- #1
Eidos
- 108
- 1
Hey guys
I'd like a steer in the right direction with this problem.
I would like to show that
[tex]P\{x_1\leq X \leq x_2\}=F_{X}(x_2)-F_{X}(x_1^{-})\quad(1)[/tex]
Where:
[tex]X[/tex] is a random variable.
[tex]F_{X}(x) \equiv P\{X \leq x \} [/tex] is its cumulative distribution function.
My notes only give an example (using dice) to show that this is true.
Generally
[tex]P\{x_1 < X \leq x_2\}=F_{X}(x_2)-F_{X}(x_1)\quad(2)[/tex]
and
[tex]P\{X = x_2\}=F_{X}(x_2)-F_{X}(x_2^{-})\quad (3)[/tex]
the latter of which is easy to prove.
I've been trying to rewrite (1) in terms of (2) & (3) but have had no success so far.
Any ideas would be most welcomed![Smile :smile: :smile:](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
I'd like a steer in the right direction with this problem.
I would like to show that
[tex]P\{x_1\leq X \leq x_2\}=F_{X}(x_2)-F_{X}(x_1^{-})\quad(1)[/tex]
Where:
[tex]X[/tex] is a random variable.
[tex]F_{X}(x) \equiv P\{X \leq x \} [/tex] is its cumulative distribution function.
My notes only give an example (using dice) to show that this is true.
Generally
[tex]P\{x_1 < X \leq x_2\}=F_{X}(x_2)-F_{X}(x_1)\quad(2)[/tex]
and
[tex]P\{X = x_2\}=F_{X}(x_2)-F_{X}(x_2^{-})\quad (3)[/tex]
the latter of which is easy to prove.
I've been trying to rewrite (1) in terms of (2) & (3) but have had no success so far.
Any ideas would be most welcomed