t-Distribution

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While we can show that the sample Variance of Normal Distribution samples is of Chi-Square Distribution, it is convenient to define the distribution for the following Random Variable

$$T_n=\frac{Z}{\sqrt{{\chi }_n^2/n}},$$

where $Z\sim \mathcal{N}(0,1)$. $T_n$ is said to have a t-distribution with $n$ degrees of freedom.

By the Law of Large Numbers, for large $n$, $T_n\approx Z\sim \mathcal{N}(0,1)$ (but always with larger Variance).

• Parameter

  • $n\in \mathbb{N}$

• Mean

  • 0

• Variance

• $\frac{n}{n-2}$

• PDF

  • $\frac{\Gamma \left(\frac{n+1}{2}\right)}{\sqrt{n\pi }\Gamma \left(\frac{n}{2}\right)}{\left(1+\frac{x^2}{n}\right)}^{-\frac{n+1}{2}}$

• MGF

  • Undefined

Examples

Normal Distribution with Unknown Variance

Suppose $\{X_i\sim \mathcal{N}(\mu ,{\sigma }^2)\}$, then

$$\frac{\overline{X}-\mu }{s/\sqrt{n}}\sim t_{n-1},$$

where $s$ is the square root of Sample Variance.

Difference of Normal Distributions with the Same Variance

Let $\{X_i\sim \mathcal{N}({\mu }_1,{\sigma }^2){\}}_{i=1}^n$ and $\{X_j\sim \mathcal{N}({\mu }_2,{\sigma }^2){\}}_{j=1}^m$. Then

$$\frac{(\overline{x}-\overline{y})-({\mu }_1-{\mu }_2)}{s_p\sqrt{\frac{1}{m}+\frac{1}{n}}}\sim t_{m+n-2},$$

where

$$s_p=\sqrt{\frac{(m-1)s_x^2+(n-1)s_y^2}{m+n-2}}.$$

Difference of Normal Distributions with Different Variances

Let $\{X_i\sim \mathcal{N}({\mu }_1,{\sigma }_1^2){\}}_{i=1}^n$ and $\{X_j\sim \mathcal{N}({\mu }_2,{\sigma }_2^2){\}}_{j=1}^m$. Then approximately, we have

$$\frac{(\overline{x}-\overline{y})-({\mu }_1-{\mu }_2)}{\sqrt{\frac{s_1^2}{n}+\frac{s_2^2}{m}}}\sim t_N,$$

where

$$N=\frac{{\left(\frac{s_1^2}{n}+\frac{s_2^2}{m}\right)}^2}{\frac{s_1^4}{n^2(n-1)}+\frac{s_2^4}{m^2(m-1)}}\gtrsim \min \{n,m\}.$$

The result follows from the Welch-Satterthwaite formula.