### Twosday, February 22, 2022

Getting to know t-distributions

This is another post in the spirit of: "get to know your friendly neighborhood distributions." Today, we consider the t-distribution, also known as the Student's t-distribution.

Can I interest you in a blog post on that?

Maybe you've seen t-distributions around but haven't gotten to know them? Maybe you've seen or used t-tests but don't have a great idea of what's so t-distributed about them? Well, stick around if you like.

The probability density function (pdf) for a standard t-distribution is given by:

$$p(x) = \frac{\Gamma((\nu + 1)/2)}{\sqrt{\nu \pi}\Gamma(\nu/2)}(1 + \frac{1}{\nu}x^2)^{-(\nu+1)/2}$$

with parameter $$\nu$$ called the degree(s) of freedom. (Note: $$\nu$$ here looks a lot like $$v$$ but is its own Greek letter "nu".)

The more general t-distribution with additional location and scale parameters $$\mu$$ and $$\sigma$$ is:

$$p(x) = \frac{\Gamma((\nu + 1)/2)}{\sqrt{\nu \pi \sigma^2}\Gamma(\nu/2)}(1 + \frac{1}{\nu}(x-\mu)^2/\sigma^2)^{-(\nu+1)/2}$$.

I'm writing both these pdfs so that you have them for reference, but you can move back and forth between them by noting that:

$$x \sim t_v(0,1) \iff (x-\mu)/\sigma \sim t_v(\mu, \sigma)$$.

If you just see $$t_\nu$$ without the additional two parameters, you can assume it's the standard t-distribution.

Here's a plot of the pdf in the standard case, when $$\mu=0, \sigma=1$$. You can adjust the degrees of freedom, as well as compare the pdf to that of a standard normal distribution if you like.

degrees of freedom Include normal pdf for comparison

It looks somewhat similar to the normal density function, in that it's generally bell-shaped. Looking a little further, we might notice that 1) the higher the degrees of freedom, the closer to normal it seems to be, and 2) compared to the normal distribution, the t-distribution has heavier tails: the distribution gives far-from-the-mean values more probability than the normal distribution.

Note that even as the degrees of freedom get large, and thus the t-distribution gets more similar to a normal distribution, the t-distribution will still have heavier tails. That is, if we go far enough from the mean, the value of the pdf of the t-distribution will be larger than that of the normal. Why? Remember that the standard normal distribution has density proportional to $$e^{-x^2}$$, so as $$x$$ gets large the density approaches zero in an exponential manner. That means it will approach zero for large $$x$$ faster than the t-distribution density, which looks roughly like $$x^{-(\nu + 1)}$$. In fact there are technical definitions for heavy-tailed; these definitions (confusingly) vary a bit depending on the community, but one definition is having tails that are not exponentially bounded. The t-distribution is heavy-tailed in this technical sense.

Here are some ways that t-distribution show up.

Normal divided by the square root of a $$\chi^2$$: If $$Z$$ distributed as a standard normal and $$Y$$ as a $$\chi^2_\nu$$, and $$X,Y$$ are independent, then the random variable $$X = \frac{Z \sqrt{\nu}}{\sqrt{Y}}$$ follows a t-distribution with $$\nu$$ degrees of freedom. Click for more justification.

We can show this using a few things about transforming probability distributions and remembering gamma integrals. Let $$g$$ be the transformation $$g(z,y) = \frac{z\sqrt{\nu}}{\sqrt{y}},y)$$ Then $$g^{-1}(x,y)=(\frac{x\sqrt{y}}{\sqrt{\nu}}, y)$$. We'll need the determinant of the Jacobian of $$g^{-1}$$, which calculate to be $$\sqrt{y/v}$$. We'll also use (without proving) the result that $$\int_0^\infty e^{-by}y^{a-1} \, dy = \Gamma(a)b^{-a}$$ (the normalizing constant for a gamma distribution). Now: \begin{aligned} p_X(x) &= \int_0^{\infty} p_{X,Y}(x,y) \, dy \\ &= \int_0^{\infty} p_{Z,Y}(\frac{x\sqrt{y}}{\sqrt{\nu}}, y) \sqrt{y/v} \, dy, & \text{(transforming the distribution} )\\ &= \int_0^{\infty} p_{Z}(\frac{x\sqrt{y}}{\sqrt{\nu}})p_Y(y) \sqrt{y/v} \, dy \, & \text{(independence of Z, Y} )\\ & \propto \int_0^{\infty} e^{-x^2y/2\nu} y^{\nu/2 - 1} e^{-y/2} \sqrt{y} \, dy \, & \text{(Z is normal, Y is } \chi^2 \text{)}\\ & \propto \int_0^{\infty} e^{-((1 + x^2/\nu)/2)y} y^ {(\nu+1)/2 - 1} \, dy & \text{(combining like)} \\ & = \Gamma(\frac{\nu+1}{2})((1 + x^2/\nu)/2)^{-\frac{\nu + 1}{2}} & \text{(using gamma integral)} \\ \end{aligned} We can recognize this last line as being proportional to the pdf for a t-distribution (with all proportionality constants above a function of $$\nu$$).

