Conditional continuous probability on a fixed point

Question

I was asked to show the following statement:

Let $X, Y$ both be equally distributed random variables on $[0,1]$, we define

$$ P(\{Y \leq y\}|\{X=x\}) = \lim_{h \downarrow 0} P(\{Y \leq y\}|\{x \leq X \leq x+h \}). $$

Show that $$ P(\{Y \leq g(X) \}|\{X=x\}) = g(x)$$.

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We haven't covered joint probability distribution functions yet. In the end I got stuck at evaluating

$$ P(\{Y \leq g(X)\} \cap \{x\leq X \leq x+h\})$$

Thanks!

Answer 1

For $x \in [0, 1]$, we have $$P(Y \leq g(X) \mid X = x) = P(Y \leq g(x) \mid X = x).$$

Assuming $X$ and $Y$ are independent, we can say further that $P(Y \leq g(x) \mid X = x) = P(Y \leq g(x))$, so $$P(Y \leq g(X) \mid X = x) = P(Y \leq g(x)).$$ Now, assuming that $Y$ is uniformly distributed on $[0, 1]$, we get $$P(Y \leq g(X) \mid X = x) = P(Y \leq g(x)) = g(x).$$

Edit: Here is the approach computing the limit $\lim_{h \to 0}P(Y \leq g(X) \mid x_0 \leq X \leq x_0 + h)$ directly: \begin{align} P(Y \leq g(X), x_0 \leq X \leq x_0 + h) &= \int_{0}^{1}\int_{0}^{1}I(y \leq g(x), x_0 \leq x \leq x_0 + h)\,dy\,dx \\ &= \int_{x_0}^{x_0 + h}\int_{0}^{1}I(y \leq g(x))\,dy\,dx \\ &= \int_{x_0}^{x_0 + h}g(x)\,dx, \end{align} and $$P(x_0 \leq X \leq x_0 + h) = h.$$ So $$\lim_{h \to 0}P(Y \leq g(X) \mid x_0 \leq X \leq x_0 + h) = \lim_{h \to 0}\frac{1}{h}\int_{x_0}^{x_0 + h}g(x)\,dx = g(x_0)$$ assuming $g$ is continuous at $x_0$.

Answer 2

When $X$ and $Y$ are independent and uniformly distributed on$[0,1]$, the probability of an event involving $X$ and $Y$ is the area of the region described by that event. If you draw out the region described by the event $\{Y \leq g(X)\} \cap \{x\leq X \leq x+h\}$, we see it is the area below the curve $y=g(x)$ between $x$ and $x+h$. Using the fact that that area is given by a certain integral, we get $$ P(\{Y \leq g(X)\} \cap \{x\leq X \leq x+h\})=\int_x^{x+h}g(t)\,\mathrm dt\tag1 $$ Therefore, $$ \lim_{h \downarrow 0} P(\{Y \leq y\}|\{x \leq X \leq x+h \})=\lim_{h\downarrow 0} \frac{P(\{Y \leq g(X)\} \cap \{x\leq X \leq x+h\})}{P(x\le X\le x+h)}=\frac1h\int_x^{x+h}g(t)\,\mathrm dt $$ This fact that this limit equals $g(x)$ is a consequence of the fundamental theorem of calculus.

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The only place where my argument lacks rigor is when I stated $(1)$. I am not sure how you are expected to justify this; it is a pretty low-level measure theoretic result.