Is it obvious that $P(X_1 < X_3 | X_1 < X_2) = \int_0^R f_{X_1}(x_1 | X_1 < X_2)P(x_1 < X_3) dx_1 dx_2dx_3$ without Bayes rule?

Question

Suppose random variables $X_1,X_2,X_3$ are IID and bounded by $[0, R]$.

Why is it obvious that

$$ P(X_1 < X_3 | X_1 < X_2) = \int_0^R f_{X_1}(x_1 | X_1 < X_2)P(x_1 < X_3) dx_1 $$ ?

I think I could get to this formula by using Bayes rule, but in my notes, it seems this result was readily known, and it's not clear to me how that is.

Is it some variant of the law of total probability applied to a conditional probability?

Answer 1

Let events be defined bt $A=(X_1\le X_3)$, $B=X_1\le X_2)$, and $C=(X_2\le X_3)$ So $P(A|B)=\frac{P(A\cap B)}{P(B)}$.

The numerator $P(A\cap B)=P(A\cap B\cap C)+P(A\cap B\cap C')=P(X_1\le X_2\le X_3)+P(X_1\le X_3\lt X_2)$$=\int\limits_0^R\int\limits_0^{x_3}\int\limits_0^{x_2}f_1(x_1)f_2(x_2)f_3(x_3)dx_1dx_2dx_3$$+\int\limits_0^R\int\limits_0^{x_2}\int\limits_0^{x_3}f_1(x_1)f_2(x_2)f_3(x_3)dx_1dx_3dx_2$

The denominator $P(B)=P(X1\le X_2)=\int\limits_0^R\int\limits_0^{x_2}f_1(x_1)f_2(x_2)dx_1dx_2$

Answer 2

Is it some variant of the law of total probability applied to a conditional probability?

Yes. It also relies on the fact that the variables are independent so the events $X_1<X_3$ and $X_1<X_2$ are conditionally independent for a given value of $X_1$ . $$\require{cancel}\color{red}{\xcancel{\begin{align}\mathsf P(X_1<X_3\mid X_1=x, X_1<X_2) &=\mathsf P(X_1<X_3\mid X_1=x)\\[1ex] &=\mathsf P(x<X_3)\end{align}}}$$

Then $$\color{red}{\xcancel{\begin{align}\mathsf P(X_1<X_3\mid X_1<X_2)&=\int_0^R f_{X_1}(x)\,\mathsf P(X_1<X_3\mid X_1=x, X_1<X_2)\,\mathrm d x\\&=\int_0^R f_{X_1}(x)\,\mathsf P(x<X_3)\,\mathrm d x\\&=\int_0^R \tfrac 1R\cdot\tfrac{R-x}{R}\,\mathrm d x&&\text{If uniformly distributed}\\&=\tfrac 12\end{align}}}$$

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Hold on, something seems wrong here. When given that $X_1$ is small, the conditional probability that $X_3$ exceeds it should be greater that the unconditional probability.

Ah, indeed. In fact the events $X_1<X_2$ and $X_1<X_3$ are not conditionally independent when given $X_1$. Rather the events $x<X_2$ and $x<X_3$ are conditionally independent when given $X_1=x$ (since the random variables are mutually independent).

$$\begin{align}\mathsf P(X_1<X_3\mid X_1<X_2)&=\dfrac{\mathsf P(X_1< X_2\cap X_1<X_3)}{\mathsf P(X_1<X_2)}\\&=2 \mathsf P(X_1<X_2\cap X_1<X_3)\\&=2\int_0^R f_{X_1}(x)\,\mathsf P(x < X_2\cap x< X_3\mid X_1=x)\,\mathrm d x\\&=2\int_0^R f_{X_1}(x)\,\mathsf P(x < X_2)\mathsf P(x< X_3)\,\mathrm d x\\&=2\int_0^R R^{-3}(R-x)^2\,\mathrm d x &&\text{If uniformly distributed}\\&=\tfrac 23\end{align}$$

So you should have been given: $$\mathsf P(X_1<X_3\mid X_1<X_2)=\dfrac{\int_0^R f_{\small X_1}(x)\,\mathsf P(x<X_2)\,\mathsf P(x<X_3)\,\mathrm d x}{\int_0^R f_{\small X_1}(x)\,\mathsf P(x<X_2)\,\mathrm d x}$$