Let's assume that we sample N times from a normal distribution with known mean and variance. With a generated sample we calculate sample mean and sample standard deviation. Then we divide sample mean by sample standard deviation (STD) to get the measure of my interest.

The question that I have: Is there an analytical expression for the distribution of the above described measure?

I know that distribution for the sample mean is well known. I have also found an expression for the distribution of sample STD. However, I cannot find the distribution that I need.

  • 5
    $\begingroup$ If "STD" is intended to be an abbreviation for standard deviation then the distribution will be a scaled version of a noncentral t distribution. A number of posts on site also discuss this distribution. If I don't locate a sufficiently close duplicate (I didn't with a quick look), and nobody posts a full answer in the meantime, I'll come back and make this more detailed. $\endgroup$
    – Glen_b
    Commented Nov 15, 2018 at 11:33
  • 1
    $\begingroup$ @Glen_b Relevant posts are stats.stackexchange.com/a/133274/919 (yours), stats.stackexchange.com/a/17288/919, and stats.stackexchange.com/a/160523/919. The last is very nearly an answer, but it does not explicitly give any expression for the distribution. $\endgroup$
    – whuber
    Commented Nov 15, 2018 at 13:22
  • $\begingroup$ @whuber I found a few others a bit like the first two but they didn't seem quite close enough to count as answers for this one (though all were relevant). $\endgroup$
    – Glen_b
    Commented Nov 15, 2018 at 14:54
  • $\begingroup$ This is just to point out that this is the reciprocal of the coefficient of variation. It's also been called signal to noise ratio, although that term doesn't appear to be uniquely defined. $\endgroup$
    – Nick Cox
    Commented Nov 22, 2018 at 0:50

2 Answers 2


Let $X_1,...,X_n \sim \text{IID N}(\mu, \sigma)$ be your data points. It is well known from Cochran's theorem that the sample mean and sample variance are independent with distributions:

$$\bar{X}_n \sim \text{N} \Big( \mu, \frac{\sigma^2}{n} \Big) \quad \quad \quad S_n^2 \sim \sigma^2 \cdot \frac{\text{Chi-Sq}(n-1)}{n-1}.$$

Hence, we can form the independent statistics:

$$Z_n \equiv \frac{\bar{X}_n - \mu}{\sigma / \sqrt{n}} \sim \text{N}(0,1) \quad \quad \quad \chi_n \equiv \frac{S_n}{\sigma} \sim \frac{\text{Chi}(n-1)}{\sqrt{n-1}}.$$

With a bit of algebra we then have:

$$\begin{equation} \begin{aligned} \frac{\bar{X}_n}{S_n} &= \frac{\bar{X}_n / \sigma}{S_n / \sigma} \\[6pt] &= \frac{\bar{X}_n / \sigma}{\chi_n} \\[6pt] &= \frac{1}{\sqrt{n}} \cdot \frac{\sqrt{n} \bar{X}_n / \sigma}{\chi_n} \\[6pt] &= \frac{1}{\sqrt{n}} \cdot \frac{\sqrt{n} (\bar{X}_n - \mu)/\sigma + \sqrt{n} \mu/\sigma}{\chi_n} \\[6pt] &= \frac{1}{\sqrt{n}} \cdot \frac{Z_n + \sqrt{n} \mu/\sigma}{\chi_n} \\[6pt] &\sim \frac{1}{\sqrt{n}} \cdot \text{Noncentral T} \big(\sqrt{n} \mu/\sigma, n-1 \big). \\[6pt] \end{aligned} \end{equation}$$

So you can see that the ratio of the sample mean on the sample standard deviation has a scaled non-central T distribution with non-centrality parameter $\sqrt{n} \mu/\sigma$ and degrees-of-freedom $n-1$. We can double-check this theoretical result empirically via simulation.

Checking the distribution by simulation: In the R code below we create a function to simulate $m$ samples of size $n$ from the IID normal model and generate the $m$ ratio statistics from these samples. We plot the kernel density of these simulated statistics against the theoretical distribution above in order to confirm that the theoretical result is correct.

