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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!

Andrei0408
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Since $Pr(X<6)=1$ (as it can only take values between $-2$ and $1$), we have that $F_{X|X>6}=1$.

Vons
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  • But, since we found the probability of X < 6, shouldn't we subtract it from 1 to get the prob of X >= 6? Isn't your solution $F_{X|X<6}$? – Andrei0408 Apr 19 '21 at 18:12
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    The definition of CDF is the probability that $X$ is less than the value given. Since we are conditioning on the fact that the value given is greater than 6, and $X$ can take values only between $-2$ and $1$, we have that $F_{X|X>6}(x)=Pr(X<x|x>6)=1$. Or did I misunderstand your question? – Vons Apr 19 '21 at 18:20
  • Yes, that's right. I was a bit confused because when I first tried doing I subtracted pr of X < 6 from 1 and got 0. I understand now – Andrei0408 Apr 19 '21 at 18:24