This fact and an analogous justification goes through even if $$Y$$ merely follows a scaled $$\chi^2_{\nu}$$ distribution — i.e. that $$cY \sim \chi^2_\nu$$ for some nonzero constant $$c$$.

That might seem like a niche fact, but hear me out. If we have a bunch of independent normal distributions then 1) their mean is also normally distributed and b) the sum of their squares follows a $$\chi^2$$ distribution. This might make us a little suspicious that $$\frac{\bar{x} - \mu}{\sqrt{s^2/n}}$$ where $$s^2$$ is the sample variance $$s^{2}=\frac {1}{n-1}\sum _{i=1}^{n}(x_{i}-\bar {x})^{2}$$ looks a little like ratio $$\frac{Z \sqrt{\nu}}{\sqrt{Y}}$$ above. That's oversimplifying - but it does turn out to be possible to show that the $$\sum _{i=1}^{n}(x_{i}-\bar {x})^{2}$$ has a scaled $$\chi^2_{n-1}$$ distribution, and that the sample mean and variance are independent (which is suprising at first glance IMO!). So in fact, $$\frac{\bar{x} - \mu}{\sqrt{s^2/n}} \sim t_{n-1}$$.

A certain continous mixture of normal distributions (that might arise in a posterior or posterior predictive): A t-distribution arises we have a mixture of normals that all have the same mean but where the variance is a random variable with scaled inverse chi-squared (equivalently, inverse-gamma) distribution

Concretely, if:
$$x | \sigma^2 \sim N(x_0, \sigma^2/\kappa_0)$$ and $$\sigma^2 \sim \text{Inv-}\chi^2(\nu_0, \sigma_0^2)$$, then marginalizing over $$\sigma^2$$ we have: $$x \sim t_{\nu_0}(x_0, \sigma_0^2/\kappa_0 )$$

Click for more justification.

We'll use below this fact about a gamma integral: $$\int_0^\infty e^{-b/y} y^{-(a+1)} \, dy = \Gamma(a)b^{-1}$$. \begin{aligned} p(x) &= \int_0^\infty p(x, \sigma^2) \, d \sigma^2 \\ &= \int_0^\infty p(x|\sigma^2) p(\sigma^2) \, d\sigma^2 \\ &= \int_0^\infty \frac{1}{\sqrt{2\pi\sigma^2/\kappa_0}} \exp{\left(-\frac{\kappa_0(x-x_0)^2}{2\sigma^2}\right)} \frac{(\nu_0/2)^{\nu_0/2}}{\Gamma(\nu_0/2)} (\sigma_0^2)^{ \nu_0} (\sigma^2)^{-(\nu_0/2+1)} \exp{\left(-\nu_0\sigma_0^2/2\sigma^2\right)} \, d \sigma^2 \\ &\propto \int_0^\infty (\sigma^2)^{-(\nu_0/2 + 1+1)}\exp{\left( -\frac{\kappa_0(x - x_0)^2 + \nu_0\sigma_0^2}{2\sigma^2}\right) }\, d\sigma^2\\ &= \Gamma(\nu_0/2 + 1)(\nu_0 \sigma_0^2 + \kappa_0(x-x_0)^2 )^{-(\nu_0/2 + 1)} \, \text{ using that gamma integral fact with } a = \nu_0/2 +1, b = (x - x_0)^2 + \nu_0 \sigma_0^2 \\ &\propto \left(1 + \frac{\kappa_0(x-x_0^2)}{\nu_0 \sigma_0^2}\right)^{-(\nu_0/2 + 1)} \end{aligned} We can recognize this last line as being proportional to the pdf for a t-distribution $$t_{\nu_n}(x_0, \sigma_0^2/\kappa_0)$$

