I'm wondering how random variables can be analyzed using parametric methods if the distribution is not normal.

For example if a variable Y is normal distributed but I'm interested only on values Y2 which are lower than a certain limit due to agreement.

enter image description here

Can I use the common statistical tools to compute estimation, standard deviation, standard errors and P-values if the sample is large?

To clarify my question I created the following theoretical population:

# 100,000 values with mean 300 and sd = 50
n <- 100000
y <- rnorm(n,300,50)

# Only y < 300 are of interest
y2 <- y[y<300]
length(y2)     # [1] 49,828

hist(y, breaks=seq(0,max(y)+10,10))
hist(y2, breaks=seq(0,max(y)+10,10))

# mean and sd of y
mean(y)        # 300.1465
median(y)      # 300.2244
sd(y)          # 49.97292

# mean and sd of the interested values (y2)
mean(y2)       # 260.1354
median(y2)     # 266.3348
sd(y2)         # 30.04559

Now, I make an experiment. I take a sample of $s=100$:

# Experiment:
y.exp <- sample(y,s)

# Only y<300 are of interest (for example due to a game rule)
y.int <- y.exp[y.exp<300]
mean(y.int)     # 259.3562
sd(y.int)       # 32.56482
length(y.int)   # 38

I'm interest in the following hypothesis:
$H_0: \mu_0 = 250$
$H_{alt}: \mu_0 \ne 250$

Assuming that the sample is large enough I do a t-test:

# t-test:
mu0 <- 250
t.test(y.int,mu=mu0)  # t = 1.7711, df = 37, p-value = 0.08478

Here are my questions:

  1. If a sample size is large enough can I use the common statistical tools like t-test, linear regression, etc. (maybe supported by the law of large numbers)?
  2. Are there alternative methods which consider special distributions (like this truncated normal distribution). I know there are non-parametric methods or truncated regression. I'm interested in parametric methods. Truncated regression answer another question: what is it the mean value of the normal distribution if the data a truncated. I'm interested in the mean value of the truncated data.
  3. Does standard deviation makes sense if the underlying distribution is not normal?
  • 2
    $\begingroup$ The central limit theorem provides approximate test procedures (your point 1). Regarding point 2, you can try to develop exact methods for any kind of distribution but in some cases you won’t be able to get rid of nuisance parameters... Finally, standard deviation is generally interpreted in the context of approximately normal data, yes, but it is still useful in general (e.g. $\sigma/\sqrt n$ is your estimate of the standard deviation of the empirical mean). $\endgroup$
    – Elvis
    Feb 3, 2017 at 10:55
  • $\begingroup$ @Elvis Thanks. My main concern about non normal residuals was that inference (through standard errors, t-value, confidence intervals, P-value) was not accurate or correct. So, obviously, thanks to central limit theorem I can use the P-values from a regression or t-test to do inference provided sample size is large enough. This is a good thing. I would be interested on "developing exact methods". Can you give me a hint (keyword, reference)? (You could put your comment as answer). $\endgroup$
    – giordano
    Feb 3, 2017 at 16:39
  • 2
    $\begingroup$ Yes, regression for example is remarkably robust to departures from the Gaussian assumption. "Developing exact methods" refers to try to reproduce what has been done with the Gaussian distribution but with other distributions... $\endgroup$
    – Elvis
    Feb 3, 2017 at 17:18

1 Answer 1


As a complement to the short answers made in remarks, I'll illustrate the behavior of linear regression with half-normal error terms.

Here is a function for generating half-normal variables with specified mean and standard deviation.

rhnorm <- function(n, mean = 0, sd = 1)
  mean + sd/sqrt(1-2/pi)*(abs(rnorm(n))-sqrt(2/pi))

Let's simulate a simple linear model and analyze it with lm

> X <- runif(1000)
> Y <- 2 + 3*X + rhnorm(length(X), sd = 2)
> summary(lm(Y~X))

