# prove this claim

• Oct 28th 2008, 01:07 PM
szpengchao
prove this claim
Let $\displaystyle B_{1},B_{2},...$ be disjoint events with $\displaystyle \cup_{n=1}^{\infty}{B_{n}}=\Omega$

Show that if A is another event and $\displaystyle P(A|B_{n})=p$ for all n then P(A)=p.

deduce that if X,Y are discrete random variables then :

X and Y are independent.
The conditional distribution of X given Y=y is independent of y.
• Oct 28th 2008, 01:30 PM
Plato
Note that $\displaystyle \begin{array}{rcl} A & = & {A \cap \Omega } \\ {} & = & {A \cap \left( {\bigcup\limits_n {B_n } } \right)} \\ {} & = & {\left( {\bigcup\limits_n {\left( {A \cap B_n } \right)} } \right)} \\ \end{array}$.

From the given: $\displaystyle \left( {\forall n} \right)\left[ {P\left( {A \cap B_n } \right) = P\left( {A|B_n } \right)P\left( {B_n } \right) = pP\left( {B_n } \right)} \right]$.

$\displaystyle P\left( A \right) = \sum\limits_n {P\left( {A|B_n } \right)P\left( {B_n } \right)} = p\sum\limits_n {P\left( {B_n } \right) = p}$ because $\displaystyle \sum\limits_n {P\left( {B_n } \right) = 1}$ (they are pairwise disjoint).