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Given the question: $$\text{Show that}\ F_X(x\mid A) = \dfrac{\Pr(A\mid X\leq x)}{\Pr(A)}\cdot F_X(x)$$

I have seen the solution via probabilities 'directly'.

My question is whether the following method was (unnecessary extra work, but nonetheless) valid:

\begin{align} & F_X(x\mid A) = \int_{-\infty}^x f_X(t\mid A)\, dt \\[8pt] = {} & \int_{-\infty}^x \dfrac{\Pr(A\mid X=t)}{\Pr(A)} f_X(t)\, dt \\[8pt] = {} & \dfrac{\Pr(A\mid X\leq x)}{\Pr(A)} \int_{-\infty}^x f_X(t)\, dt \\[8pt] = {} & \dfrac{\Pr(A\mid X\leq x)}{\Pr(A)}\cdot F_X(x) \end{align}

I suppose my doubts lie in the original formulation as integral of $f_X(x\mid A)$, and also the extraction of $\Pr(X=x)$ to be $\Pr(X\leq x)$ outside the integral - which seems to intuitively make sense, but I'm not sure that's actually sound?

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  • $\begingroup$ Thanks for the edit @michael-hardy - you ask a math question, and learn some better tex for free! :) $\endgroup$
    – OJFord
    May 17, 2015 at 16:52
  • $\begingroup$ I see you are trying to use Bayes rule on the conditional PDF. Some quick thoughts on this: 1. What's your definition of $f_X(t~|~A)$? I would think of something like: $\frac{f_X(t)\mathbf{1}_A(t)}{\int_{A}f(\tau)d\tau}$. 2. Since $Pr(X=t)=0$, what is the definition of $Pr(A~|~X=t)$? $\endgroup$
    – Aprilius
    May 17, 2015 at 20:14
  • $\begingroup$ @Aprilius : I would not take that to be a definition. The conditional density (which I prefer to denote as $x\mapsto f_{X\mid A}(x)$) is characterized by the fact that $\displaystyle \Pr(X\in S\mid A) = \int_S f_{X \mid A}(x)\,dx$ for every Borel set $S$. It is also characterized by the seemingly-but-demonstrably-not-really-weaker statement that for every number $x_0$, $\displaystyle\Pr(X\le x_0\mid A) = \int_{-\infty}^{x_0} f_{X\mid A}(x)\,dx$. Those can be taken to be a definition. ${}\qquad{}$ $\endgroup$ May 17, 2015 at 22:54
  • $\begingroup$ The more substantial question is what one means by $\Pr(A\mid X=x)$. One cannot define it as a quotient $\Pr(A\ \&\ X=x)/\Pr(X=x)$ when $X$ has a continuous distribution because that amounts to zero over zero. Nonetheless there is a reasonable way to define it. Maybe I'll post an answer below. ${}\qquad{}$ $\endgroup$ May 17, 2015 at 22:57

1 Answer 1

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@Michael Hardy, you are definitely correct. Also, I found this question interesting since it kind of connects the undergraduate notions of conditional probability with the measure theoretical definitions. This is too long for a comment so I just posted it as an answer. See if I got this correct:

Given a probability space $(\Omega,~\mathcal{F},P)$, define the sub-$\sigma$ algebra of $\mathcal{F}$ as

$$ \mathcal{G}=\{\emptyset,~A,~A^c,~\Omega\}, $$ for some set $A\in\mathcal{F}$. Then, we have $E[1_B~|~\mathcal{G}]$ satisfies $$ E[1_B~|~\mathcal{G}]\int_Af_X(t)dt =\int_A1_B(t)f_X(t)dt,~\textrm{on}~A,~~~~(1) $$ where $1_B(\cdot)$ is the indicator function. We didn't take quotient for the reason which is commented by @Michael Hardy above. This can be readily verified by computing

$$ E[E[1_B~|~\mathcal{G}];~A]=\int_AE[1_B~|~\mathcal{G}]f_X(t)dt =\int_A1_B(t)f_X(t)dt=E[1_B;~A]. $$ For simplicity, we will denote $E[1_B~|~\mathcal{G}]$ on $A$ as $E[1_B~|~A]$.

Similarly, it can be verified that $E[1_A~|~X\leq x]$ satisfies

$$ E[1_A~|~X\leq x]\int_{-\infty}^xf_X(t)dt =\int_{-\infty}^x 1_A(t)f_X(t)dt.~~~~(2) $$

With these preparations, I can work out a similar proof as OP but in a more strict way: \begin{align} F_X(x~|~A)=&E[1_{\{X\leq x\}}~|~A]\\ =&\frac{\int_A1_{\{X\leq x\}}(t)f_X(t)dt}{\int_Af_X(t)dt}\\ =&\frac{\int_{-\infty}^x1_{A}(t)f_X(t)dt}{\int_Af_X(t)dt}\\ =&\frac{E[1_A~|~X\leq x]\int_{-\infty}^xf_X(t)dt}{\int_Af_X(t)dt}\\ =&\frac{Pr(A~|~X\leq x)F_X(x)}{Pr(A)}, \end{align}
where the second equality follows from $Pr(A)>0$ and (1), and the fourth equality follows from (2).

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