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$X$ and $Y$ are real-valued random variables such that the distribution of $(X,Y)$ is absolutely continuous with density function $p$ and let $p_x$ denote the marginal density function of $X$. Suppose that there exists a point $x_0$ such that $p_x(x_0) > 0$, $p_x$ is continuous at $x_0$, and for almost all $y$, $p(\cdot,y)$ is continuous at $x_0$. Let $A$ denote a subset of $\mathbb{R}$. For each $\epsilon \gt 0$, let

$$d(\epsilon)=\Pr(Y \in A | x_0 \leq X \leq x_0 + \epsilon).$$

Show that $\Pr[Y \in A|X = x_o]$ = $\lim_{\epsilon \to 0} d(\epsilon).$

Attempt:

$$\eqalign{ \lim_{\epsilon \to 0} d(\epsilon) &= \lim_{\epsilon \to 0}\Pr(Y \in A|x_0≤ X≤ x_0+\epsilon)\\ &= \Pr(Y∈A|x_0≤X≤x_0+0)\\ &= \Pr(Y \in A|x_0≤ X ≤x_0)\\ &= \Pr(Y \in A|X = x_0). }$$

I realize this attempt is wrong, but I'm not sure how to approach this problem.

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  • $\begingroup$ At some point--probably several points--you will need to invoke and exploit a definition of conditional probability. How you proceed depends on what definition(s) you know. What are they? $\endgroup$
    – whuber
    Commented Jun 23, 2014 at 12:19

1 Answer 1

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Recalling the definition of conditional density we have $$ \Pr(Y\in A\mid X=x_0) = \int_A f_{Y\mid X}(y\mid x_0)\,dy = \frac{1}{f_X(x_0)} \int_A f_{X,Y}(x_0,y)\,dy \, . $$ For $\epsilon>0$, consider $$ \Pr(Y\in A\mid x_0\leq X\leq x_0+\epsilon) = \frac{\Pr(Y\in A, x_0\leq X\leq x_0+\epsilon)}{\Pr(x_0\leq X\leq x_0+\epsilon)} \, . $$ Very informally (drawing a figure may help), if $\epsilon$ is "sufficiently small" and, for fixed $y$, $f_{X,Y}(\,\cdot\,,y)$ "varies slowly" inside the interval $[x_0,x_0+\epsilon]$, we can approximate $$ \begin{align} \Pr(Y\in A, x_0\leq X\leq x_0+\epsilon) &= \int_{\{(x,y):x_0\leq x\leq x_0+\epsilon, y\in A\}} f_{X,Y}(x,y)\,dxdy \\ &\approx \epsilon \int_A f_{X,Y}(x_0,y)\, dy \, . \end{align} $$ Analogously, $$ \Pr(x_0\leq X\leq x_0+\epsilon) \approx \epsilon\,f_X(x_0)\, . $$ Hence, $$ \lim_{\epsilon\to 0} \Pr(Y\in A\mid x_0\leq X\leq x_0+\epsilon) = \Pr(Y\in A\mid X=x_0) \, . $$

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    $\begingroup$ (+1) It wouldn't take much to make this rigorous by appealing to the Intermediate Value Theorem for the integral approximations (which would replace the $x_0$ in the integrands with some value $x^\prime$ in the interval $[x_0, x_0+\epsilon]$) and then invoking the continuity of $p_x$ to take the limit as $\epsilon\to 0$. $\endgroup$
    – whuber
    Commented Jun 23, 2014 at 14:33

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