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In the report http://www.stat.berkeley.edu/~breiman/wald2002-3.pdf Breiman says:

Three decades ago many statisticians and quantitative social scientists were enamored of multilinear regression and its theory of hypothesis testing on the coefficients. Every statistical package had a regression program variable selection program based on F-to delete and F to enter. It was almost impossible to get a paper published unless you showed that a certain coefficient was signifigant at the 5% level. This was regardless of how well the linear model fit the data and little effort was made to find out. Many conclusions were undoubtedly wrong, and I don't think statisticians now-a-days dispute the error of these ways

I would like to have an example obtained with simulated data where a linear regression show that a coefficient is significant while it is not true. And I would like that the error will arise from the fact that the linear model is not fitting well the data.

EDIT: R or python code is welcome

EDIT: The example should make use of at least 1000 samples

EDIT: I found a possible example with simulated data.

      n = 1000
x1 = seq(-10, 10, length.out = 1000)
x2 = sin(x1)/x1 + rnorm(n = n, mean = 0, sd = 2)
y = sin(x1)/x1 + rnorm(n = n, mean = 0, sd = .1)
lr = glm(y~x1+x2) #wrong model
summary(lr)
z = sin(x1)/x1
lr = glm(y~z+x2) #real model
summary(lr)

plot(x1, y, col='blue')
points(x1, x2)
points(x1, z)
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  • $\begingroup$ "The example should make use of at least 1000 samples"... so is this homework? Otherwise, why 1000 as a specific lower limit? $\endgroup$
    – Glen_b
    Commented Mar 4, 2015 at 15:12
  • $\begingroup$ @Glen_b not it is not homework. It is a way to avoid example that are not if interest for me $\endgroup$
    – Donbeo
    Commented Mar 4, 2015 at 15:13
  • $\begingroup$ Why would 1000 be of interest, but 900 not be of interest? $\endgroup$
    – Glen_b
    Commented Mar 4, 2015 at 15:14
  • $\begingroup$ The paragraph starting with I would like to... is not completely clear, judging from the discussion in the comments below my answer. Maybe you could specify more exactly what you are looking for? $\endgroup$ Commented Mar 4, 2015 at 15:14
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    $\begingroup$ The logic of this request is mysterious, because it is problematic to ascribe a truth value to the "significance" of a coefficient in a linear regression which is understood to be an incorrect model. Suppose the linear regression indicates a coefficient is significant. That could--and often does--happen because the corresponding variable is capturing, albeit imperfectly, some real form of variation in the data. We should not be so hasty to equate "imperfect" and "not true," however. $\endgroup$
    – whuber
    Commented Mar 4, 2015 at 17:51

2 Answers 2

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Try a regression of $y=e^{2x}$ on $x$ for $x$ distributed uniformly in the interval $[0,1]$.

In R it can be coded as follows:

set.seed(1)
x=runif(10^3)
y=exp(2*x)
lm1=lm(y~x)
summary(lm1)

The true model is nonlinear, the estimated model is linear but the coefficient is highly significant.

Update: An example of spurious correlation between integrated processes:

set.seed(1); y=cumsum(rnorm(10^3))
set.seed(2); x=cumsum(rnorm(10^3))
lm1=lm(y~x)
summary(lm1)

The two variables $y$ and $x$ are unrelated but there is a spurious correlation between them, and the regression coefficient is highly significant. Many real world examples can be found here.

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  • $\begingroup$ Thanks for answering. In this case the linear model is able to correctly classify x as a relevant variable even if the relation of x and y is not linear. I would like a situation where x seems relevant while it is not or when it does not seem relevant while it is. $\endgroup$
    – Donbeo
    Commented Mar 4, 2015 at 14:40
  • $\begingroup$ Then spurious correlations from unrelated integrated time series is the answer. Here are bunches of examples where completely irrelevant regressors would be statistically significant. $\endgroup$ Commented Mar 4, 2015 at 14:44
  • $\begingroup$ I am familiar with these examples. The spurious correlation there is caused mainly by the lack of data. While I am looking for error due to the use of the wrong model. $\endgroup$
    – Donbeo
    Commented Mar 4, 2015 at 14:46
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    $\begingroup$ I am afraid spurious correlations between integrated processes are not due to lack of data. t-statistics have different distributions there so the phenomenon remains even in the asymptotics. $\endgroup$ Commented Mar 4, 2015 at 14:54
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    $\begingroup$ I am curious about why someone down voted this answer.. $\endgroup$ Commented Mar 4, 2015 at 15:18
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I found a possible example but it is a bit artificial and maybe it would be difficult to find something similar in real life

