[機率論] 一類條件機率取極限的問題

Theorem: 令 $X,Y$ 為 jointly continuous，
\[
\lim_{h \to 0} P((X,Y) \in A| x < X \leq x +h) = \int_{-\infty}^\infty 1_A(x,y) f_{Y|X}(y|x)dy
\]其中 $1_A(\cdot)$ 為 indicator function

Proof: 首先觀察
\begin{align*}
  P((X,Y) \in A|x < X \leqslant x + h) &= \frac{{P(\left\{ {(X,Y) \in A} \right\},\left\{ {x < X \leqslant x + h} \right\})}}{{P(x < X \leqslant x + h)}} \hfill \\
   &= \frac{{P(\left\{ {\left( {X,Y} \right) \in A} \right\} \cap \left\{ {X \in \left( {x,x + h} \right]} \right\})}}{{P(x < X \leqslant x + h)}} \hfill \\
   &= \frac{{P(\left\{ {\left( {X,Y} \right) \in A} \right\} \cap \left\{ {\left( {X,Y} \right) \in \left( {x,x + h} \right] \times \mathbb{R}} \right\})}}{{P(x < X \leqslant x + h)}} \hfill \\
   &= \frac{{P(\left( {X,Y} \right) \in A \cap \left\{ {\left( {x,x + h} \right] \times \mathbb{R}} \right\})}}{{P(x < X \leqslant x + h)}} \hfill \\
   &= \frac{{\iint\limits_{A \cap \left\{ {\left( {x,x + h} \right] \times \mathbb{R}} \right\}} {{f_{XY}}\left( {t,s} \right)dtds}}}{{\int_x^{x + h} {{f_X}\left( t \right)dt} }}
\end{align*} 由於 $1_{A \cap B} = 1_A \cdot 1_B$ 我們有
\begin{align*}
  P((X,Y) \in A|x < X \leqslant x + h) &= \frac{{\iint\limits_{} {{1_A}\left( {t,s} \right){1_{\left( {x,x + h} \right] \times \mathbb{R}}}\left( {t,s} \right){f_{XY}}\left( {t,s} \right)dtds}}}{{\int_x^{x + h} {{f_X}\left( t \right)dt} }} \hfill \\
   &= \frac{{\int_{ - \infty }^\infty  {\int_x^{x + h} {{1_A}\left( {t,s} \right){f_{XY}}\left( {t,s} \right)dtds} } }}{{\int_x^{x + h} {{f_X}\left( t \right)dt} }} \hfill \\
   &= \frac{{\int_x^{x + h} {\int_{ - \infty }^\infty  {{1_A}\left( {t,s} \right){f_{XY}}\left( {t,s} \right)dsdt} } }}{{\int_x^{x + h} {{f_X}\left( t \right)dt} }} \hfill \\
\end{align*} 上下同乘 $1/h$ 我們得到
\[P((X,Y) \in A|x < X \leqslant x + h) = \frac{{\frac{1}{h}\int_x^{x + h} {\int_{ - \infty }^\infty  {{1_A}\left( {t,s} \right){f_{XY}}\left( {t,s} \right)dsdt} } }}{{\frac{1}{h}\int_x^{x + h} {{f_X}\left( t \right)dt} }}\]讓 $h \to 0$ 我們有
\begin{align*}
  P((X,Y) \in A|x < X \leqslant x + h) &= \frac{{\frac{1}{h}\int_x^{x + h} {\int_{ - \infty }^\infty  {{1_A}\left( {t,s} \right){f_{XY}}\left( {t,s} \right)dsdt} } }}{{\frac{1}{h}\int_x^{x + h} {{f_X}\left( t \right)dt} }} \hfill \\
   &= \frac{{\int_{ - \infty }^\infty  {{1_A}\left( {x,s} \right){f_{XY}}\left( {x,s} \right)ds} }}{{{f_X}\left( x \right)}} \hfill \\
   &= \int_{ - \infty }^\infty  {{1_A}\left( {x,s} \right)\frac{{{f_{XY}}\left( {x,s} \right)}}{{{f_X}\left( x \right)}}ds}  \hfill \\
   &= \int_{ - \infty }^\infty  {{1_A}\left( {x,s} \right){f_{Y|X}}\left( {s|x} \right)ds} 
\end{align*} 上述等式即為所求，至此證畢。 $\square$