries of the inverse $(X^TX)^{-1}$ will be lower.

Hence, even if there is a lot of noise, we can still reach good estimates $\hat{\beta_i}$ of the parameters if we increase the sample size $n$.

I hope this helps.

Reference: Section 7.3 on Least squares: Cosentino, Carlo, and Declan Bates. Feedback control in systems biology. Crc Press, 2011.

This builds on @Alecos Papadopuolos' answer.

Recall that the result of a least-squares regression doesn't depend on the units of measurement of your variables. Suppose your X-variable is a length measurement, given in inches. Then rescaling X, say by multiplying by 2.54 to change the unit to centimeters, doesn't materially affect things. If you refit the model, the new regression estimate will be the old estimate divided by 2.54.

The $X'X$ matrix is the variance of X, and hence reflects the scale of measurement of X. If you change the scale, you have to reflect this in your estimate of $\beta$, and this is done by multiplying by the inverse of $X'X$.