If I standardize a standard normal random variate , will it be still standard normal ? That is, if $X\sim N(0,1)$ , then can I do $$X^*=\frac{x-\bar x}{sd(x)}$$ ? and will $X^*\sim N(0,1)$ ?
In R code :
x <- rnorm(5)
scale(x)
It seems to me I am standardizing a standard normal, sounds like double standardization. Also I don't know whether it will retain the standard normal distribution.
If $X_i$ are iid Normal(0,1), then a sample from it won't have sample mean 0 or sample standard deviation 1 just due to random variation.
Now consider what happens when we do $Z=\frac{X-\overline{X}}{s_X}$
While we do now have sample mean 0 and sample standard deviation 1, what we don't have is $Z$ being normally distributed.
In small to moderate sample sizes, it has short tails, and substantially smaller kurtosis than a standard normal, Indeed from simulation for samples of size n=10 it looks pretty similar to a scaled beta(4,4) (that has been scaled to lie in (-3,3) ):
(The x-axis is a random sample of B(4,4) scaled to (-3,3). Of course this doesn't mean the distribution shape is a beta(4,4). -- Edit: as Henry points out, it is in fact a Beta(4,4).)
The values in res were generated as follows:
res=replicate(100000,scale(rnorm(10)))
For samples of size 5, the result looks rather like a scaled beta(3/2,3/2).
Further, the values in each sample are no longer independent, since they sum to 0 and their squares sum to $n-1$
We have that
$$X_i^* = \frac{X_i}{s} - \frac{\bar X}{s}$$
The sample variance from a normal sample follows an exact distribution,
$$(n-1)s^2/\sigma^2\sim\chi^2_{n-1} \implies s^2 \sim \frac{1}{n-1}\chi^2_{n-1} \implies s \sim \frac{1}{\sqrt{n-1}}\chi_{n-1}$$
i.e. $s$ follows the square root of a chi-square divided by its degrees of freedom.
But even if this means that $\frac{X_i}{s}$ is the ratio of a standard normal over the square root of a chi-square divided by its degrees of freedom, the numerator is not independent of the denominator, and so we cannot say that the ratio follows a Student's $t$-distribution (and personally I do not know its distribution).
As for the second term, it is known that the sample mean and the sample variance are independent random variables if and only if the sample consists of independent normals, which is the case here.
Furthermore, the sample mean follows a zero-mean normal distribution with variance here $1/n$, so $\sqrt{n}\bar X$ follows a standard normal.
So we have that $$\frac{\sqrt {n} \bar X}{s} \sim t \implies \frac{\bar X}{s} \sim \frac{1}{\sqrt {n}}t $$
i.e. the second term of $X_i^*$ follows a scaled student's $t$-distribution
So in all
$$X^*_i = \frac{Z_i}{\sqrt{\chi^2_{n-1}/(n-1)}} - \frac{1}{\sqrt {n}}t$$
where I have used the symbol $Z$ to denote a random variable following a standard normal. The first term is not a Student's $t$, and moreover, it is not independent from the second term. Put together it doesn't look much of a normal or of a Student's distribution either.
The original standard normal variables have TRUE mean 0 (E(X) = 0) and are independent. By taking a set of them and dividing them by their standard deviation, you DO standardize them, but the result, ironically, isn't standard normal. They are dependent (because they share the denominator) and actually have t-distributions. So if you want standard normal, just stick with rnorm(5).
plot(c(0,2.8),c(0,4),type="n"); jk=replicate(10000, {x=rnorm(5);num=x-mean(x);points(sd(x),num[1])})
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Just did some experiments. It seems after scale again, you are closer to get some data with $\mu=0$ and $\sigma=1$.
set.seed(123)
x <- rnorm(1000,0,1)
mean(x)
sd(x)
y<-scale(x)
mean(y)
sd(y)
Results:
> mean(x)
[1] 0.01612787
> sd(x)
[1] 0.991695
> y<-scale(x)
> mean(y)
[1] -8.235085e-18
> sd(y)
[1] 1
There are already some general answers that cover the question, but personally I find the following reasoning most easy to follow.
Your definition of $X^*$ is as follows
$$X^*=\frac{x-\bar x}{sd(x)}$$
Because the sample size is 1, we have $\bar x = x$, so for any $x$ the expression reduces to
$$X^*=\frac{\bar x-\bar x}{sd(x)} =\frac{0}{0}$$
As $X^*$ is clearly not normally distributed for sample size 1, it can definitely not have a standard normal distribution in general.