Where might this come up? The normal scaled inverse-$$\chi^2$$ (or, an alternate parametrization of the normal inverse gamma) distribution is a conveniently conjugate prior for normally distributed data. Specifically, the set up is the following: \begin{aligned} \mu | \sigma ^2 &\sim N(\mu_0, \sigma^2/\kappa_0) \\ \sigma^2 &\sim \text{Inv-}\chi^2(\nu_0, \sigma_0^2)\\ y_i &\sim N(\mu, \sigma^2), i =1,...,n \text{(i.i.d)} \end{aligned} We can use the previous result about an inverse-$$\chi^2$$ mixture of normals with common mean to conclude that $$\mu | \sigma \sim t_{\nu_0}(\mu_0, \sigma_0^2/\kappa_0)$$.

Next, we can note that since the bunch of algebra that goes into the conjugate analysis gives us that $$\mu| y, \sigma^2$$ is normal and $$\sigma^2 |y$$ is inverse-$$\chi^2$$, then when we marginalize over $$\sigma^2$$ we have that $$\mu | y$$ is t-distributed. That is:

\begin{aligned} &\mu | y, \sigma^2 \sim N(\mu_n, \sigma^2/\kappa_n) \\ &\sigma^2 | y \sim \text{Inv-}\chi^2(\nu_n, \sigma_n^2)\\ & \implies \mu \sim t_{\nu_n}(\mu_n, \sigma_n^2/\kappa_n) \end{aligned}

And, we can find one more t-distribution if we note that the posterior predictive distribution for one more data point given $$\sigma^2$$ is also normal, so when we marginalize we again get a t-distribution. \begin{aligned} &\tilde{y} | y, \sigma^2 \sim N(\mu_n, \sigma^2 (\kappa_n + 1)/\kappa_n \\ &\sigma^2 | y \sim \text{Inv-}\chi^2(\nu_n, \sigma_n^2)\\ &\implies \tilde{y} | y \sim t_{\nu_n}(\mu_n, \sigma_n^2(\kappa_n + 1)/\kappa_n) \end{aligned}

Show parameter definitions

\begin{aligned} \nu_n \sigma_n^2 &= \nu_0\sigma_0^2 + \sum(\bar{y}-y_i)^2 + \frac{\kappa_0 n}{\kappa_0 + n}(\bar{y} - \mu_0)^2 \\ \kappa_n &= \kappa_0 + n \\ \nu_n &= \nu_0 + n \\ \mu_n &= \frac{\kappa_0\mu_0 + n \bar{y}}{\kappa_0 + n} \end{aligned} See, e.g. BDA ch. 3 for derivations of these values.

Cauchy distributions: A t-distribution with $$\nu =1$$ is a Cauchy distribution. That's it. That's the fact.

A t test is any statistical test where the test statistic is (at least approximately) t-distributed.

Let's say we have a bunch of observations $$x_1,...,x_n$$ from a normal distribution whose standard deviation we don't know, and that we think the mean of that distribution might be some particular value $$\mu_0$$ but we're not sure. One way to proceed might be to calculate something from the observations (this something is the "test statistic") that we can compare to some known distribution for that test statistic in order to say something about how likely or not it would be to see the kind of data we observed if the mean were $$\mu_0$$.

It turns out that if $$x_1,...,x_n$$ are from a normal distribution with mean $$\mu_0$$ and fixed but unknown standard deviation $$\sigma^2$$, then $$\bar{x} - \mu_0$$ (and so also $$\frac{\sqrt{n}(\bar{x} - \mu_0)}{\sigma^2}$$ ) is normally distributed, $$\frac{\sum_{i=1}^n(\bar{x} - x_i)}{\sigma^2} \sim \chi^2_{n-1}$$, and those two quantities are independent. Therefore (referring back to our first place to find t-distributions in the wild, above), $$\frac{(\bar{x} - \mu_0)\sqrt{n}\sqrt{n-1}}{\sqrt{\sum_{i=1}^n(\bar{x} - x_i)^2}} \sim t_{n-1}$$.

In this case if, for instance, $$n$$ is 10 and our test statistic $$\frac{(\bar{x} - \mu_0)\sqrt{n}\sqrt{n-1}}{\sqrt{\sum_{i=1}^n(\bar{x} - x_i)^2}} \sim t_{n-1}$$ turns out to be, say, 4, we might say that the data is pretty suprising in some way, since values as extreme as 4 arise only rarely as a draw from a $$t_9$$ distribution (see plot below).

