Find conditional CDF

Question

The PDF of the random variable X is:
$$f_X(x)=\begin{cases} 1\over3 & , -2\leq x\leq 1\\ 0 & , \text{otherwise}
\end{cases}$$
I've observed that this is a uniform random variable (-2,1), I calculated the CDF:
$$F_X(x)=\begin{cases} 0 & , x ⋹ (-\infty,-2]\\ x+2\over3 & , x ⋹ (-2, 1] & \\1 &, \text{otherwise}
\end{cases}$$
and now I want to calculate the conditional CDF of $F_{X|X>6}$. First, I thought I would calculate the conditional PDF with the formula $f_{X|X>6} = \begin{cases} f_X(x)\over P(X>6) & , x>6\\ 0 & , \text{otherwise}\end{cases}$, but this gave me 0 since $P(X>6) = 1 – P(X\leq6)$. Am I doing something wrong? How would the conditional CDF be found if I manage to find a valid conditional PDF? Thank youy!

Answer 1

Since $Pr(X<6)=1$ (as it can only take values between $-2$ and $1$), we have that $F_{X|X>6}=1$.