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In an exercise, I have to compute $\mathbb P(g(X,Y)\leq a)$ where $g:\mathbb R^2\to \mathbb R$ and $(X,Y)$ has joint distribution $\mu_{X,Y}$. I know that $$\mathbb P\{X\in A,Y\in B\}=\int_{A}\mathbb P(Y\in B\mid X=x)\mu_X(dx).$$ But I was wondering : does $$\mathbb P(g(X,Y)\leq a)=\int_{\mathbb R}\mathbb P(g(x,Y)\leq a\mid X=x)\mu_X(dx)\ \ ?$$

I wouldn't be surprise, but I'm not so sure. What I tried is : if $(X,Y)$ has density $f_{X,Y}$, then $$\mathbb P(g(X,Y)\leq a)=\iint_{\{g(x,y)\leq a\}}f_{X,Y}(x,y)dxdy=\int_{\mathbb R}dx\int_{\{y\mid g(x,y)\leq a\}}f_{X,Y}(x,y)dxdy=\int_{\mathbb R}f_{X}(x)dx\int_{\{y\mid g(x,y)\leq a\}}\underbrace{\frac{f_{X,Y}(x,y)}{f_X(x)}}_{=f_{Y|X=x}(y)}dy=\int_{\mathbb R}\mathbb P(g(x,Y)\leq a\mid X=x)f_X(x)dx.$$

So, I strongly suspect that the result is true in the more general case, but how can I prove it ?

Walace
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  • It is unclear what level of rigor would convince you. At the basic level, what you are trying to prove is simply the Law of Total Probability, continuous version. At the rigorous level, this likely requires measure theory and going back to the definition of $P(\cdot \mid X = x)$. Maybe see this for a start and see if that is helpful? – antkam Apr 25 '20 at 19:25

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