# Thread: probability condition F (x | y)

1. ## probability condition F (x | X<K)

I know that maybe this issue can be pre-college, but hey, an apology in advance.
This is the problem:

"The random variable $\displaystyle X$ is defined on the interval [0,2]. It is known that:
$\displaystyle P (X = 1) = \frac{1}{4}$
$\displaystyle F (x | X <1) = x^2$
$\displaystyle F (x | X> 1) = x-1$
$\displaystyle E (X) = 1$

calculate: $\displaystyle F (1)$."

then, i did this:

(i) $\displaystyle F(x|X<1)=\sum\limits_{0}^{2} p(x|X<1)$ .... I apply the Baye’s theorem: $\displaystyle F(x|X<1)=\sum\limits_{0}^{2}\frac{p(X<1|x) *p(x)}{P(X<1)}$

on the other hand: $\displaystyle p(X<1)=p(X=0)$ cuz $\displaystyle X \rightsquigarrow [0,2] \Longrightarrow F(x|X<1)=\sum\limits_{0}^{2}\frac{p(X=0|x) *p(x)}{P(X=0)} = x^2$

(ii) in the case of $\displaystyle F(x|X>1)=x-1$ :
$\displaystyle F(x|X>1)=\sum\limits_{0}^{2} p(x|X>1)$ …. I apply the Baye’s theorem: $\displaystyle F(x|X>1)=\sum\limits_{0}^{2} \frac{p(X>1|x) *p(x)}{P(X>1)}$

on the other hand: $\displaystyle p(X>1)=p(X=2)$ cuz $\displaystyle X \rightsquigarrow [0,2] \Longrightarrow F(x|X>1)=\sum\limits_{0}^{2} \frac{p(X=2|x) *p(x)}{P(X=2)} = x-1$

(iii) $\displaystyle p(X=0)=\frac 3 4 - p(X=2)$

$\displaystyle p(X=2)=\frac 3 4 - p(X=0)$

Note: notice the difference between $\displaystyle X$ and $\displaystyle x$ (uppercase and lowercase)

then I think I'm wrong from the foundation, we would appreciate if someone guide me in this problem, and another thing, is $\displaystyle P (X = 1) = P (1)$?

2. is this zero again?
I have no idea what F (x | X <1) means.
is that P(X<x|X<1)?

3. Originally Posted by matheagle
is this zero again?
I have no idea what F (x | X <1) means.
is that P(X<x|X<1)?
ready, so I edited in code "LaTex" I hope it is better understood

4. Originally Posted by killertapia
ready, so I edited in code "LaTex" I hope they will understand better
I'm strugglng to understand what you're trying to say here.

It appears that X is a discrete random variable that can take on the value 0, 1 and 2 with non-zero probability.

You know Pr(X = 1) = 1/4.
From E(X) = 1 it follows that Pr(X = 2) is 3/8.
From Pr(X = 0) + Pr(X = 1) + Pr(X = 2) = 1 it follows that Pr(X = 0) = 3/8.

So the distribution is now completely defined. I have no idea what F is meant to be but whatever it is, F(1) can surely be calculated from what I have said above.