2018111373 최인렬
문제
{X_t} 는 포아송 과정.
- E(X_{10}~|~X_{11})
- Var(X_{11}|X_{10})
- Cov(X_1,~X_{10}|X_9)
1번
P(X_{10}=x_{10}~|~X_{11}=x_{11}) = {P(X_{10}=x_{10},~X_{11}=x_{11})
\over P(X_{11} = x_{11})}
\\[15pt]
={P(X_{10}=x_{10,}~
X_{11}-X_{10}=x_{11}-x_{10})
\over
P(X_{11} = x_{11})}
\\[15pt]
= {P(X_{10}=x_{10})~
P(X_{11}-X_{10}=x_{11}-x_{10})
\over P(X_{11} = x_{11})}
={\dfrac{e^{-10\lambda}~{(10\lambda)}^{x_{10}}}{(x_{10})!}
\cdot
\dfrac{e^{-\lambda}~
{\lambda}^{x_{11}-x_{10}}}
{(x_{11}-x_{10})!}
\over
\dfrac{e^{-11\lambda}~
{(11\lambda)}^{x_{11}}}
{(x_{11})!}
}
= {x_{11}! \over x_{10}!
~(x_{11}-x_{10})!} ~\cdot~
{{(10\lambda)}^{x_{10}}~~
\lambda^{x_{11}-x_{10}}
\over
{(11\lambda)}^{x_{10}}~~
{(11\lambda)}^{x_{11}-x_{10}}}
\\[10pt]
= {x_{11}! \over x_{10}!
~(x_{11}-x_{10})!} ~\cdot~
\bigg(\dfrac{10}{11}\bigg)^{x_{10}}
~\bigg(\dfrac{1}{11}\bigg)
^{x_{11} - x_{10}}
\therefore~X_{10}~|~X_{11} = x_{11} \sim binominal
\bigg(x_{11},~{10\over 11}\bigg)
\\[10pt]
E(X_{10}~|~X_{11}) = {10\over 11}
\cdot x_{11}
2번
P(X_{11}~|~X_{10}) =
{P(X_{10}=x_{10},~X_{11}=x_{11})
\over P(X_{10} = x_{10})}
\\[15pt]
={P(X_{10}=x_{10,}~
X_{11}-X_{10}=x_{11}-x_{10})
\over
P(X_{10} = x_{10})}
\\[15pt]
= {P(X_{10}=x_{10})~
P(X_{11}-X_{10}=x_{11}-x_{10})
\over P(X_{10} = x_{10})}
= P(X_{11}-X_{10}=x_{11}-x_{10})
= P(X_{1} = x_{11}-x_{10})
\\[15pt]
\\[15pt]
\therefore~
P(X_{11}~|~X_{10}) =
\dfrac{e^{-\lambda}~
{\lambda}^{x_{11}-x_{10}}}
{(x_{11}-x_{10})!}
3번
Cov(X,Y~|~Z) = E(XY| Z) - E(X | Z)~E(Y|Z) 를 이용한다.
Cov(X_{1},~X_{10}|X_{9}) = E(X_{1}X_{10}|X_{9}) - E(X_{1}|X_{9})~
E(X_{10}|X_{9})
E(X_{1}|X_{9})를 구해보자.
P(X_{1}=x_{1}~|~X_{9}=x_{9}) =
{P(X_{1}=x_{1},~X_{9}=x_{9})
\over P(X_{9} = x_{9})}
\\[15pt]
={P(X_{1}=x_{1},
~X_{9}-X_{1}=x_{9}-x_{1})
\over P(X_{9} = x_{9})}
\\[15pt]
={P(X_{1}=x_{1})~
P(X_{9}-X_{1}=x_{9}-x_{1})
\over P(X_{9} = x_{9})}
\\[15pt]
={P(X_{1}=x_{1})~
P(X_{8}=x_{9}-x_{1})
\over P(X_{9} = x_{9})}
={\dfrac{e^{-\lambda}~{\lambda}^{x_{1}}}{(x_{1})!}
\cdot
\dfrac{e^{-8\lambda}~
{8\lambda}^{x_{9}-x_{1}}}
{(x_{9}-x_{1})!}
\over
\dfrac{e^{-9\lambda}~
{(9\lambda)}^{x_{11}}}
{(x_{9})!}
}
= {x_{9}! \over x_{1}!
~(x_{9}-x_{1})!} ~\cdot~
{{(\lambda)}^{x_{1}}~~
{(8\lambda)}^{x_{9}-x_{1}}
\over
{(9\lambda)}^{x_{1}}~~
{(9\lambda)}^{x_{9}-x_{1}}}
\\[10pt]
= {x_{9}! \over x_{1}!
~(x_{9}-x_{1})!} ~\cdot~
\bigg(\dfrac{1}{9}\bigg)^{x_{1}}
~\bigg(\dfrac{8}{9}\bigg)
^{x_{9} - x_{1}}
\therefore~X_{1}~|~X_{9} = x_{9} \sim binominal
\bigg(x_{9},~{1 \over 9}\bigg)
\\[10pt]
E(X_{1}~|~X_{9}) = {1 \over 9}
\cdot x_{9}
이번에는 E(X_{10}|X_{9})를 구해보자.
P(X_{10}~|~X_{9}=x_{9}) ~=~
{P(X_{9}=x_{9},~X_{10}=x_{10})
\over P(X_{9} = x_{9})}
\\[15pt]
={P(X_{9}=x_{9},
~X_{10}-X_{9}=x_{10}-x_{9})
\over P(X_{9} = x_{9})}
\\[15pt]
={P(X_{9}=x_{9})~
P(X_{10}-X_{9}=x_{10}-x_{9})
\over P(X_{9} = x_{9})}
= P(X_{1}=x_{10}-x_{9})
=~\dfrac{e^{-\lambda}~
{\lambda}^{x_{10}-x_{9}}}
{(x_{10}-x_{9})!}
\\[15pt]
\therefore~X_{10}~|~X_{9}=x_9 \sim
poisson(\lambda).
