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I asked a question here about why if the ratios of two variables which are drawn from a normal distribution come out as a Student t-distribution. I was provided with some excellent explanations and resources.

Now I come to some more practical questions, say I repeat my ratio many times the resultant distribution will look like a Student t-distribution -- specifically a Cauchy as my degrees of freedom, $\nu$ , are one. So if I wanted to calculate the mean and the error in this mean from my data set, how would I do it?

With the usual $$\hat\mu_N = \frac{1}{N} \sum_{i=1}^{i=N} x_i$$ the error in this value coming from the standard deviation, $$\hat\sigma_N=\sqrt{\frac{1}{N}\sum_{i=1}^{i=N} \left( x_i - \mu \right)^2 }$$ divided by the square-root of the number of points -- so $\hat\mu_N \pm \hat\sigma_N/\sqrt{N}$.

Are there any special considerations for calculating the mean for a set of data that is Student t-distributed? I ask as I have read that no standard deviation or mean exists for the Cauchy distribution. Does this refer to the analytic expression of a mean as in $$\mu = E[X] = \int_{-\infty}^{+\infty} x f(x) dx$$ and $$\sigma = \sqrt{V(X)} = \sqrt{\int_{-\infty}^{+\infty} (x-\mu)^2 f(x) dx}$$ where $f(x)$ is the PDF?

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Q.P.
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    When the $X_i$'s are Cauchy rv's neither the empirical mean nor the empirical variance converge. The confidence interval has no validity either. – Xi'an Mar 06 '20 at 17:28
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    You would calculate the mean the same as always. But the "error of the mean" is undefined if you intend to use the sample mean to estimate the mean of the underlying distribution, simply because the distribution has no mean. (It's not even infinite: it is undefined.) But if you are using the sample mean to estimate some other property of the distribution, such as its median (which *is* defined), you are using a truly poor procedure for the reasons given by @Xi'an. A better question to ask, then, would be to explain *why* you are doing this analysis so we can propose workable procedures. – whuber Mar 06 '20 at 18:52
  • @whuber I'll vote to close this and ask a new one centred around the topic you suggest. – Q.P. Mar 07 '20 at 00:01
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    Sample means from a standard Cauchy have the same distribution as the original population of individual values. – Glen_b Mar 07 '20 at 10:46
  • For instance, [$\hat\sigma^2_N$ grows like $\text{O}_\text{P}(N)$.](https://www.jstor.org/stable/25050525?seq=1#metadata_info_tab_contents) – Xi'an Mar 07 '20 at 13:16

1 Answers1

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Here’s an R simulation to try to see that the mean doesn’t converge.

set.seed(2020)
Ns <- seq(2,10002,100)
xbars <- rep(NA, length(Ns))
for (i in 1:length(Ns)){
    xbars[i] <- mean(rt(Ns[i]), 1)
}
plot(Ns, xbars)

Perhaps simulate even larger samples.

What you should observe is that, even for large samples, the mean can be way different from.

The way I think about it, $t_1$ has so much density far out in the tails that the usual business about many observations offsetting an extreme value does not apply, because even as we have many observations, having many observations increases the chance of getting one that’s so far away from 0 that it wrecks what the others are doing. We have an equal chance of getting another extreme observation with the same sign as we do of getting an extreme observation with the opposite sign, so we can end up with either large positive sample means or large negative sample means.

This illustrates why Cauchy has no expected value, which makes its second central moment not exist, either: no mean or variance.

Dave
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