If I add a control variable to my regression, this changes the variances of the parameter estimates. Why is this the case? Is this because SSE(=explained sum of squares) increases and therefore the SSR(=residual sum of squares) decreases? So basically as long as I avoid multicollinearity adding more control variables will always improve the statistical significance of the parameter estimates?

  • $\begingroup$ The SSE (sum of squares of errors) can never increase when additional variables are added to a regression. That is because you are using least squares: estimating zero for the new coefficient and keeping the original coefficient estimates will produce the original SSE, which therefore must be an upper bound for whatever is the new least sum of squares. $\endgroup$
    – whuber
    Feb 12 '16 at 14:01
  • $\begingroup$ Sorry for the confusion: SSE= explained sum of squares, SSR=residual sum of squares, SST=total sum of squares (as in Wooldridge (2012)) $\endgroup$
    – Peter
    Feb 19 '16 at 13:50

The question is what happens to the variance of a parameter estimates ($\hat \beta_i$) with the introduction of a new control variable in a regression model?

The variance of the estimated parameter will increase.

The more regressors we add, the lower the $\text{RSS}$ (residual sum of squares) and the higher the $R^2$. In fact, if we just add columns of pure noise (rnorm()) to our data, and run OLS regressions after every addition, we will see a marked, monotonic decrease in the $\text{RSS}$. I tested this with an example:

enter image description here

The idea is that in OLS the hat matrix $X (X^TX)^{-1}X^T$ is a projection matrix of the vector of observed $y$ values onto the column space of the model matrix. The higher the dimensions of the column space (the more vectors to form a basis), the closer $\hat y$ will be to $y$. But at a price!

The estimation of the variance of $\hat \beta_i$ is given by:

$Var[\hat\beta_i]= \sigma^2(X^TX)^{-1}$ with $\sigma^2$ corresponding to the variation of the observations around the predicted values, ($Var[\epsilon|X] = \sigma^2I_n$). The estimation of $\sigma^2$ from the sample is $s^2=\frac{e^Te}{n-p}$ or $\text{MSE}$ (mean squared error). The denominator is the number of observations, $n$ minus the number of parameters, $p$, counting the intercept. It is also referred to as the error or residual degrees of freedom ($\small\text{no. observations−no. indepen't variables−1}$). Alternatively, the formula for the $\sigma^2$ estimation, can be expressed as $\text{MSE = RSS/df}$ with $\text{RSS}$ being the same as $\text{SSE}$ (sum of squared errors), $\sum_1^n(y_i - \hat y)^2$.

Therefore the estimation of the variance of the parameter $\hat\beta_i$ is $s^2(X^TX)^{-1}$ or $\text{MSE}\times(X^TX)^{-1}$.

And I think this may be a source of confusion - just because adding another regressor decreases the $\text{MSE}$ by decreasing the $\text{RSS}$ - if it actually does at all, because of the change in the degrees of freedom in the denominator, it can't be said that the variance for the estimated parameter decreases.

In a parallel OLS simulation (here), I found, quite anecdotally, a $\small 3.5\%$ increase in $\text{MSE}$ when adding an extra regressor, with a minimal decrease in $\text{RSS}$.

What controlled the increased variance for the estimate of $\hat\beta_i$ turned out to be the entry for $\hat\beta_i$ in the matrix of cofactors involved in the calculation of the inverse of $(X^TX)^{-1}$, explaining a variance for the estimate of the parameter $\hat\beta_i$ $2.8$ times higher in the presence of the control variable.

An alternative formula for the variance not applicable to the intercept is:

$\large \text{var}[\hat\beta_i]= \frac{\sigma^2}{n \times var[X_i]\times(1\,-\,R_i^2)}$. The key here is that $R_i^2$ is the $R$-square of the regression of the corresponding variable for the parameter $\hat\beta_i$, or $X_i$ against all the other regressors. Therefore, the better the regression model of $X_i \sim \text{control variable}$, the higher the estimated $\text{var}[\hat\beta_i].$

  • 1
    $\begingroup$ Did you undelete an old answer? I didn't see this when answering. Anyway, this is a thoughtful post. Just two small notes. 1a. The $p$-value is not guaranteed to increase (that's why no +1 yet). 1b. In the last sentence: The increase in estimator SE does not necessitate a decrease in importance. The change that caused the increased SE might also increase the actual value of the estimator. 2. In my answer I did not talk about $t$-values because I did not want to introduce degrees of freedom; you did introduce DFs so I think it is natural to include them when going from $t$-values to $p$-values. $\endgroup$
    – usεr11852
    Feb 13 '16 at 5:09
  • $\begingroup$ To show when this can happen check what I wrote in the final sentence of my post. If you rerun lm2 with x2 = x2 - mean(x2) you will see that the $p$-value of $\beta_0$ decreased from $0.461$ to $0.454$. The SE of it is higher (so yes, you had an increase in variance) but its actual value also increase so the $t$-value is larger and thus even by losing that 1DF the $p$-value was lower. (Clearly from an inferential point of view $0.461$ vs $0.454$ is still useless.) $\endgroup$
    – usεr11852
    Feb 13 '16 at 5:15
  • $\begingroup$ @usεr11852 Thank you very much for your comments. Yes, I had undeleted my post. It is an attempt at getting these concepts right by simulation and playing with the linear algebra. I get lost with Hessian matrices at this point, though. After your comments, I erased the comments about the p-values. I would like to get to a point where I know the statements in the post are correct to proceed from there to explore the more advanced treatment in your post. $\endgroup$ Feb 13 '16 at 5:34
  • $\begingroup$ (+1) The Hessian matrix is just the matrix of second derivatives of your log-likelihood function. Think of getting the MLE as exploring a 2D surface. The MLE is found at a local optimum $\beta_{opt}$ where your first derivatives are (usually) very close to zero. (If they were it would not be an optimum). The second derivatives at that point encapsulate how "steep" that optimum is. That's why the smaller values (ie. flatter) result to higher variances (remember you invert that). Practically all regression problems can be formulated as optimization problems; exploit that! :) $\endgroup$
    – usεr11852
    Feb 13 '16 at 6:04
  • $\begingroup$ Thank you! Is the post now correct in the statements about the matrix of cofactors, for example? $\endgroup$ Feb 13 '16 at 15:56

