## t Distributions

The Student's t-Distribution (more simply the t-distribution) is the distribution of sample means when the population standard deviation is unknown. Estimates of a population mean based on a sample will use this distribution. The distribution received the name "Student's t" because its discoverer, William Gosset, published his work under the pseudonym "Student".

### The Formulas

If the random variable $T$ has an t-distribution over the interval   $(-\infty, \infty)$,   with $n$ degrees of freedom, then the PDF of the distribution is given by the following formula.

 $f_T(t) = \dfrac{ \Gamma\left(\dfrac{n+1}{2}\right) }{ \Gamma\left(\dfrac{n}{2}\right) \sqrt{n \pi} \left(1 + \dfrac{t^2}{n} \right)^{(n+1)/2}}$

### Computing with the t-Distribution

As with many important distributions, computations involving the t-distribution are typically left to available technology. The Texas Instruments calculator syntax for the CDF is $\operatorname{tcdf}(x_1,x_2,n)$.

• For the t-distribution with 7 degrees of freedom, the area under the curve between 2.5 and 7.3 is $P(2.5 < t < 7.3) = \operatorname{tcdf}(2.5,7.3,7) = 0.0204$.
• For the t-distribution with 15 degrees of freedom, the area under the curve less than 1.25 is $P(t < 1.25) = \operatorname{tcdf}(-\infty, 1.25, 15) = 0.8848$.

### Derivation of the t Distribution

Standard scores for sample means are normally distributed with   $z = \dfrac{\bar{x} - \mu}{\sigma/\sqrt{n}}$.   However, the population standard deviation $\sigma$ is most often unknown, and the sample standard deviation $s$ is used in its place. In that case, we have   $t = \dfrac{\bar{x} - \mu}{s/\sqrt{n}}$,   a slightly different formula, and we cannot expect it to be normally distributed as $z$ was.

Let the random variable $S$ represent the sample standard deviation (whose value depends on the random sample obtained), and consider the random variable $T^2$. Algebraically, we find:

\begin{equation} T^2 = \dfrac{(\bar{x} - \mu)^2}{\dfrac{S^2}{n}} \dfrac{\dfrac{n}{\sigma^2}}{\dfrac{n}{\sigma^2}} = \dfrac{\dfrac{(\bar{x} - \mu)^2}{\sigma^2/n}}{\dfrac{S^2}{\sigma^2} \dfrac{n-1}{n-1}} = \dfrac{ Z^2 / 1}{ \left[ (n-1)\dfrac{S^2}{\sigma^2} \right] / (n-1)} \end{equation}

We know that squared standard scores have a chi-square distribution with one degree of freedom. We also know that the random variable $\dfrac{(n-1)S^2}{\sigma^2}$ has a chi-square distribution with   $n-1$   degrees of freedom. And we have found that the random variable $T^2$ is the ratio of two chi-square distributions, each divided by their degrees of freedom. Therefore, $T^2$ has an F-distribution with degrees of freedom 1 and   $n-1$.

Since probabilities of each distribution can be found with its CDF, we have the following result.

\begin{align} F_{T^2}(u) &= P(T^2 < u) \\ &= P( -\sqrt{u} < T < \sqrt{u} ) \\ &= F_T(\sqrt{u}) - F_T(-\sqrt{u}) \end{align}

The PDFs of these distributions can be found by taking derivatives.

\begin{equation} f_{T^2}(u) = \dfrac{1}{2\sqrt{u}} \left[ f_T(\sqrt{u}) + f_T(-\sqrt{u}) \right] \end{equation}

But the $T$ distribution is symmetric, because of the underlying symmetry of the normal distributions from which it was formed. Therefore   $f(t) = f(-t)$,   and we have:

\begin{align} f_{T^2}(u) &= \dfrac{1}{\sqrt{u}} f_T(\sqrt{u}) \\ \sqrt{u} f_{T^2}(u) &= f_T(\sqrt{u}) \end{align}

Making the substitution   $u = t^2$   and swapping sides of the equation, we obtain:

\begin{equation} f_T(t) = |t| f_{T^2}(t^2) \end{equation}

The left side of this equation is the PDF of the random variable $T$, while the right hand side gives us a formula in terms of the PDF of $T^2$, which has an F-distribution. Using that formula produces the PDF of a t-distribution with   $n-1$   degrees of freedom.

\begin{align} F_T(t) &= (t^2)^{1/2} \dfrac{ \Gamma\left(\dfrac{n}{2}\right) \left(\dfrac{1}{n-1}\right)^{1/2} (t^2)^{1/2-1} }{ \Gamma\left(\dfrac{n-1}{2}\right) \Gamma\left(\dfrac12\right) \left(1 + \dfrac{1}{n-1} t^2 \right)^{n/2} } \\ &= \dfrac{ \Gamma\left(\dfrac{n}{2}\right)}{ \Gamma\left(\dfrac{n-1}{2}\right) \sqrt{(n-1)\pi} \left(1 + \dfrac{t^2}{n-1} \right)^{n/2} } \end{align}

Replacing $n$ by   $n+1$   in this result will produce the PDF of a t-distribution with $n$ degrees of freedom.