1

How do I prove the following law of total probability? Any help is appreciated.

Let $X,Y:\Omega\to\mathbb{R}$ be jointly continuous. Then, for $B\subseteq\mathbb{R}^{2}$: \begin{equation} \mathbb{P}\left((X,Y)\in B\right)=\int_{-\infty}^{\infty}\mathbb{P}\left((x,Y)\in B\mid X=x \right)f_{X}(x)\mathrm{d}x. \end{equation}

Miski123
  • 307

2 Answers2

1

Here are two different ways. I like the second best.

Iterated expectations:

For events $A$ and random variables $X$ we can define: $$ P[A|X] = E[1_A|X]$$ where $1_A$ is an indicator function that is 1 if event $A$ holds, and 0 else. Then iterated expectations gives $$ P[A] = E[E[1_A|X]] $$ For the event $A = \{(X,Y) \in B\}$ we get \begin{align} P[(X,Y)\in B] &= E[E[1_{\{(X,Y) \in B\}}|X]] \\ &= \int_{-\infty}^{\infty} f_X(x) P[(X,Y)\in B|X=x] dx \\ &= \int_{-\infty}^{\infty} f_X(x) P[(x,Y)\in B|X=x] dx \\ \end{align} where the last line is using the idea in the Graham Kemp answer.

Another way:

For events $A$ and random variables $X$ with $f_X(x)>0$ we can define: $$P[A|X=x] = \frac{f_{X|A}(x)P[A]}{f_X(x)}$$ and you can verify that $$ P[A] = \int_{-\infty}^{\infty} P[A|X=x] f_X(x)dx$$ For the event $A = \{(X,Y) \in B\}$ we get \begin{align} P[(X,Y) \in B] &= \int_{-\infty}^{\infty} P[(X,Y) \in B| X=x] f_X(x)dx\\ &=\int_{-\infty}^{\infty} P[(x,Y) \in B| X=x] f_X(x)dx\\ \end{align} where again the last line is using the idea in the Graham Kemp answer.

Michael
  • 23,905
0

Under the condition that $X=x$, the events of $\langle X,Y\rangle\in B$ and $\langle x,Y\rangle\in B$ will clearly be identical.

$$\{\omega:\langle X(\omega),Y(\omega)\rangle\in B\land X(\omega)=x\}=\{\omega:\langle x,Y(\omega)\rangle\in B\land X(\omega)=x\}$$

Everything else is just the definition for conditional probability over a zero measure condition.

Graham Kemp
  • 129,094