The variances of the parameter estimates change when one adds more explanatory variables to a regression model because this action affects the overall model fit. To this why this is the case remember that:

  1. You are concerned about the maximum likelihood estimates (MLE) of your linear model.
  2. The Hessian of the log-likelihood evaluated at the ML estimates is the observed Fisher information.
  3. The estimated standard errors are the square roots of the diagonal elements of the inverse of the observed Fisher information matrix.
  4. ML estimates and least squares (LS) estimates are identical given you have a response variable coming from an exponential family.

OK, let's say you have already $m_1$ model such that $y = \beta_0 + \beta_{1} x_{1} + \epsilon$ and you augment it with an additional explanatory variable $x_{2}$. Let's also assume that $x_{2}$ is not orthogonal to $y$ or $x_{1}$, (ie. $y \not\perp x_{2}$ and $x_{1} \not\perp x_{2}$ ). This is crucial because it means that when adding $x_{2}$ in the model $m_1$ we added some additional information about $y$. This increased amount of information will cause the sum-of-squares-error (SSE) to get smaller. (Geometrically speaking by incorporating $x_2$ in $m_1$ we add an additional non-orthogonal plane to project the sample $y$, thus the projections $\hat{y}$ are closer to original data $y$ than before; see this excellent answer by silverfish here for more details.)

Let's interpreter getting a smaller SSE means in terms of LS and then ML estimates: A smaller SSE means that the optimum (the vector of $\hat{\beta}$'s) of our LS function is different than before. This means that ML estimates are different (point 4). Which means that unless the Hessian of the log-likelihood function at this new solution (point 2) are exactly the same as before, the standard errors will be different too (point 3). Therefore adding a new variable $x_2$ is extremely unlikely to result the same standard errors.

Your second assessment about improved statistical significance is directly related to the above. The wording: "As long as I avoid multicollinearity adding more control variables will always improve the statistical significance of the parameter estimates." is slightly over-optimistic. The significance of a predictor will always get smaller if you have another predictor that is collinear with it. You can also say that the significance of the others predictors will never get smaller if the newly predictor is not collinear. Nevertheless, this last statement downplays how hard is to avoid even the slightest collinearity between two predictors while at the same time making sure that they are related to $y$. That could possibly be the case if one used PCA scores as predictors variables but this really escapes the scope of this answer.

OK, how about a code example in R. We will define a model m1 and then we will add we will add a variable x2 to it and see if this all holds.

N = 1000;
x1 = runif(N);
e = rnorm(N);
y = 3*x1 + e
m1 <- lm(y~x1) 

#  898.0496

x2 = runif(N) 
cor(x2,y)  # 0.02983 linear correlation 
cor(x2,x1) # 0.02598 linear correlation 

m2 <- lm(y~x1+x2) 
#   897.7952

So one immediately sees that even adding a variable $x_2$ that is pure noise, exactly because of some small spurious correlations he got a smaller SSE (which was what @whuber's comment essentially was about). Clearly the improvement in the SSE is marginally (after all we added just noise) but it is still there. What about the standard errors? Are they the same?

summary(lm1)$coefficients["x1", 'Std. Error']    # [1] 0.103062  
summary(lm2)$coefficients["x1", 'Std. Error']    # [1] 0.103134
summary(lm1)$coefficients["(Intercept)", 'Std. Error']    # [1] 0.0602753
summary(lm2)$coefficients["(Intercept)", 'Std. Error']    # [1] 0.0783329

Again, while the difference is marginal, the difference is there. The standard errors for $\hat{\beta_0}$ ((Intercept)) and $\hat{\beta_1}$ (x1) are different between the two models and that is because as the location of the optimum changed almost certainly the Hessian of the likelihood at the new optimum point is also different. In this example we also see that the standard errors are larger. This was not certain but it was likely. What about the statistical significance though?

summary(lm2)$coefficients["x1", 'Pr(>|t|)'] # [1] 9.3e-128
summary(lm1)$coefficients["x1", 'Pr(>|t|)'] # [1] 5.1e-128

summary(lm1)$coefficients["(Intercept)", 'Pr(>|t|)'] # [1] 0.4606
summary(lm2)$coefficients["(Intercept)", 'Pr(>|t|)'] # [1] 0.8192

Well the statistical significance in terms of $\beta_1$ is the same (differences of $3e-128$ are inconceivable). The statistical significance of the intercept $\beta_0$ is less and this exactly because the expected value of $x_2$ ($E\{x_2\}$) is not exactly $0$. As such some tiny variation around the intercept was explained because of $x_2$. If we had centred $x_2$ we would have the same or even higher significance for $\beta_0$ because the actual estimate of $\beta_0$ would be somewhat higher (assuming it's SE did not change proportionally too.)


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