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.

2 thoughts on “t distribution as a mixture of normals

  1. Pingback: Negative binomial distribution as a mixture of Poissons | Statistical Odds & Ends

  2. Pingback: Laplace distribution as a mixture of normals | Statistical Odds & Ends

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