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Let $X$ be a random variable on a probability space,
and let $\alpha \in \mathbb R$. Show that $\alpha X$ is a random variable.

So this is what I did. Let $x \in \mathbb R$ be given.
Since $X$ is a random variable, $\{\omega \in \Omega : X(w) \le {x\over \alpha}\} \in \mathscr F$.
$\{\omega \in \Omega : \alpha X(\omega) \le x \} \in \mathscr F$.
$\alpha X$ is a random variable.

Now, show that $X - X = 0$, and the random variable always takes the value $0$.
How do you do that?
I have seen many examples that demonstrate how
the random variable gives a numerical value to an event.
But I think that's exactly why I am not grasping it.
How are you doing numerical operations on events?

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How about if you factor $X$ out of the statement you need to show, which gives you $(1-1)X = 0$? –  Trevor Alexander Jun 10 '13 at 2:18
    
I'd state explicitly that the set $\{\omega\in\Omega : \alpha X(\omega)\le x\}$ is the same set as that mentioned on the line before it. –  Michael Hardy Jun 10 '13 at 2:27
    
@ Hardy: Yes, the two sets are the same. I don't know; I can't even verbalize my question .... $x$ is an arbitrary real number, and now the question is asking why the random variable $X - X$ always returns $0$. By definition, the range of $X(\omega) \subset \mathbb R$. So, $X - X$ has to be $0$. –  Andy Tam Jun 10 '13 at 2:35

1 Answer 1

up vote 0 down vote accepted

With regards to your last question, I think maybe what they are trying to coax out of you is to explicitly state that by definition of random variable, there does not exist $\omega \in \Omega$ such that $X(\omega) = \pm \infty$--that is, $X(\omega)$ is finite for all $\omega$. For if that were not the case, then there would exist some $\omega_0$ such that, say, $X(\omega_0)-X(\omega_0) = \infty-\infty$ which is indeterminate.

Anyways, that's just my two cents. It doesn't seem like a problem so much as an exercise in being precise with your language.

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After more thinking, this is what I did. As indicated, since $X \in \mathbb R$, $X - X = 0$. Let $t$ be given. If $t \lt 0$, $\{\omega : X - X(\omega) \le t\} = \{\}$. If $t \ge 0$, $\{\omega : X - X(\omega) \le t\} = \Omega$. Both sets belong to $\mathscr F$. So $X-X$ is a random variable. –  Andy Tam Jun 11 '13 at 16:05

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