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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm self studying probability theory and I'm stuck in the following problems

1) Prove the following for a random variable $X$ with cdf $F$

$$E(x)=\int_0^\infty (1-F(x)) dx - \int_\infty^0 F(x) dx$$

$$\text{Var}(x)=2 \int \!\!\! \int_{-\infty<x<y<+\infty}\ F(x)(1-F(y)) dxdy$$

Find the equivalent type for $\mathbb{E}(x^n)$ and $\mathbb{E}(|x|^n)$

Hint:use the Tonelli-Fubini theorem.

2)Let $\mathbb{X}_n$,$\mathbb{X}$ r.v. with $\mathbb{X}_n\geq0$,$\mathbb{X}_n\rightarrow\mathbb{X}$ and $\mathbb{E}(X)\leq c<\infty$. Prove that $\mathbb{X}$ is integrable and $\mathbb{E}(X)\leq c$

3)Let $\mathbb{X}$,$\mathbb{Y}$,$\mathbb{Z}$ and $\mathbb{X}_n$,$\mathbb{Y}_n$,$\mathbb{Z}_n$ r.v. with $\mathbb{X}_n\rightarrow\mathbb{X}$,$\mathbb{Y}_n\rightarrow\mathbb{Y}$,$\mathbb{Z}_n\rightarrow\mathbb{Z}$, $\mathbb{X}_n\leq\mathbb{Y}_n\leq\mathbb{Z}_n$, $\mathbb{E}(X_n)\rightarrow\mathbb{E}(X)$ and $\mathbb{E}(Z_n)\rightarrow\mathbb{E}(Z)$.

Prove that if $\mathbb{E}(Z)$ and $\mathbb{E}(X)$ exist then $\mathbb{E}(Y)$ exists and $\mathbb{E}(Y_n)\rightarrow\mathbb{E}(Y)$.

share|cite|improve this question
Welcome to Math SE. I have improved the formatting a little. Please check that I didn't change the meaning of anything in your question. – Martin Argerami Dec 11 '12 at 2:39
Thank you very much.It looks much better. – user52561 Dec 11 '12 at 2:50
@user52561 Maybe you should ask these as separate questions and explain for each one of them where you are stuck. – Learner Dec 11 '12 at 2:59
up vote 3 down vote accepted

I will answer the first question only.

Write $X = X^+ - X^-$ where $X^+ = \max \left( X, 0 \right)$ and $X^- = \max \left( -X, 0 \right)$. Assume that for integer $p \geqslant 1$, $E \left[ \left| X \right|^p \right] < \infty$. Let $f$ be the density (just to make the derivations here simpler to write). By integration by part (and truncating at the point $m > 0$) \begin{eqnarray*} E \left[ \left( X^+ \right)^p 1 \left( X^+ < m \right) \right] & = & \int_0^m x^p f \left( x \right) \mathrm{d} x\\ & = & \left[ x^p \left( F \left( x \right) - 1 \right) \right]^m_0 + p \int_0^m x^{p - 1} \left( 1 - F \left( x \right) \right) \mathrm{d} x\\ & = & - m^p P \left[ X^+ > m \right] + p \int_0^m x^{p - 1} \left( 1 - F \left( x \right) \right) \mathrm{d} x \end{eqnarray*} By Markov inequality $P \left[ X^+ > m \right] \leqslant \frac{E \left[ \left( X^+ \right)^p \right]}{m^p}$ which converges to 0 as $m \rightarrow \infty$, implying $$ E \left[ \left( X^+ \right)^p \right] = p \int_0^{\infty} x^{p - 1} \left( 1 - F \left( x \right) \right) \mathrm{d} x $$ Replace $p$ by 1 and 2 respectively to get $E \left[ X^+ \right] = \int_0^{\infty} \left( 1 - F \left( x \right) \right) \mathrm{d} x$ and $E \left[ \left( X^+ \right)^2 \right] = 2 \int_0^{\infty} x \left( 1 - F \left( x \right) \right) \mathrm{d} x$. