Bayesian t-test¶

We want to test the hypothesis that $\mu \ne \mu_0$ foro some known value $\mu_0$ (ofthen 0), given values $x_i \sim N(\mu, \sigma^2)$. This is called a two-sided one sample t-test. We perform this test by checking if $\mu_0 \in I_{0.95}(\mu|D)$. If not than we can be 95% sure that $\mu \ne \mu_0$. A more common scenario is when we want to test if two paried samples have the same mean.

\[\begin{split}y_i \sim N(\mu_1, \sigma^2) \\ z_i \sim N(\mu_2, \sigma^2)\end{split}\]

We want to determine if $\mu = \mu_1 - \mu_2 \ge 0$ using $x_i = y_i - z_i$

This can be evaluted as:

\[ p(\mu \ge \mu_0|D) = \int_{\mu_0}^{\infty}p(\mu|D)du \]

This is called one sided paired t-test.

To find $p(\mu|D)$ we specify an uninformative prior the posterior marginal on $\mu$ is T distributed:

\[ p(\mu|D) = T(\mu| \bar{x}, \frac{s^2}{N}, N - 1) \]

Now we define the t-statistics as:

\[ t \triangleq \frac{\bar{x} - \mu_0}{s/\sqrt{N}} \]

$s/\sqrt{N}$ is the standard error of the mean

Thus the posterior: $$ p(\mu|D) = 1 - F_{N-1}(t) $$

$F_v(t)$ is the CDF of the standard Student t distribution $T(0,1,v)$

Connection to frequentist¶

Using uninformative prior gives us the same result as using frequentist methods.

Frequentist $$\frac{\mu - \bar{x}}{\sqrt{s/N}}| D \sim t_{N-1}$$

Bayesian The sampling distribution of the MLE:

\[ \frac{\mu - \bar{X}}{\sqrt{s/N}}| \mu \sim t_{N-1} \]

The reason is that the Student T distribution is symmetric in its first two arguments

\[ T(\bar{x}| \mu , \sigma^2, v) = T(\mu| \bar{x}, \sigma^2, v) \]

The results are similar but have different interpretation. In the Bayesian approach $\mu$ is unknown and $\bar{x}$ is fixed. In frequentist $\bar{X}$ is unkown and $\mu$ is fixed.

study-notes

Bayesian t-test¶

Connection to frequentist¶