## Two-side composite hypothesis test

$$X_1,\dotsc,X_n \sim N(\mu,1)$$

$$H_0: \mu = \mu_0$$

$$H_1: \mu \neq \mu_0$$

$\Lambda = \frac{L(\mu_0|x)}{L(\overline{x}|x)}$

given that the sample mean $$\overline{x}$$ is the MLE of $$\mu$$.

$\Lambda = \frac{exp \left[ - \sum_{i=1}^{n} (x_i - \mu_0)^2 / 2 \right] }{exp \left[ - \sum_{i=1}^{n} (x_i - \overline{x})^2 / 2 \right] } = exp \left[ \left( - \sum_{i=1}^{n} (x_i - \mu_0)^2 + \sum_{i=1}^{n} (x_i - \overline{x})^2\right) / 2 \right] = exp \left[ \frac{- n(\overline{x}-\mu_0)^2}{2} \right]$

$R = \{ x: \Lambda(x) < c \} = \{ x: |\overline{x} - \mu_0 | > \sqrt{-2(\log{c}) / n} \}$

## Simple hypothesis test

$$X_1,\dotsc,X_n \sim N(\mu,1)$$

$$H_0: \mu = \mu_0$$

$$H_1: \mu = \mu_1$$

$\Lambda = \frac{L(\mu_0|x)}{L(\mu_1|x)}$

$\Lambda = \frac{exp \left[ - \sum_{i=1}^{n} (x_i - \mu_0)^2 / 2 \right] }{exp \left[ - \sum_{i=1}^{n} (x_i - \mu_1)^2 / 2 \right] } = exp \left[ \left( - \sum_{i=1}^{n} (x_i - \mu_0)^2 + \sum_{i=1}^{n} (x_i - \mu_1)^2\right) / 2 \right]$

### Example

If we have just one random variable $$X \sim N(\mu,1)$$ and

$$H_0: \mu = 0$$

$$H_1: \mu = d'$$

then

$\Lambda = exp \left( - d' \left( x - d' / 2\right)\right)$

and

$\log{\Lambda} = -d' \left( x - d' / 2\right)$

Thus

$R = \{x: \Lambda(x) < c \} = \{x: \log \Lambda(x) < \log c \} = \{x: \log \Lambda(x) < \lambda \}$

$\text{If }x > \frac{\lambda}{d'} + \frac{d'}{2} \text{ then }a=a_1$ $\text{If }x < \frac{\lambda}{d'} + \frac{d'}{2} \text{ then }a=a_0$

In this case, the value of the random variable can be directly compared to a criterium.