Independent Event Example 1 - Solution.

Example 1.) Mathematical Statistics by K.Knight. Chapter 1, 1.2
Suppose that A and B are independent events. Determine which of the following pairs of events are always independent and which are always disjoint.
(a) A and $B^c$
(c) $A^c$ and $B^c$  


Solution
(a) A and $B^c$
We need to check whether P(A \cap B^c)=P(A)\cdot P(B^c) or not.
$P(A)=P(A \cap B) \cup (A \cap B^c) \Rightarrow P(A \cap B^c)=P(A)-P(A \cap B) $
         $=P(A)\left \{ 1-P(B^c) \right \} = P(A) \cdot P(B^c)$
Therefore, it's true! They are independent.

(c) $A^c$ and $B^c$  
We need to check whether $P(A^c \cap B^c)=P(A^c) \cdot P(B^c)$ or not.
We know $(A^c \cap B^c)=(A \cup B)^c$, $\Rightarrow P( (A \cup B)^c) = 1 - P(A\cup B)=1 - P(A)-P(B)+P(A)\cdot P(B)$
     $= 1 - P(A)+P(B)(P(A)-1)=(1-P(A))(1-P(B))=P(A^c) \cdot P(B^c)$
Therefore, it's true! They are independent.  

Independent

Independent Event
Definition) Two events A, and B are independent if and only if
$P(A\cap B)=P(A)\cdot P(B)$
$P(A|B)= \frac{ P(A \cap B)}{P(B)}=P(a)$, where P(B) > 0
 
Example 1.) Mathematical Statistics by K.Knight. Chapter 1, 1.2
Suppose that A and B are independent events. Determine which of the following pairs of events are always independent and which are always disjoint.
(a) A and $B^c$
(c) $A^c$ and $B^c$  
Solution??!!
 

Hypothesis Testing - Likelihood Ratio Test Example (1)

Example) Mathematical Statistics by Keith Knight, Chapter 7-19

Let $X_{1}, X_{2},...X_{n}$ be iid N$(\mu,\sigma^2)$. Both are unknown. 
We want to test $H_{0}: \mu=\mu_{0}$ vs. $H_{1}: \mu \neq \mu_{0}$.   

$\triangleright$ Solution.
As we know the MLE of $\mu$ and $\sigma ^2$ are follwing: $\widehat{\mu}=\bar{X}$, and $\widehat{\sigma^2}=\frac{1}{n}\sum(X_{i}-\bar{X})^2$.
Under the $H_{0}$, the MLE of $\widehat{\sigma^2_{0}}= \frac{1}{n}\sum(X_{i}-\mu_{0})^2$.

$f(x_{1},... x_{n};\mu, \sigma^2)= \prod_{i=1}^{n} \frac{1}{\sigma\sqrt{2\pi}}\exp[-\frac{(x_{i}-\mu)^2}{2\sigma^2}]= (2\pi\sigma^2)^{-\frac{n}{2}} \exp[-\frac{\sum(X_{i}-\mu)}{2\sigma^2}]$

* We know the Likelihood Ratio test will reject $H_{0}$ when $\Lambda$ $\geq$ k,
where $\Lambda = (\frac{\widehat{\sigma^2_{0}}}{\widehat{\sigma^2}})^\frac{n}{2}$, and k is chosen based on the level of the test is specifed $\alpha$.

$\frac{\widehat{\sigma^2_{0}}}{\widehat{\sigma^2}}=\frac{\sum(X_{i}-\mu_{0})^2}{\sum(X_{i}-\bar{X})^2} = 1+\frac{n(\bar{x}-\mu_{0})}{\sum(X_{i}-\bar{x})^2}$

* Why? $\sum ((X_{i}-\bar{x})+(\bar{x}-\mu_{0}))^2=\sum(X_{i}-\bar{x})^2+n(\bar{x}-\mu_{0})^2$ !!
* But Wait!! We can use the T distribution, where $T=\frac{\sqrt{n}(\bar{x}-\mu_{0})}{S}$, where $S^2= \frac{1}{n-1}\sum(X_{i}-\bar{x})^2$!
  
Therefore, we can rearrange the EQTN; $=1+\frac{1}{n-1}(\frac{n(\bar{x}-\mu_{0})}{s^2})=1+\frac{T^2}{n-1}$
Therefore, when $H_{0}$ is true, T has student T distribution with n-1 degree of freedom.

Now we know the distribution of Lambda, we can define the rejection region.