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It is well known that if $x \sim \mathcal{N}(0, \sigma^2)$ and we have a sample size of $n$ observations of $x$, the distribution of $\frac{x}{\hat{\sigma}}$, where $\hat{\sigma}$ is the sample variance, will follow the Studnent-t law with $n-1$ degrees of freedom.

But what if the ML estimator of $\sigma$ is functionally penalized, i.e. $\sigma = f(\theta)$, and the estimator of $\sigma$, $\hat{\sigma}(\theta)$, can be only obtained numerically by means of MLEs of $\theta$? In that case, is there anything we can say about the distribution of $\frac{x}{\hat{\sigma}(\theta)}$?

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  • $\begingroup$ Could you elaborate on what you mean by "functionally penalized" and how that might affect $\hat\sigma$ compared to the usual MLE of $\sigma$? $\endgroup$
    – whuber ♦
    Commented Nov 15 at 15:49
  • $\begingroup$ What is $\theta$? What is $n$? (You haven't defined things in such a way that $n$ occurs.) $\endgroup$
    – Christian Hennig
    Commented Nov 15 at 15:56
  • $\begingroup$ Is $x$ one of the observations in the sample, say $X_1$? If so, it is not true that $X_1$ divided by the sample standard deviation is Student-t (since $X_1$ is not independent of $\sum_{i=1}^n (X_i -\bar X)^2$). $\endgroup$
    – Jarle Tufto
    Commented Nov 15 at 17:17
  • $\begingroup$ $f(\theta)$ is any function. By penalization I meant that estimates of $\sigma$ that agree with sample variances might no be obtainable with $f(\theta)$ (i.e. there is no such $\theta$ that $f(\theta)$ equals sample-variance). $\endgroup$
    – Geo
    Commented Nov 15 at 17:43
  • $\begingroup$ If indeed $f$ is "any function," then we can't say anything whatsoever about $\hat\sigma(\theta),$ and we are still left wondering how $x$ might be related to the sample, if at all. $\endgroup$
    – whuber ♦
    Commented Nov 15 at 21:21

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First, what follows a t-distribution is not $x/\hat\sigma$ (which would be a random vector, which cannot value a univariate distribution), but, for $\mu=0$, $$ \sqrt{n}\frac{\bar x}{\hat\sigma} $$ Second, it does, under normality follow a t-distribution with $n-1$ degrees of freedom, not $n$.

Third, see e.g. Proof that the coefficients in an OLS model follow a t-distribution with (n-k) degrees of freedom for a proof of why the t-statistic follows a t-distribution under the stated assumptions (you can specialize the proof for the linear regression case to the present case by considering a regression on a constant).

What "drives" that proof is, among other things, that, under multivariate normality, uncorrelatedness and independence are the same thing. Now, uncorrelatedness is, as far as I can see, a property quite specific to least squares based statistics, so that I would not jump to the conclusion that the result goes through for other estimators of the variance.

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    $\begingroup$ Your link goes to this same question! $\endgroup$
    – kjetil b halvorsen ♦
    Commented Nov 15 at 13:09
  • $\begingroup$ Was this answer written by ChatGPT? Just in case it is not: 1, 2. Thank you for your valuable corrections. However, I would like to point out that the distribution of $\sqrt(n) \bar(x)/\hat(\sigma)$ is the same distribution as $x/\hat(\sigma)$. OK, I did not mention that the data is iid., but merely specified a marginal, which is a very common abuse of a notation. 3. I know the derivation of the t-distribution. I also never did a jump the conclusion you mention. The question is exactly about that: can we hope, in a general case, to derive a distribution of the $x/\hat{\sigma}(\theta)$ $\endgroup$
    – Geo
    Commented Nov 15 at 15:26
  • $\begingroup$ Thanks, Kjetil, evidently I copied the link from the wrong tab; edited. Geo, I am not sure the strategy of conversing with people who donate their time like this will prove productive on this site. Also please refer to the comments of Christian and Jarle who raise points as to why your abuse of notation is not really as innocuous here as you argue. I also cannot read the typesetting in your comment. As to your question, I meant to say that the same exact finite sample distribution will likely be unavailable, and under the generality you specify, I very much doubt you can find any other. $\endgroup$
    – Christoph Hanck
    Commented Nov 15 at 18:31
  • $\begingroup$ In large samples, if the variance estimator is consistent, a Slutzky argument may be available to get asymptotic normality of the t statistic $\endgroup$
    – Christoph Hanck
    Commented Nov 15 at 18:31

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