Index: The Book of Statistical ProofsProbability Distributions ▷ Univariate continuous distributions ▷ Normal distribution ▷ Relationship to t-distribution

Theorem: Let $X_1, \ldots, X_n$ be independent random variables where each of them is following a normal distribution with mean $\mu$ and variance $\sigma^2$:

$\label{eq:norm} X_i \sim \mathcal{N}(\mu, \sigma^2) \quad \text{for} \quad i = 1, \ldots, n \; .$

Define the sample mean

$\label{eq:mean-samp} \bar{X} = \frac{1}{n} \sum_{i=1}^{n} X_i$

and the unbiased sample variance

$\label{eq:var-samp} s^2 = \frac{1}{n-1} \sum_{i=1}^{n} \left( X_i - \bar{X} \right)^2 \; .$

Then, subtracting $\mu$ from the sample mean, dividing by the sample standard deviation and multiplying with $\sqrt{n}$ results in a qunatity that follows a t-distribution with $n-1$ degrees of freedom:

$\label{eq:norm-t} t = \sqrt{n} \, \frac{\bar{X}-\mu}{s} \sim \mathrm{t}(n-1) \; .$

Proof: Note that $\bar{X}$ is a linear combination of $X_1, \ldots, X_n$:

$\label{eq:X-bar-lincomb} \bar{X} = \frac{1}{n} X_1, + \ldots + \frac{1}{n} X_n \; .$ $\label{eq:X-bar-dist} \bar{X} \sim \mathcal{N}\left( \frac{1}{n} \, n \mu, \left(\frac{1}{n}\right)^2 n \sigma^2 \right) = \mathcal{N}\left( \mu, \sigma^2/n \right) \; .$

Let $Z = \sqrt{n} \, (\bar{X}-\mu)/\sigma$. Because $Z$ is a linear transformation of $\bar{X}$, it also follows a normal distribution:

$\label{eq:Z-dist} Z = \sqrt{n} \frac{\bar{X}-\mu}{\sigma} \sim \mathcal{N}\left( \frac{\sqrt{n}}{\sigma} (\mu - \mu), \left(\frac{\sqrt{n}}{\sigma}\right)^2 \sigma^2/n \right) = \mathcal{N}\left( 0, 1 \right) \; .$

Let $V = (n-1) \, s^2/\sigma^2$. We know that this function of the sample variance follows a chi-squared distribution with $n-1$ degrees of freedom:

$\label{eq:V-dist} V = (n-1) \frac{s^2}{\sigma^2} \sim \chi^2(n-1) \; .$

Observe that $t$ is the ratio of a standard normal random variable and the square root of a chi-squared random variable, divided by its degrees of freedom:

$\label{eq:t-Z-V} t = \sqrt{n} \, \frac{\bar{X}-\mu}{s} = \frac{\sqrt{n} \frac{\bar{X}-\mu}{\sigma}}{\sqrt{(n-1) \frac{s^2}{\sigma^2}/(n-1)}} = \frac{Z}{\sqrt{V/(n-1)}} \; .$

Thus, by definition of the t-distribution, this ratio follows a t-distribution with $n-1$ degrees of freedom:

$\label{eq:norm-t-qed} t \sim \mathrm{t}(n-1) \; .$
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Metadata: ID: P234 | shortcut: norm-t | author: JoramSoch | date: 2021-05-27, 08:10.