How to Derive the T Distribution

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How to Derive the T Distribution

Table of Contents


Context / Definitions

Before diving into the step by step proof of how to derive the T distribution it is necessary to briefly define some concepts that will be used in the proof.


Z (aka Standard Normal) Distribution

A “standard normal” or Z distribution is a normal distribution with a mean 0 and standard deviation 1. Any normal distribution can be transformed into standard normal by subtracting its own mean and dividing by its own standard deviation. Therefore, the standard normal distribution, through the Z statistic formula, is an analogue for all other normal distributions.
\[\large
\begin{align}
&X \sim \text{Nor}(\mu_X,\sigma_X) & && & Z = \frac{X\;-\;\mu_X}{\sigma_X} & && &Z \sim \text{Nor}(0,1) &
\end{align}
\]
Most often in the context of hypothesis testing, the Z distribution comes up in relation to the distribution of a sample mean.

Let \(X\) be a random variable with a finite second moment. Given a sufficiently large sample of independently generated values from \(X\), the sample mean, \(\widehat{\mu}_X\), will be approximately normal (according to the second fundamental theorem of probability).
\[\large
\begin{align}
&X \sim \text{R.V.} & && &\sigma_X < \infty &\\ \\
&\widehat{\mu}_X = \frac{1}{n}\sum_{i=1}^n X_i & && &\lim_{n \rightarrow \infty} \widehat{\mu}_X = \text{Nor}(\mu_X,\sigma_X) &
\end{align}
\]


T Distribution

If the population standard deviation is not known (and cannot be assumed) it can be replaced in the Z value formula by the sample standard deviation: \(\widehat{\sigma}_X^2\), which yields the T value formula.

The T distribution is a distribution which approximates, but is not exactly equal to, the Z distribution. The T distribution accepts a single parameter, degrees of freedom, and just as sample variance becomes less biased with a larger sample, the T distribution becomes more normal as the degrees of freedom increase.
\(\widehat{\sigma}_X\).
\[\large
\begin{align}
& T = \frac{X-\mu_X}{\widehat{\sigma}_X} & && & \lim_{k\rightarrow \infty} T(k) = Z &
\end{align}
\]
The T distribution is used for hypothesis tests in several contexts including to test the coefficients of a linear regression.


Chi-Squared Distribution

A chi squared distribution is created by the sum of one or more z values squared. It accepts one parameter, the degrees of freedom of the summation.
\[\large
\begin{align}
V\sim\chi^2(k) = \sum_{i=1}^k Z_i^2 && f_V(v) = \frac {v^{k/2-1}}{2^{k/2}\Gamma(k/2)}e^{-v/2}
\end{align}
\]


Variable Change

Through the fundamental theorem of calculus \(\frac{\partial }{\partial y}F(y) = f(y)\) along with the chain rule \(\frac{\partial}{\partial x} f(g(x)) = \frac{\partial f}{\partial g}\frac{\partial g}{\partial x}\) it is possible to use a variable equivalency to perform a “variable change” on a function. such that its integral before and after the variable change evaluates to the same value.
\[\large{
\begin{align}
y = g(x) && \int_y f(y)\partial y = \int_x f(g(x))\frac{\partial g}{\partial x}\partial x
\end{align}
}\]


How to Derive the T Distribution Probability Function

Let \(X\) be a normally distributed random variable and \(\widehat{\sigma}_X\) be the estimator random variable for its standard deviation. Then \(\frac{X – \mu_X}{\widehat{\sigma}_X}\) creates a T distribution with probability function \(f_T(t;k) = \frac{\Gamma(\frac{k+1}{2})}{\sqrt{\pi k}\;\Gamma(k/2)}(1+t^2/k)^{-\frac{k+1}{2}}\) with the parameter \(k\) being the degrees of freedom from the standard deviation estimator random variable.

Start with two independent random variables \(Z\) and \(V\). \(Z\) is a standard normal distribution. \(V\) is a Chi-squared distribution expressed as \(V = \sum_{i=1}^n(\frac{x_i-\widehat{\mu}_X}{\sigma_X})^2\) with \(k = n – f\) degrees of freedom (\(f\) being the number of factors in the estimator \(\widehat{\mu}_X\)).

\(V\) can be expressed in terms of a sample variance estimator through simple factoring.
\[\large{
\begin{align}
V &= \frac{k}{k}\sum_{i=1}^n\bigg(\frac{X_i-\widehat{\mu}_X}{\sigma_X}\bigg)^2 && \normalsize (1) \\ \\
&= \frac{k}{\sigma_X^2}\sum_{i=1}^n\frac{(X_i-\widehat{\mu}_X)^2}{k} \\ \\
&= \frac{k\widehat{\sigma}_X^2}{\sigma_X^2}
\end{align}
}\]
We can now establish an equivalency between \(T\), \(V\), and \(Z\).
\[\large{
\begin{align}
T &= \frac{X-\mu_X}{\widehat{\sigma}_X} \sqrt{\frac{k\sigma_X^2}{k\sigma_X^2}} && \normalsize (2)\\ \\

