# t distribution as a mixture of normals

In class, the $t$ distribution is usually introduced like this: if $X \sim \mathcal{N}(0,1)$ and $Z \sim \chi_\nu^2$ are independent, then $T = \dfrac{X}{\sqrt{Z / \nu}}$ has $t$ distribution with $\nu$ degrees of freedom, denoted $t_\nu$ or $t_{(\nu)}$.

Did you know that the $t$ distribution can also be viewed as a (continuous) mixture of normal random variables? Specifically, let $W$ have inverse-gamma distribution $\text{InvGam}\left(\dfrac{\nu}{2}, \dfrac{\nu}{2} \right)$, and define the conditional distribution $X \mid W = w \sim \mathcal{N}(0, w)$. Then the unconditional distribution of $X$ is the $t$ distribution with $\nu$ degrees of freedom.

The proof follows directly from computing the unconditional (or marginal) density of $X$:

\begin{aligned} f_X(x) &= \int_0^\infty f_{X \mid W}(x) f_W(w) dw \\ &\propto \int_0^\infty \frac{1}{\sqrt{w}} \exp \left( -\frac{x^2}{2w} \right) \cdot w^{-\nu/2 - 1} \exp \left( - \frac{\nu}{2w} \right) \\ &= \int_0^\infty w^{-\frac{\nu + 1}{2} - 1} \exp \left( - \frac{x^2 + \nu}{2w} \right) dw. \end{aligned}

Note that the integrand above is proportional to the PDF of the inverse-gamma distribution with $\alpha = \dfrac{\nu + 1}{2}$ and $\beta = \dfrac{x^2 + \nu}{2}$. Hence, we can evaluate the last integral exactly to get

$f_X(x) \propto \Gamma \left( \dfrac{\nu + 1}{2} \right) \left(\dfrac{x^2 + \nu}{2}\right)^{-\frac{\nu + 1}{2}} \propto\left( x^2 + \nu \right)^{-\frac{\nu + 1}{2}} \propto \left( 1 + \dfrac{x^2}{\nu} \right)^{-\frac{\nu + 1}{2}},$

which is proportional to the PDF of the  $t_\nu$ distribution.

Sources for the information above:

1. Student-t as a mixture of normals, John D. Cook Consulting.