# Mean and predicted response

(Redirected from Mean 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, εi is the random error, and $\alpha$ and $\beta$ are parameters. The predicted response value for a given explanatory value, xd, 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 xd, 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 xd. 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 εi 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

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

## References

• Draper, N.R.; Smith, H. (1998). Applied Regression Analysis (3rd ed.). John Wiley. ISBN 0-471-17082-8.