Trying to understand why in some problems I have seen regarding conditional probability, the section after the 'and' disappears without an explanation.
For example I have been looking at this problem:
$Pr(T-t>u|T>t) = \frac{Pr(T>t+u,T>t)}{Pr(T>t)} = \frac{Pr(T>t+u)}{Pr(T>t)}$
I do not understand why the $T > t$ on the numerator seems to disappear.
Why has $Pr(A|B)$ essentially been reduced to $\frac{Pr(A)}{Pr(B)}$ here?
First, to generalize the concept lets have $A \subset B$, and we want $\mathbb{P}(A | B)$. We can write: $$ \mathbb{P}(A | B) = \frac{\mathbb{P}(A \cap B)}{\mathbb{P}(B)}$$ And remember here $A, B = A \cap B$ its just different notation for clarity of the concept. Now, if $A \subset B$ then $A \cap B = A$, which you may remember from discrete math*. We can now write: $$\mathbb{P}(A | B) = \frac{\mathbb{P}(A \cap B)}{\mathbb{P}(B)} = \frac{\mathbb{P}( A)}{\mathbb{P}(B)}$$ Or in your notation: $$\mathbb{P}(A | B) = \frac{\mathbb{P}(A ,B)}{\mathbb{P}(B)} = \frac{\mathbb{P}( A)}{\mathbb{P}(B)}$$ Which should clear up the general idea. Pause for a moment here and try to think to your example as to why we can apply this concept to your problem.
If $T > t + u$, then $T > t$, since $u > 0$. Hence, if we call $B$ the set of events such that $T < t$, and $A$ the set of events such that $T > t + u$, any event in $A$ is also in $B$, thus $A$ is a subset of $B$. That's why we can forget about the second part in your problem. Hope this helps!
*Proof for $A \subset B \Rightarrow A \cap B = A$. Let $A$ and $B$ be two sets such that $A$ is a subset of $B$. Then iff $x \in A \cap B$ then $x \in A$ and $x \in B$, but by definition of subset, $x \in A \Rightarrow x \in B$. Thus our condition can be shortened to iff $x \in A \cap B$ then $x \in A$, and hence $A \cap B = A$.