&= \frac{X-\mu_X}{\sigma_X} \sqrt{\frac{k\sigma_X^2}{k\widehat{\sigma}_X}} \\ \\

&= Z \sqrt{\frac{k}{V}}
\end{align}
}\]
Some simple rearrangement yields the equivalency in terms of \(Z\).
\[\large{
\begin{align}
Z &= T\sqrt{\frac{V}{k}} && \normalsize (3)
\end{align}
}\]
\(Z\) being standard normal, has a probability density function given below.
\[\large{
\begin{align}
f_Z(z) = \frac{e^{-\frac{z^2}{2}}}{\sqrt{2\pi}} && \normalsize (4)
\end{align}
}\]
Let the relationship from (3) define a function \(z = g(t) = t\sqrt{\frac{k}{v}}\) and use that to perform a variable change on the probability function of \(Z\) from \(z\) to \(t\). Since the variable change is only with respect to \(t\), but \(v\) is also part of the equivalency the resulting function will be a marginal probability function.
\[\large{
\begin{align}
&f_Z(z) & &= \frac{e^{-\frac{z^2}{2}}}{\sqrt{2\pi}} && \normalsize (5) \\ \\
&f_{T|V}(t|v) & &= \frac{e^{-\frac{t^2k}{2v}}}{\sqrt{2\pi}}\cdot\sqrt{\frac{k}{v}}
\end{align}
}\]

**NOTE**: If it is unclear why the variable change produces a marginal probability function consider that the integral must evaluate to 1 before and after the variable change (because all probability function evaluate to 1). T is the only variable in the probability function, but random variable v is present as a parameter, hence the function is a marginal probability function.

The chi-squared variable can be integrated out of the marginal probability function. As a first step, simply combine like terms and move all constants outside of the integral.
\[\large
\begin{align}

f_T(t) & = \int_0^\infty f_{T|V}(t|v)f_V(v)\partial v && \normalsize (6)\\ \\

&= \int_0^\infty \frac{1}{\sqrt{2\pi}}e^{-\frac{t^2v}{2k}}\sqrt{\frac{v}{k}} \frac{v^{k/2-1}}{2^{k/2}\Gamma(k/2)} e^{-v/2} \partial v \\ \\

&= \frac{2^{-\frac{k+1}{2}}}{\sqrt{\pi k}\;\Gamma(k/2)}\int_0^\infty v^{\frac{k+1}{2}-1}e^{-\frac{v}{2(1+t^2/k)^{-1}}}\partial v
\end{align}
\]

Next lets multiply by a constant 1, but expressed in terms of the denominator of the exponent of the Euler constant and raised to same power as the \(v\) term.
\[\large
\begin{align}
&= \frac{2^{-\frac{k+1}{2}}}{\sqrt{\pi k}\;\Gamma(k/2)}\int_0^\infty v^{\frac{k+1}{2}-1}e^{-\frac{v}{2(1+t^2/k)^{-1}}}\partial v\cdot\bigg(\frac{2(1+t^2/k)^{-1}}{2(1+t^2/k)^{-1}}\bigg)^{\frac{k+1}{2}-1} && \normalsize (7)\\ \\

&= \frac{2^{-\frac{k+1}{2}}2^{\frac{k-1}{2}}(1+t^2/k)^{-\frac{k-1}{2}}}{\sqrt{\pi k}\;\Gamma(k/2)}\int_0^\infty \bigg(\frac{v}{2(1+t^2/k)^{-1}}\bigg)^{\frac{k+1}{2}-1}e^{-\frac{v}{2(1+t^2/k)^{-1}}} \partial v
\end{align}
\]
The above expression may seem messy, but notice that within the integral is the kernel of a gamma function. Let \(v = g(h) = 2(1+t^2/k)^{-1}h\), and then perform a variable change to complete the gamma function.
\[\require{cancel}\large{
\begin{align}
&= \frac{2^{-\frac{k+1}{2}}2^{\frac{k-1}{2}}(1+t^2/k)^{-\frac{k-1}{2}}}{\sqrt{\pi k}\;\Gamma(k/2)} \int_0^\infty h^{\frac{k+1}{2}-1} e^{-h} \cdot\frac{\partial g}{\partial h} \partial h && \normalsize (8) \\ \\
&= \frac{\cancel{2^{-\frac{k+1}{2}}}\cancel{2^{\frac{k-1}{2}}}(1+t^2/k)^{-\frac{k-1}{2}}}{\sqrt{\pi k}\;\Gamma(k/2)}\cdot \cancel{2}(1+t^2/k)^{-1}
\end{align}
}\]
Finally, cancel and combine terms to complete the probability function of a T distribution
\[\large{
\begin{align}
f_T(t) &= \frac{\Gamma(\frac{k+1}{2})}{\sqrt{\pi k}\;\Gamma(k/2)}(1+t^2/k)^{-\frac{k+1}{2}} && \normalsize \text{(QED)}
\end{align}
}\]


Summary

The proof of the T distributions probability density function is not complicated, though the math can appear somewhat messy. Begin by defining a standard normal and a Chi-squared random variable, which are independent of one another. Establish that a Chi-squared random variable can be expressed as the ratio of a standard deviation estimator multiplied by its own degrees of freedom and divided by the population standard deviation. This alternate expression reveals an equivalency between a T random variable and a Z random variable. Perform a variable change on the probability function of a Z random variable using the equivalency of T to Z, which produces the marginal probability function for a T random variable given a Chi-Squared random variable. Integrate out the Chi-squared component to finally arrive at the probability function for a T random variable.

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