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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.

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