Computing probabilities by conditioning

Question

Given continuous random variables $X$ and $Y$, it is assumed by definition: $$f_{Y|X}(y|x) = \frac{f_{XY}(x,y)}{f_X(x)}$$ $$P(Y\le y|X=x) = \int_{-\infty}^{y}f_{Y|X}(t|x)dt$$ In this context the formula $P(g(X,Y)\le z|X=x) = P(g(x,Y)\le z|X=x)$ was used very liberally to calculate probabilities. Although it is intuitivelly clear where it comes from, I can't give a formal justification of this step using the definition provided. Can someone help?

Answer 1

This should work.

From your definition it's not difficult to get: $$P[g(X,Y) ≤ z | X = x] = lim_nP[g(X,Y) ≤ z | x < X ≤ x+1/n]$$

If $g(x,y)$ is continuous wrt $x$ and $0 < X – x ≤ 1/n$, for each $ε > 0$ exists $n_ε ∈ ℕ$ such that: $$n ≥ n_ε ⇒ g(x,Y) - ε < g(X,Y) < g(x,Y) + ε ⇒ g(X,Y) - ε < g(x,Y) < g(X,Y) + ε.$$

For $n ≥ n_ε$ you can write: $$g(X,Y) ≤ z ⇒ g(x,Y) < g(X,Y) + ε ≤ z + ε ⇒ P[g(X,Y) ≤ z | x < X ≤ x+1/n] ≤ P[g(x,Y) ≤ z + ε | x < X ≤ x+1/n]$$ $$g(x,Y) ≤ z – ε ⇒ g(X,Y) < g(x,Y) + ε ≤ z ⇒ P[g(x,Y) ≤ z – ε | x < X ≤ x+1/n] ≤ P[g(X,Y) ≤ z | x < X ≤ x+1/n]$$

Taking the limit for $n → ∞$ you get: $$P[g(x,Y) ≤ z – ε | X = x] ≤ P[g(X,Y) ≤ z | X = x] ≤ P[g(x,Y) ≤ z + ε | X = x]$$

Now taking the limit for $ε → 0$ and using the continuity laws for $P$ you get the wanted result.