This a Continuous Conditional Probability question from my actuarial exam p study manual. I am having trouble understanding the solution.
Question: $X$ has pdf $f(x)=e^{-x}$, for $x>0$. If $a>0$ and $A$ is the event that $X>a$, find $f_{X|A}(x|x>a)$, the density of the conditional distribution of $X$ given that $X>a$. \ \
The Solution: $P(A)=P(X>A)=\int_{a,\infty}e^{-x}=e^{-a}$
$f_{X|A}(x|x>a)=\frac{f_{X}(x)}{P(A)}=\frac{e^{-x}}{e^{-a}}$.
My problem: wouldn't $f_{X|A}(x|x>a)=\frac{P({x>0}\cap x>a )}{P(A)}$
And the set $\{x:x>0 \cap x>a>0 \}=\{x:x>a\}$.
So by that logic, wouldnt $f_{X|A}(x|x>a)=\frac{P(x>a)}{P(x>a)}=1$
For $x>0$ we find: $$P\left(X>x\right)=\int_{x}^{\infty}f\left(u\right)\;du=\int_{x}^{\infty}e^{-u}\;du=\left[-e^{-u}\right]_{x}^{\infty}=e^{-x}$$
Then for $x>a>0$ we find: $$P\left(X>x\mid X>a\right)=\frac{P\left(X>x\wedge X>a\right)}{P\left(X>a\right)}=\frac{P\left(X>x\right)}{P\left(X>a\right)}=\frac{e^{-x}}{e^{-a}}=e^{a-x}$$
This reveals that the CDF of $X$ under condition $X>a$ is the function: $$F_{X\mid X>a}\left(x\right)=P\left(X\leq x\mid X>a\right)=\begin{cases} 1-e^{a-x} & \text{if }x>a\\ 0 & \text{otherwise} \end{cases}$$
Finally we find the PDF of $X$ under condition $X>a$ by taking the derivative of the CDF of $X$ under condition $X>a$ and end up with: $$f_{X\mid X>a}\left(x\right)=\frac{dF_{X\mid X>a}\left(x\right)}{dx}=\begin{cases} e^{a-x} & \text{if }x>a\\ 0 & \text{otherwise} \end{cases}$$