chapter2_theory.lyx

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\begin_body

\begin_layout Section*
2.1 Independent Events
\end_layout

\begin_layout Standard

\series bold
Definition 2.2.1 
\series default
Independent Events.
 Two events 
\begin_inset Formula $A$
\end_inset

 and 
\begin_inset Formula $B$
\end_inset

 are independent if 
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
P(A\cap B)=P(A)P(B)
\]

\end_inset


\end_layout

\begin_layout Standard
Suppose that 
\begin_inset Formula $P(A)>0$
\end_inset

 and 
\begin_inset Formula $P(B)>0$
\end_inset

.
 Then it follows easily from the definitions of independence and conditional
 probability that 
\begin_inset Formula $A$
\end_inset

 and 
\begin_inset Formula $B$
\end_inset

 are independent if and only if 
\begin_inset Formula $P(A|B)=P(A)$
\end_inset

 and 
\begin_inset Formula $P(B|A)=P(B)$
\end_inset

.
\end_layout

\begin_layout Subsubsection*
Example 2.2.3 
\series medium
is good example to review, which shows that two events 
\begin_inset Formula $A$
\end_inset

 and 
\begin_inset Formula $B$
\end_inset

, which are physically related, can, nevertheless, satisfy the definition
 of independence.
 
\end_layout

\begin_layout Standard

\series bold
Theorem 2.2.1 
\series default
If two events 
\begin_inset Formula $A$
\end_inset

 and 
\begin_inset Formula $B$
\end_inset

 are independent, then the events 
\begin_inset Formula $A$
\end_inset

 and 
\begin_inset Formula $B^{c}$
\end_inset

 are also independent.
\end_layout

\begin_layout Subsubsection*
Definition 2.2.2 
\series medium
(Mutually) Independent Events.
 The 
\begin_inset Formula $k$
\end_inset

 events 
\begin_inset Formula $A_{1},...,A_{k}$
\end_inset

 are independent (or mutually independent) if, for every subset 
\begin_inset Formula $A_{i_{1}},...,A_{i_{j}}$
\end_inset

 of 
\begin_inset Formula $j$
\end_inset

 of these events (
\begin_inset Formula $j=2,3,...,k$
\end_inset

),
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
P(A_{i_{1}}\cap...A_{i_{j}})=P(A_{i_{1}})...P(A_{i_{j}})
\]

\end_inset


\end_layout

\begin_layout Standard
As an example, in order to for three events 
\begin_inset Formula $A,B$
\end_inset

 and 
\begin_inset Formula $C$
\end_inset

 to be independent, the following four relations must be satisfied:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
P(A\cap B)=P(A)P(B);P(A\cap C)=P(A)P(B);P(B\cap C)=P(B)P(C)
\]

\end_inset

(2.2.1)
\end_layout

\begin_layout Standard
and 
\begin_inset Formula 
\[
P(A\cap B\cap C)=P(A)P(B)P(C).
\]

\end_inset

(2.2.2)
\end_layout

\begin_layout Standard
It is possible that Eq.
 (2.2.2) will be satisfied, but one or more of the three reations (2.2.1) will
 not be satisfied.
 On the other hand, maybe sometimes it is also possible that each of the
 three relations (2.2.1) will be satisfied but Eq.
 (2.2.2) will not be satisfied.
 (like 
\series bold
example 2.2.4
\series default
)
\end_layout

\begin_layout Standard
The results of example 2.2.4 says that the events 
\begin_inset Formula $A,B,C$
\end_inset

 are 
\emph on
pairwise independent, 
\emph default
but all three events are not indepedent.
\end_layout

\begin_layout Subsubsection*
Theorem 2.2.2 
\series medium
Let 
\begin_inset Formula $A_{1},...,A_{k}$
\end_inset

 be events such that 
\begin_inset Formula $P(A_{1}\cap...\cap A_{k})>0$
\end_inset

