Random Variable over probability space

[Francobati]

Hello. Can you help me solve it? ($F$ is a $\sigma $ algebra).
Let $X$ be a rv over $(\Omega ,F,P)$. Set $Y:= min\left \{ 1,X \right \}$. What statement is TRUE?

(1): $\left \{ Y=X \right \}\neq \Omega $;
(2): $F_Y(x)=F_X(x)$ for every $x\epsilon \Re $;
(3): $Y\leqslant X$ for every outcome;
(4): $E(Y)=\int_{-\infty}^1xd F_{X}(x)$;
(5): $E(Y)=\int_{\Re }max\left \{ 1,x \right \}d F_{X}(x)$.

 

[Euge]

What are your thoughts on this problem? Can you eliminate any of the answer choices?

 

[Francobati]

I have to find the true answer among these five.

 

[Euge]

That's clear, but you haven't mentioned which answer choices you know can be eliminated.

 

[Francobati]

I have no idea. What information do I need to resolve it? Help me, please.

 

[Jameson]

Ok, let's start with what is the definition of $E[X]$ and $F(x)$? What do these two concepts represent in general for random variables?

 

[Francobati]

$E(X)=:x_{1}P(A_{1})+...+x_{k}P(A_{k})\geqslant 0$
$F(x)=P(X\leqslant x)$

 

[Euge]

By the very definition of $Y$, isn't it the case that $Y(\omega) \le X(\omega)$ for all $\omega\in \Omega$?

 

[Francobati]

Yes, the answer is 3. But as I explain in an exhaustive way?

 

[Euge]

For all pairs of real numbers $a$ and $b$, $\min\{a,b\} \le a$ and $\min\{a,b\} \le b$. So, for all $\omega\in \Omega$, $Y(\omega) \le 1$ and $Y(\omega) \le X(\omega)$. In particular, for every $\omega\in \Omega$, $Y(\omega) \le X(\omega)$.

For the other cases, you can explain why they are false using counterexamples. Take for instance statement (1). If we let $\Omega = [0,1]$ and $X$ the indicator of $[0,1/2]$, then $X\le 1$ and hence $Y$ is identical to $X$; therefore, the event $(Y = X)$ is equal to the whole sample space $\Omega$.

 

[Francobati]

Ok. Many thanks.