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 $\displaystyle P (X = 1) = P (1)$?