#Simulate m values of the ratio statistic for samples of size n
SIMULATE <- function(m, n, mu, sigma) { 
              X <- array(rnorm(n*m, mean = mu, sd = sigma), dim = c(m,n));
              R <- rep(0, m);
              for (i in 1:m) { R[i] <- mean(X[i,])/sd(X[i,]); }
              R; }

#Plot the density of the simulated values against theoretical
PLOTSIM  <- function(m, n, mu, sigma) { 
              require(stats); require(ggplot2);
              RR    <- SIMULATE(m, n, mu, sigma);
              DENS  <- density(RR);
              DENS$yy <- dt(DENS$x*sqrt(n), df = n-1, ncp = sqrt(n)*mu/sigma)*sqrt(n);
              DATA <- data.frame(x = DENS$x, y = DENS$y, yy = DENS$yy);
              ggplot(data = DATA, aes(x = x)) +
                geom_line(aes(y = y), size = 1.2, colour = 'black') +
                geom_line(aes(y = yy), size = 1.2, colour = 'red', linetype = 'dotted') +
                theme(plot.title    = element_text(hjust = 0.5, size = 14, face = 'bold'),
                      plot.subtitle = element_text(hjust = 0.5)) +
                ggtitle('Density plot - Simulated Data') +
                labs(subtitle = 
                     paste0('(Sample size = ', n, ', Simulation size = ', m, ')')) +
                xlab('Sample Mean / Sample Standard Deviation') +
                ylab('Density'); }

#Generate example plot
m     <- 10^4;
n     <- 100;
mu    <- 12;
sigma <- 6;

PLOTSIM(m, n, mu, sigma);

enter image description here

  • $\begingroup$ thanks as usual @Ben! I ran the same simulation in Python and got identical results. Could you suggest the analytical formula for $mean(\frac{\mu}{\sigma})$ and $stdev(\frac{\mu}{\sigma})$ as a function of $\mu$ and $\sigma$ ? I am not familiar with the Noncentral T-distribution, I couldn't figure it out myself. $\endgroup$
    – elemolotiv
    Commented Nov 25, 2018 at 20:43
  • 1
    $\begingroup$ Just have a look at the moments of the noncentral T distribution. To get the moments for the above ratio, all you need to do is substitute the degrees-of-freedom $n-1$ and the noncentrality parameter$\sqrt{n} \mu / \sigma$. Good luck! $\endgroup$
    – Ben
    Commented Nov 25, 2018 at 21:42
  • $\begingroup$ @elemolotiv, would it be possible for you to share your Python code? I am asking since I am also working in Python. $\endgroup$
    – Roman
    Commented Nov 26, 2018 at 8:53

@Roman, here is my Python code, of @Ben's example above

  • it produces a file "walks.tsv" with N realisations of a random walk with mean=μ and stdev=σ.

  • for each realisation there is a row, with the sample mean, sample stdev, and sample mean/stdev.

  • I loaded "walk.tsv" in Excel and used Excel to aggregate and plot the data.

Hope it helps 🙂

import math, numpy.random

class RandomWalk:

    def __init__(self, step_mean, step_stdev, steps_per_walk):
        self.step_mean = step_mean
        self.step_stdev = step_stdev
        self.steps = steps_per_walk
        self.log = open("walk.tsv","w+")

    def realize(self):
        total = 0
        total_sqr = 0
        for i in range(self.steps):
            step = float(numpy.random.normal(self.step_mean, self.step_stdev, 1))
            total += step
            total_sqr += step * step
        self.sample_mean = total / self.steps
        self.sample_stdev = math.sqrt(total_sqr / self.steps - self.sample_mean * self.sample_mean)

    def log_names(self):

    def log_values(self):
        self.log.write("{:0.2f}\t".format(self.step_mean / self.step_stdev))
        self.log.write("{:0.2f}\n".format(self.sample_mean / self.sample_stdev))

def simulate(step_mean, step_stdev, steps_per_walk, walks):
    walk = RandomWalk(step_mean, step_stdev, steps_per_walk)
    for i in range(walks):

simulate(step_mean = 12, step_stdev = 6, steps_per_walk = 100, walks = 10000)

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