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I would like to test a linear restriction in R. Instead of the usual $\beta_i=0$, I want to test if $\beta_k=0.5$ and $\beta_j=-0.5$.

Is there a way to do this using lm command, and just writing a different formula for the models that lm command uses?

I was thinking of writing $y-0.5x_k+0.5x_j=\sum_{i\neq j,k} x_i\beta_i+(\beta_k-0.5)x_k+(\beta_j+0.5)x_j$. However, I'm not sure how to code this reparametrization...

P.S:Would writing $y−I(−0.5x_k+0.5x_j)$ ~ $\sum x_i$ as the formula to be run on lm, work as 2 t-tests?

Any help would be appreciated.

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I think your current parameterization is sufficient to test either coefficient restriction (via the t-tests for each coefficient), but not both restrictions at the same time. To do that one can conduct an F-test between the unrestricted and the restricted model (since it is nested).

Example below in R using the car package.

library(car)
set.seed(10)
x1 <- rnorm(100)
x2 <- rnorm(100)
x3 <- rnorm(100)
y <- 5*x1 + 0.5*x2 - 0.5*x3 + rnorm(100)
m1 <- lm(y~x1+x2+x3)
linearHypothesis(m1,c("x2 = 0.5","x3 = -0.5"),test="F")

Which then spits out a table with the RSS for each model and the degrees of freedom, needed to calculate the F-statistic.

Linear hypothesis test

Hypothesis:
x2 = 0.5
x3 = - 0.5

Model 1: restricted model
Model 2: y ~ x1 + x2 + x3

  Res.Df    RSS Df Sum of Sq      F Pr(>F)
1     98 107.21                           
2     96 107.10  2   0.10752 0.0482  0.953

You answered you own question in terms of testing individual coefficients, you can transform the variable on the left hand side. To replicate the linearHypothesis function I trick R here by making the transformed variable:

#equivalent F test - can trick R by making y a new variable
y <- y -0.5*x2 +0.5*x3
m2 <- lm(y~x1)
anova(m2,m1)

Which reproduces the earlier table. You can use this same trick to get the default summary t-test's in the regression output to test against the alternative hypotheses:

#test for either coefficient restriction (with updated y)
m3 <- lm(y~x1+x2+x3)
summary(m3)
#you can see m3 is equivalent to m1 - just changes the location of the test
#anova(m1,m3)

Which produces for coefficient estimates:

Coefficients:
             Estimate Std. Error t value Pr(>|t|)    
(Intercept)  0.208282   0.107351   1.940   0.0553 .  
x1           5.045149   0.112973  44.658   <2e-16 ***
x2           0.033042   0.110693   0.299   0.7660    
x3          -0.004875   0.110122  -0.044   0.9648   

To test with the original values, user2864849 answers that question. You just subtract out the null from the point estimate and use the same standard error. So using the coefficients from m1 you can see below t_est reproduces the t-statistic in model 3.

#to see how it is just the original t-test with the location changed for x2
pt_est <- summary(m1)$coefficients[3,1] #point estimate x2
se_est <- summary(m1)$coefficients[3,2] #standard error x2
t_est <- (pt_est - 0.5)/se_est          #t-stat for null of 0.5
t_est
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  • $\begingroup$ Andy, thanks for your answer. If I was interested in doing the t-tests, how would I do it in R? I do not know how to code that part as a formula for the lm command. This is what I'm more interested about. $\endgroup$ – An old man in the sea. May 9 '16 at 14:22
  • $\begingroup$ Andy, should I write '$y-I(-0.5x_k+0.5x_j) $ ~ $ \sum x_i$' as the formula to be run on lm? $\endgroup$ – An old man in the sea. May 9 '16 at 15:56
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    $\begingroup$ Yep - you already answered your own question if that is the case. $\endgroup$ – Andy W May 9 '16 at 16:54
  • $\begingroup$ Hi, is there a restriction test for time series models such as GARCH model? I cannot apply this 'linearHypothesis' onto my GARCH model at all. $\endgroup$ – Eric Mar 5 '18 at 22:38
  • $\begingroup$ I presume you can construct whatever hypothesis test manually using the parameter estimates and the variance/covariance matrix as the other answers have illustrated. $\endgroup$ – Andy W Mar 6 '18 at 12:32
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Let's say you want to test the hypothesis of the form $R\beta = r$. In your case: $$\underbrace{\left[ \begin{array}{cccc} \ldots & 1& 0 & \ldots \\ \ldots & 0& 1 & \ldots \end{array}\right]}_R \underbrace{\left[ \begin{array}{c}\ldots \\ \beta_j \\ \beta_k \\ \ldots \end{array} \right]}_\beta = \underbrace{\left[ \begin{array}{c}-.5 \\ .5 \end{array} \right]}_r $$

Let $\hat{\beta}$ be your regression estimates of $\beta$. Under the condition that $\hat{\beta} \sim \mathcal{N}\left(\beta, \Sigma \right)$ conditional on data $X$ (eg. $\hat{\beta}$ asymptotically normal with mean $\beta$ and covariance matrix $\Sigma$) then the linear restrictions $R \beta = r$ can be tested with a $\chi^2$ test. Observe that $R\hat{\beta}-r$ would be normal with $Var(R \hat{\beta} - r\mid X) = R \Sigma R'$. Then we would have: $$(R\hat{\beta} - r)' \left( R \Sigma R' \right)^{-1} (R \hat{\beta} - r) \sim \chi^2_{\#r}$$ Where $\#r$ is the number of restrictions (in your case two).

The most classic way though is to do an F-test. This would take the form: $$(R\hat{b} - r)' \left(R \Sigma R' \right)^{-1} (R\hat{b} - r) / (\#r)\sim F(\#r, n- k) $$ where $\Sigma$ is estimate of $Var(\hat{\beta} \mid X)$ and $\#r$ is the number of restrictions (note rank of R should be $\#r$.)

See Fumio Hayashi Econometrics p. 65-66

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You can test the hypothesized value of $\beta_k$ much in the same way that it would be tested for $\beta_k = 0$. You can get a t-value on n-2 df with the formula:

$$T = \frac{\beta_i- hypothesezed~~ \beta_i}{Standard~~ Error~~ \beta_i}$$

In many cases this simplifies to beta over standard error because making the hypothesis be zero removes it from the equation.

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