Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $X$ be a real valued random varaible on a probability space ($\Omega , \mathscr F ,\mathsf P)$ Carefully stating any result you use explain why $\exp(c|X|)$ is a real valued random variable for $c>0.$

Assume $\mathsf E[\exp(c|X|)]< \infty $ for some $c>0$ by the same reasons as above, $\exp(tX)$ is a real valued random variable for every real number $t$, show $\exp(tX)$ is integrable on $(\Omega , \mathscr F , \mathsf P)$ for all $t\in (-c,c).$

For the first part, the bit I am struggling with is the modulus, I know that a continuous function on a probability space is a random variable but I though |X| was not continuous?

For the second part, is it correct to say that |exp(c|X|)| $\geq$ |exp(tX)| for t$\in $(-c,c) so therefore the integral |exp(c|X|)| $\geq$ Integral |exp(tX)| and since the integral of the LHS is finite because we know exp(c|X|) is intregable this means that the integral of the RHS is finite too, thus exp(tX) is integrable ?

This is not a homework question, it is a question from a mock paper to which there are no solutions.

share|improve this question
    
The function $f(x)=|x|$ is continuous. –  André Nicolas May 21 '12 at 14:46
    
Dear Rosie, next time you visit this chat, don't forget to upvote an answer which you've found useful (arrow up to the left) and to accept it whenever it completely answers your question (check sign in the same place) –  Ilya May 23 '12 at 20:30

1 Answer 1

up vote 0 down vote accepted

First of all $\exp(c|X|)$ is real-valued because $X$ is real-valued. That $X$ is a random variable means that $X$ is $(\mathscr{F},\mathcal{B}(\mathbb{R}))$-measureable. Now $\exp(c|X|)$ is just the composition $g(X)$, where $g: \mathbb{R}\to \mathbb{R}$ is given by $g(x)=\exp(c|x|)$, which is obviously $(\mathcal{B}(\mathbb{R}),\mathcal{B}(\mathbb{R}))$-measureable because it is continuous. Now compositions preserves measureability and hence $g(X)$ is $(\mathscr{F},\mathcal{B}(\mathbb{R}))$-measureable, i.e. it is a random variable.

What you did concerning the integrability of $\exp(tX)$ is correct.

share|improve this answer
    
thanks very much. –  Rosie May 21 '12 at 15:45

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.