Mean and predicted response

In linear regression mean response and predicted response are values of the dependent variable calculated from the regression parameters and a given value of the independent variable. The values of these two responses are the same, but their calculated variances are different.

Straight line regression

In straight line fitting, the model is

$y_i=\alpha+\beta x_i +\epsilon_i\,$

where $y_i$ is the response variable, $x_i$ is the explanatory variable, ε_i is the random error, and $\alpha$ and $\beta$ are parameters. The predicted response value for a given explanatory value, x_d, is given by

$\hat{y}_d=\hat\alpha+\hat\beta x_d ,$

while the actual response would be

$y_d=\alpha+\beta x_d +\epsilon_d \,$

Expressions for the values and variances of $\hat\alpha$ and $\hat\beta$ are given in linear regression.

Mean response is an estimate of the mean of the y population associated with x_d, that is $E(y | x_d)=\hat{y}_d\!$. The variance of the mean response is given by

$\text{Var}\left(\hat{\alpha} + \hat{\beta}x_d\right) = \text{Var}\left(\hat{\alpha}\right) + \left(\text{Var} \hat{\beta}\right)x_d^2 + 2 x_d\text{Cov}\left(\hat{\alpha},\hat{\beta}\right) .$

This expression can be simplified to

$\text{Var}\left(\hat{\alpha} + \hat{\beta}x_d\right) =\sigma^2\left(\frac{1}{m} + \frac{\left(x_d - \bar{x}\right)^2}{\sum (x_i - \bar{x})^2}\right).$

To demonstrate this simplification, one can make use of the identity

$\sum (x_i - \bar{x})^2 = \sum x_i^2 - \frac{1}{m}\left(\sum x_i\right)^2 .$

The predicted response distribution is the predicted distribution of the residuals at the given point x_d. So the variance is given by

$\text{Var}\left(y_d - \left[\hat{\alpha} + \hat{\beta}x_d\right]\right) = \text{Var}\left(y_d\right) + \text{Var}\left(\hat{\alpha} + \hat{\beta}x_d\right) .$

The second part of this expression was already calculated for the mean response. Since $\text{Var}\left(y_d\right)=\sigma^2$ (a fixed but unknown parameter that can be estimated), the variance of the predicted response is given by

$\text{Var}\left(y_d - \left[\hat{\alpha} + \hat{\beta}x_d\right]\right) = \sigma^2 + \sigma^2\left(\frac{1}{m} + \frac{\left(x_d - \bar{x}\right)^2}{\sum (x_i - \bar{x})^2}\right) = \sigma^2\left(1+\frac{1}{m} + \frac{\left(x_d - \bar{x}\right)^2}{\sum (x_i - \bar{x})^2}\right) .$

Confidence intervals

The $100(1-\alpha)\%$ confidence intervals are computed as $y_d \pm t_{\frac{\alpha }{2},m - n - 1} \sqrt{\text {Var}}$. Thus, the confidence interval for predicted response is wider than the interval for mean response. This is expected intuitively – the variance of the population of $y$ values does not shrink when one samples from it, because the random variable ε_i does not decrease, but the variance of the mean of the $y$ does shrink with increased sampling, because the variance in $\hat \alpha$ and $\hat \beta$ decrease, so the mean response (predicted response value) becomes closer to $\alpha + \beta x_d$.

This is analogous to the difference between the variance of a population and the variance of the sample mean of a population: the variance of a population is a parameter and does not change, but the variance of the sample mean decreases with increased samples.

General linear regression

The general linear model can be written as

$y_i=\sum_{j=1}^{j=n}X_{ij}\beta_j + \epsilon_i\,$

Therefore since $y_d=\sum_{j=1}^{j=n} X_{dj}\hat\beta_j$ the general expression for the variance of the mean response is

$\text{Var}\left(\sum_{j=1}^{j=n} X_{dj}\hat\beta_j\right)= \sum_{i=1}^{i=n}\sum_{j=1}^{j=n}X_{di}M_{ij}X_{dj},$

where M is the covariance matrix of the parameters, given by

$\mathbf{M}=\sigma^2\left(\mathbf{X^TX}\right)^{-1}$.

See also

• Prediction interval