suppose $y = e^{x_1}+err$ and $x_2 =e^{x_1}+ERR$ where $ERR$ and $err$ are two measurement errors and $ERR>>err$. In this case the model $y=x1+x_2$ will consider $x_2$ as relevant variable while the model $y=e^{x_1}+x_2$ will consider $x_1$ as relevant.

> n = 1000
> x1 = seq(-10, 10, length.out = 1000)
> x2 = exp(x1) + rnorm(n = n, mean = 0, sd = 4)
> y = exp(x1) + rnorm(n = n, mean = 0, sd = 1) #y is a function of x1 only
> lr = glm(y~x1+x2) #the wrong model is used
> summary(lr)

Call:
glm(formula = y ~ x1 + x2) 

Deviance Residuals: 
     Min        1Q    Median        3Q       Max  
-11.9787   -2.8482    0.0355    2.7221   11.9990  

Coefficients:
              Estimate Std. Error   t value Pr(>|t|)    
(Intercept) -2.347e-01  1.383e-01    -1.697    0.090 .  
x1          -1.056e-02  2.610e-02    -0.404    0.686    
x2           1.000e+00  4.521e-05 22116.688   <2e-16 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

(Dispersion parameter for gaussian family taken to be 16.60931)

    Null deviance: 1.1126e+10  on 999  degrees of freedom
Residual deviance: 1.6559e+04  on 997  degrees of freedom
AIC: 5652.8

Number of Fisher Scoring iterations: 2

> lr = glm(y~exp(x1)+x2)
> summary(lr)

Call:
glm(formula = y ~ exp(x1) + x2) #the real model is used

Deviance Residuals: 
    Min       1Q   Median       3Q      Max  
-3.2881  -0.6651   0.0234   0.6947   3.1845  

Coefficients:
             Estimate Std. Error t value Pr(>|t|)    
(Intercept) -0.003736   0.033282  -0.112    0.911    
exp(x1)      0.990101   0.007912 125.147   <2e-16 ***
x2           0.009886   0.007911   1.250    0.212    
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

(Dispersion parameter for gaussian family taken to be 0.99421)

    Null deviance: 1.1126e+10  on 999  degrees of freedom
Residual deviance: 9.9123e+02  on 997  degrees of freedom
AIC: 2837.1

Number of Fisher Scoring iterations: 2

> 

EDIT:

A very similar but possibly better example

n = 1000
x1 = seq(-10, 10, length.out = 1000)
x2 = sin(x1)/x1 + rnorm(n = n, mean = 0, sd = 2)
y = sin(x1)/x1 + rnorm(n = n, mean = 0, sd = .1)
lr = glm(y~x1+x2) #wrong model
summary(lr)
z = sin(x1)/x1
lr = glm(y~z+x2) #real model
summary(lr)

plot(x1, y, col='blue')
points(x1, x2)
points(x1, z)
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  • $\begingroup$ This example does not satisfy the requirement in the original post. Finding a significant $x_2$ in the regression $y=\beta_0+\beta_1 x_1+\beta_2 x_2+\epsilon$ makes perfect sense as $y=0+0 \cdot x_1+1 \cdot x_2+\epsilon$ where $\epsilon=err-ERR$. Thus you actually have a correct model with one extra irrelevant regressor which is found to be insignificant just as it should be. This violates coefficient is significant while it is not true if I decipher the it is not true correctly. However, it is not true should be treated with care (see e.g. a valid comment by whuber under the OP). $\endgroup$ Commented Mar 4, 2015 at 21:19

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