\kern{10pt}
E(X_{10}|X_{9}) = \lambda
마지막으로 E(X_{1}X_{10}|X_{9})를 구해보자.
\begin{align*}
&P(X_1 = x_{1,}~
X_{10} = x_{10}~|~X_9 = x_9)
\\[10pt]
&= P(X_1 = x_{1,}~
X_{10} = x_{10,}~X_9 = x_9|~X_9 = x_9)
\\[10pt]
&= {P(X_{1}=x_{1},
~X_{8}=x_{9}-x_{1},
~X_{1}=x_{10}-x_{9})
\over P(X_{9} = x_{9})}
\\[10pt]
&={P(X_{1}=x_{1})
~P(X_{8}=x_{9}-x_{1})
~P(X_{1}=x_{10}-x_{9})
\over P(X_{9} = x_{9})}
\\[20pt]
&=~~{
\dfrac{e^{-\lambda}
~\lambda^{x_1}}
{x_1!}
\cdot
\dfrac{e^{-8\lambda}~
{(8\lambda)}^{x_{9}-x_{1}}}
{(x_{9}-x_{1})!}
\cdot
\dfrac{e^{-\lambda}~
{\lambda}^{x_{10}-x_{9}}}
{(x_{10}-x_{9})!}
\over
\dfrac{e^{-9\lambda}~
{(9\lambda)}^{x_{9}}}
{(x_{9})!}
}
\\[30pt]
&= {
\dfrac{e^{-\lambda}
~\lambda^{x_1}}
{x_1!} \cdot
\dfrac{e^{-8\lambda}~
{(8\lambda)}^{x_{9}-x_{1}}}
{(x_{9}-x_{1})!}
\over
\dfrac{e^{-9\lambda}~
{(9\lambda)}^{x_{9}}}
{(x_{9})!}
} \cdot
\dfrac{e^{-\lambda}~
{\lambda}^{x_{10}-x_{9}}}
{(x_{10}-x_{9})!}
\\[30pt]
&= \dfrac{x_{9}!}
{x_1!~(x_{9}-x_{1})!}~
\bigg({1\over9}\bigg)^{x_1}
\bigg({8\over9}\bigg)^{x_9-x_1}
\cdot
\dfrac{e^{-\lambda}~
{\lambda}^{x_{10}-x_{9}}}
{(x_{10}-x_{9})!}
\end{align*}
\begin{align*}
&E(X_1X_{10}|X_9) =
\sum_{X_1=0}^{x_9}~
\sum_{X_{10}=x_{9}}^{\infty}~
x_1\cdot x_{10} \cdot
P(X_1 = x_{1,}~
X_{10} = x_{10}~|~X_9 = x_9)
\\[20pt]
&=
\sum_{X_1=0}^{x_9}~x_1 \cdot
\dfrac{x_{9}!}
{x_1!~(x_{9}-x_{1})!}~
\bigg({1\over9}\bigg)^{x_1}
\bigg({8\over9}\bigg)^{x_9-x_1}~
\Bigg[
\sum_{X_{10}=x_{9}}^{\infty}~x_{10}\cdot \dfrac{e^{-\lambda}~
{\lambda}^{x_{10}-x_{9}}}
{(x_{10}-x_{9})!}
\Bigg]
\\[30pt]
&= E(X_1|X_9) \times
\Bigg[
\sum_{X_{10}=x_{9}}^{\infty}~
(x_{10}-x_{9}+x_{9})
\cdot \dfrac{e^{-\lambda}~
{\lambda}^{x_{10}-x_{9}}}
{(x_{10}-x_{9})!}
\Bigg]
\\[30pt]
&= E(X_1|X_9) \times \Bigg[
\sum_{X_{10}=x_{9}}^{\infty}
(x_{10}-x_9)\cdot \dfrac{e^{-\lambda}~
{\lambda}^{x_{10}-x_{9}}}
{(x_{10}-x_{9})!}
+
\sum_{X_{10}=x_{9}}^{\infty}(x_{9})\cdot \dfrac{e^{-\lambda}~
{\lambda}^{x_{10}-x_{9}}}
{(x_{10}-x_{9})!}\Bigg]
\\[30pt]
&= {x_9 \over 9} \cdot \Bigg[
\sum_{X_{10}=x_{9}}^{\infty}
\dfrac{e^{-\lambda}~
{\lambda}^{x_{10}-x_{9}}}
{(x_{10}-x_{9}-1)!}
+ x_9\Bigg]
\\[30pt]
&= {x_9 \over 9} \cdot \Bigg[
\sum_{k=0}^{\infty}
\dfrac{e^{-\lambda}~
{\lambda}^{k}}
{(k-1)!}
+ x_9\Bigg]
~=~ {x_9 \over 9} \cdot \Bigg[
\sum_{k=1}^{\infty}
\dfrac{e^{-\lambda}~
{\lambda}^{k}}
{(k-1)!}
+ x_9\Bigg]
\\[30pt]
&= {x_9 \over 9} \cdot
\big(\lambda + x_9 \big)
\end{align*}
따라서 공분산을 구해보면 다음과 같다.
Cov(X_{1}X_{10}|X_{9}) = E(X_{1}X_{10}|X_{9}) - E(X_{1}|X_{9})~E(X_{10}|X_{9})
\\[15pt]
= {8 \over 9}
\cdot (x_{9}-x_{1}) - {1 \over 9}
\cdot x_{9} \cdot \lambda