.
 Then 
\begin_inset Formula $A_{1},...,A_{k}$
\end_inset

 are independent if and only if, for every two disjoint subsets 
\begin_inset Formula $\{i_{1},...,j_{m}\}$
\end_inset

 and 
\begin_inset Formula $\{j_{1},...,j_{l})$
\end_inset

 of 
\begin_inset Formula $\{1,...,k\}$
\end_inset

 we have
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
P(A_{i_{1}}\cap...\cap A_{i_{m}}|A_{j_{1}}\cap...\cap A_{j_{l}})=P(A_{i_{1}}\cap...\cap A_{i_{m}})
\]

\end_inset


\end_layout

\begin_layout Standard
Theorem 2.2.2 says that 
\begin_inset Formula $k$
\end_inset

 events are independent if and only if learning that some of the events
 occur does not change the probability that any combination of the other
 events occurs.
\end_layout

\begin_layout Paragraph*
The meaning of Independent
\end_layout

\begin_layout Standard
When deciding whether to model events as independent, try to answer the
 following question: 
\begin_inset Quotes eld
\end_inset

If I were to learn that some of these events occurred, would I change the
 probabilities of any of the others?
\begin_inset Quotes erd
\end_inset

 If we feel that we already know everything that we could learn from these
 events about how likely the others should be, we can safely model them
 as independent.
 If, on the other hand, we feel that learning some of these events could
 change our minds about how likely some of the others are, then we should
 be more careful about determining the conditional probabilities and not
 model the events as independet.
\end_layout

\begin_layout Standard

\series bold
To more specific, 
\series default
Suppose that we begin by thinking that the probability is 0.08 for (In example
 2.2.5) that an item will be defective.
 If we observe one or zero defective items in the first nine, we might not
 make much revision to the probability that the 10th item is defective.
 On the other hand, if we observe eight or nine defectives in the first
 nice items, we might be uncomfortable keep the probability at 0.08 that
 the 10th item will be defective.
\end_layout

\begin_layout Paragraph*
Theorem 2.2.3 
\series medium
Let 
\begin_inset Formula $n>1$
\end_inset

 and let 
\begin_inset Formula $A_{1},...,A_{n}$
\end_inset

 be events that are mutually exclusive.
 The events are also mutually independent if and only if all the events
 except possibly one of them has probability 0.
\end_layout

\begin_layout Standard

\series bold
Conditional Independent
\end_layout

\begin_layout Paragraph*
Definition 2.2.3 
\series medium
Conditional Independence.
 We say that events 
\begin_inset Formula $A_{1},..,A_{k}$
\end_inset

 are 
\emph on
conditionally independent given 
\begin_inset Formula $B$
\end_inset

 if, 
\emph default
for every subcollection 
\begin_inset Formula $A_{i_{1}},...,A_{i_{j}}$
\end_inset

 of 
\begin_inset Formula $j$
\end_inset

 of these events 
\begin_inset Formula $(j=2,3,...,k)$
\end_inset

,
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
P(A_{i_{1}}\cap...\cap A_{i_{j}}|B)=P(A_{i_{1}}|B)....P(A_{i_{j}}|B)
\]

\end_inset


\end_layout

\begin_layout Paragraph*
Theorem 2.2.4 
\series medium
Suppose that 
\begin_inset Formula $A_{1},A_{2}$
\end_inset

 and 
\begin_inset Formula $B$
\end_inset

 are events such that 
\begin_inset Formula $P(A_{1}\cap B)>0$
\end_inset

.
 Then 
\begin_inset Formula $A_{1}$
\end_inset

 and 
\begin_inset Formula $A_{2}$
\end_inset

 are conditionally independent given 
\begin_inset Formula $B$
\end_inset

 if and only if 
\begin_inset Formula $P(A_{2}|A_{1}\cap B)=P(A_{2}|B)$
\end_inset


\end_layout

\end_body
\end_document