When continuous random variable is in both the conditional and the conditioned Let $A,B$ be continuous random variables. Let $E,G,K$ be events. Let $t \in image(B)$.
(I forgot if any 2 continuous random variables necessarily have a well-defined joint pdf. If not, then assume joint pdf is well-defined whenever necessary.)
Question 1: Can we say $P(B \le A|B=t) = P(t \le A|B=t)$?
Question 2: If so, then how? If not then why?
What I've tried: I think we're doing something like $P(E|K)=P(E \cap K|K)$, and I get it when $P(G|K)$ is defined for $P(K)>0$. What is being done here when technically $P(K)=0$? I mean, of course, in the 1st place when we say like '$P(\cdot|B=t)$', this is notational, we're not really conditioning on the $P$-null event $\{B=t\}$. But I still don't get exactly what's being done here. Well...
As I understand, $$RHS = P(t \le A|B=t) := \frac{\int_{t \le A} f_{A,B}(a,t) da}{\int_{\mathbb R} f_{A,B}(a,t) da} = \frac{\int_{t}^{\infty} f_{A,B}(a,t) da}{\int_{\mathbb R} f_{A,B}(a,t) da}$$
$$= \frac{\int_{t}^{\infty} f_{A,B}(a,t) da}{f_{B}(t)}= \int_{t}^{\infty} \frac{f_{A,B}(a,t)}{f_{B}(t)} da,$$ where/whence $f_{A|B=t}(a) := \frac{f_{A,B}(a,t)}{f_{B}(t)}$.
That's all I got.
Not really sure how to evaluate LHS. Perhaps LHS := RHS, i.e. LHS is defined as RHS?
Update: Wait I think I have an idea how to do LHS
$$P(B \le A|B=t) = \int \int_{b \le a} f_{(A,B)|B=t}(a,b) da db$$
Here,

*

*'$f_{(A,B)|B=t}(a,b)$' is I guess meant like $f_{(A,B)|T=t}(a,b)$ where $T$ is a random variable that is (not just almost surely but really) surely equal to $B$.


*'$f_{(A,B)|T=t}(a,b)$' is meant like 2 random variables conditioned on a 3rd random variable. As I understand multivariate conditional joint distributions, we have this is $f_{(A,B)|T=t}(a,b) := \frac{f_{A,B,T}(a,b,t)}{f_T(t)}$ and then...


*$$P(B \le A|B=t) = \int \int_{b \le a} f_{(A,B)|B=t}(a,b) da db = \int \int_{b \le a} \frac{f_{A,B,T}(a,b,t)}{f_T(t)} da db$$
$$= \int \int_{b \le a} \frac{f_{A,B,T}(a,b,t)}{\int \int_{(a,b) \in \mathbb R^2} f_{A,B,T}(a,b,t) da db} da db$$
$$= \frac{\int \int_{b \le a} f_{A,B,T}(a,b,t) da db}{\int \int_{(a,b) \in \mathbb R^2} f_{A,B,T}(a,b,t) da db} $$


*And then I don't know. Apparently we might not be able to have a joint pdf if 2 of the random variables are surely equal (maybe even for almost surely):

*

*4.1. joint distribution of x with..itself


*4.2. https://stats.stackexchange.com/questions/47162/what-is-the-joint-probability-distribution-of-two-same-variables
 A: Here is an argument that uses no measure theory that you might like.  A measure theory argument that runs parallel to this one can be made using conditional expectation $E[1\{B \leq A\}|B]$ and by slightly modifying the claim.

Assume the density of $B$ is continuous at point $t$ and $f_B(t)>0$. Under this assumption, for an event $C$ that satisfies either $P[C]=0$ or $f_{B|C}(x)$ exists and is continuous at $x=t$, it can be shown the following limits exist and are equal:
$$ \lim_{\delta \rightarrow 0^+} P[C|B \in [t, t+\delta]]=\lim_{\delta \rightarrow 0^+} P[C|B \in [t-\delta, t]]$$
So under these assumptions we can define
$$ P[C|B=t] = \lim_{\delta \rightarrow 0^+} P[C|B \in [t, t+\delta]]$$

Now for any $\delta>0$ we have
$$ P[B \leq A|B \in [t, t+\delta]] \leq P[t \leq A | B \in [t, t+\delta]] $$
Taking $\delta \searrow 0$ and assuming we converge appropriately gives
$$ P[B \leq A | B=t] \leq P[t \leq A |B=t]$$
For the reverse inequality, observe that for any $\delta>0$ we have
$$ P[B\leq A | B \in [t-\delta, t]] \geq P[t \leq A | B\in [t-\delta, t]] $$
Taking limits and assuming the limits converge appropriately gives
$$  P[B \leq A | B =t] \geq P[t \leq A |B=t]$$
A: Here is a measure theory argument that runs parallel to my non-measure theory argument.

Let $A, B$ be random variables. Suppose $B$ has a density $f_B(x)$ that is continuous at the point $x=t$, and $f_B(t)>0$.   Fix $t \in \mathbb{R}$.  Let $h:\mathbb{R}\rightarrow \mathbb{R}$ and $g:\mathbb{R}\rightarrow\mathbb{R}$ be bounded and measurable functions that define versions of the following conditional expectations:
\begin{align}
&h(B) = E[1\{B \leq A\}|B]\\
&g(B) = E[1\{t \leq A\}|B]
\end{align}
Claim:   If $h(x)$ and $g(x)$ are continuous at $x=t$ then $h(t)=g(t)$.
Proof: Fix $\delta>0$. Observe that
$$ 1\{B \leq A\}1\{B \in [t, t+\delta]\} \leq 1\{t \leq A\}1\{B \in [t, t+\delta]\}$$
Thus
$$ E[1\{B \leq A\}1\{B \in [t, t+\delta]\}] \leq E [1\{t \leq A\}1\{B \in [t, t+\delta]\}]$$
and so, by the definition of a conditional expectation:
$$ E[h(B)1\{B \in [t, t+\delta]\}] \leq E[g(B)1\{B \in [t, t+\delta]\}]$$
Thus
$$ \int_{t}^{t+\delta} h(x)f_B(x)dx \leq \int_t^{t+\delta}g(x)f_B(x)dx$$
This holds for all $\delta>0$ and since $h, g$ are both continuous at $x=t$, and $f_B(x)$ is continuous at $x=t$ and $f_B(t)>0$, we get
$$ h(t)\leq g(t)$$
A similar argument shows the reverse inequality.
A: Here is a third approach that assumes the joint PDF $f_{A,B}(a, b)$ exists.
Define $Z=B-A$.   We have the PDF transformation:
$$ f_{Z,B}(z,b) = f_{A,B}(b-z,b)$$
Then, assuming $f_B(t)>0$:
\begin{align}
P[B\leq A | B=t] &= P[Z\leq 0|B=t]\\
&=\int_{-\infty}^0 f_{Z|B}(z|t)dz\\
&=\int_{-\infty}^0 \frac{f_{Z,B}(z,t)}{f_B(t)} dz\\
&=\int_{-\infty}^0 \frac{f_{A,B}(t-z,t)}{f_B(t)}dz\\
&\overset{(a)}{=}\int_t^{\infty} \frac{f_{A,B}(u,t)}{f_B(t)}du\\
&=\int_t^{\infty} f_{A|B}(u|t)du\\
&=P[A\geq t|B=t]
\end{align}
where (a) uses the change of variable $u=t-z$ and $du=-dz$.