Indeed, if we were in a null hypothesis significance testing setup, we might conclude that it was probably not true that the mean of the distribution that the $$x_i$$ were drawn from actually was $$\mu_0$$ - we might "reject" that hypothesis that the mean was $$\mu_0$$. (Big note: Do I love this framework? Not really. Bayesian statistics is more my home base, so I'd tend think of that mean that we're so curious about as a random variable and go from there. But it's worth understanding how t-tests often show up in practice.)

Medium-sized note: what do we mean by "as extreme as 4" above? Are we talking about values 4 and above (that is, far from the mean in one direction)? Or values above 4 and below -4 (that is far from the mean in either direction). Which notion you choose here depends on your context, and leads to the "one-tailed" or "two-tailed" t-tests respectively.

Here's a set of plots illustrating how we get from normally distributed data to our particular t-distributed test statistic in the example above.

Robustness? Ok, but what if your data is not quite normally distributed? We leaned on the $$\bar{x}$$ being normally distributed, the $$\frac{\sum_{i=1}^n(\bar{x} - x_i)}{\sigma^2}$$ being appropriately $$\chi^2$$ distributed, and their independence. The central limit theorem might give you some hope that $$\bar{x}$$ be approximately normally distributed even if the i.i.d $$x_i$$ are not. You might still be nervous about the denominator losing its $$\chi^2$$ distribution and not having independence if we leave the special case of normal $$x_i$$. That's very much a reasonable nervousness to have, although it turns out that even when we deviate some amount from meeting those conditions, the test statistic we gave above can stick pretty close to t-distributed (if you want, a little exploration is here).

Different kinds of t-testsThere are multiple situations where t-distributions arise as a test statistic, and therefore there are other kinds of t-tests. Above, we gave an example of a t-test for testing whether the mean of the population that the data came from was equal to a particular value $$\mu_0$$, which is what is called a one-sample t-test.

A minor variation on that scheme is what's called a paired t-test, when we start with some pairs of values $$(y_{11}, y_{12}),...(y_{n1}, y_{n2})$$ and perform a one-sample t-test on the $$n$$ differences $$x_1 = y_{12} - y_{11},...,x_n = y_{n2} - y_{n1}$$. The pair $$(y_{i1}, y_{i2})$$ might be something like measurements from the ith subject before and after some intervention, and typically we'd be testing $$\mu_0 = 0$$, which corresponds to the idea of the differences being zero on average.

In another variation, we have samples from two different populations: $$x_1,...,x_{n_1}$$ and $$y_1,...,y_{n_2}$$, and want to test whether the population means are equal to each other. If the $$x_i$$ and $$y_i$$ are each i.i.d. normal random variables, from two normal distributions with equal (but unknown) variances, then: $$\frac{(\bar{x} - \bar{y})(\sqrt{n_1 + n_2 - 2})}{\sqrt{\sum_{i=1}^{n_1} (\bar{x} - x_i)^2 + \sum_{i=1}^{n_2} (\bar{y} - y_i)^2} \sqrt{1/n_1 + 1/ n_2}} \sim t_{n_1 + n_2 - 1 }.$$ So, we can calculate the test statistic on the left hand side above, and compare it to a $$t_{n_1 + n_2 -2}$$ distribution to get an idea of how extreme a test statistic that would be in the case where the $$x_i$$ and $$y_i$$ really do come from populations with the same mean. (Note: ANOVA adapts a two-sample t-test to more than two groups.)

Again, the core procedure in these variations on the t-test is we calculate a quantity (the test statistic) from data where the test statistic would be t-distributed if some assumptions and hypotheses held. Interesting!

And Guinness beer? Ah, others tell it better than I do!

I guess I'll end with the caveat that I included information about t-tests because they're around, not because I'm aiming at a ringing endorsement of the current state and practice of null hypothesis significant testing in science. Whether you need to use them or are trying to get away from them, you may want to understand them.

Thanks for reading and good luck with your t-distribution endeavors! I write these posts for fun and with the hope that they'll help someone - if you read this far, hopefully the post helps you.

• Gelman, A., Carlin, J.B., Stern, H.S., Dunson, D.B., Vehtari, A. and Rubin, D.B., 2013. Bayesian data analysis. Chapman and Hall/CRC.