    Min      1Q  Median      3Q     Max 
-2.6706 -1.6689 -0.4588  1.1634  8.4628 

                Estimate Std. Error t value Pr(>|t|)    
(Intercept)   2.0377     0.1251   16.29   <2e-16 ***
X             2.9564     0.2230   13.26   <2e-16 ***
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 2.032 on 998 degrees of freedom
Multiple R-squared:  0.1497,    Adjusted R-squared:  0.1489 
F-statistic: 175.7 on 1 and 998 DF,  p-value: < 2.2e-16

As you see, the coefficients are well estimated, as well as the residual error, and of course the residuals. The histogram of the residuals is skewed, which would wrongly lead many users to consider that they should not use linear regression (and many reviewers to tell research paper authors that they should not!).

enter image description here

Let's explore further with simulation. First, repeat the above simulation $10^5$ times. Theory tells that the estimate of the X coefficients should be normal, with mean 3 and standard deviation $2/(\sqrt n \sigma_X)$, in our case $0.22$.

> a <- replicate( 1e5, { Y <- 2 + 3*X + rhnorm(length(X), sd = 2);
                         summary(lm(Y~X))$coefficients[2,]  } )

> mean( a[1,] )
[1] 2.999933
> sd( a[1,] )
[1] 0.2186882
> qqnorm(a[1,]); abline(3, .22, col="red")

enter image description here

I fail to see any problem! We can do similar simulations under the null hypothesis to be sure that there’s no gross type I error issue.

> b <- replicate( 1e5, { Y <- 2 + rhnorm(length(X), sd = 2);
                         summary(lm(Y~X))$coefficients[2,]  } )
> mean( b[1,] )
[1] -0.0002721078
> sd( b[1,] )
[1] 0.2186087

Let's so do a qq-plot of the p-values, in normal scale and in $(-\log_ {10})$-scale to have a better view on small p-values.

> par(mfrow=c(1,2))
> plot(ppoints( 1e5) , sort(b[4,]))
> plot(-log10(ppoints( 1e5)) , -log10(sort(b[4,]))); abline(0,1,col="red")

enter image description here

Definitely nothing’s wrong happening.

I am a lecturer, and I have tried this with my students with several distribution of the error terms. It seems that the only way to get serious problems is to use pathologically heavy tailed error terms, such as $t(1)$ (or $t(2)$ but I am not even sure of this one — try if you’re curious!).

PS Thinking about it again a little later... we have $\hat\beta = (X'X)^{-1}X'Y$. As long as $X'Y$ is approximately (multivariate) normal, everything is ok; however if some columns of $X$ are indicator functions, this could be fail e.g. if there only a few ones in the column, or if a linear combination of the columns has the same property.

  • 2
    $\begingroup$ nice answer (+1). Just an observation, why using plot(... instead than directly qqplot(runif(10^5), b[4, ]) to produce the penultimate plot? $\endgroup$
    – DeltaIV
    Feb 4, 2017 at 11:54
  • 1
    $\begingroup$ @DeltaIV I want theoretical quantiles on the x-axis, not empirical quantiles from another uniform sample... this would not change the result much, though, but still. An other solution would be to use car::qqPlot but I took the habit to use ppoints. $\endgroup$
    – Elvis
    Feb 4, 2017 at 12:19
  • $\begingroup$ @Elvis I was stucked in doing residuals analysis. Interestingly, residual analysis with large number of observations would never met the assumptions (normal distributed residuals, constant variance). This is not necessary thanks to central limit theorem. Would be interesting where are the threshold. Thanks for this simulation and your statement about users and reviewers. It makes me feel more self-confident. I suppose that this is valid also for mixed models. $\endgroup$
    – giordano
    Feb 4, 2017 at 15:32
  • $\begingroup$ @Elvis I tried with t-distribution. Here the results (se from summary/sd(b[1,])): df=2: 0.317/ 0.451; df=1: 1.956/ 4986.8; df=3: 0.189/ 0.190. Obviously the long-tailed distribution (df=1 and 2) does not fulfill central limit theorem. $\endgroup$
    – giordano
    Feb 5, 2017 at 15:17
  • 2
    $\begingroup$ @giordano Well yes, to use central limit theorem you need a distribution with a finite variance... Everything will be fine as long as X'Y is approximately normal in fact. $\endgroup$
    – Elvis
    Feb 5, 2017 at 15:18

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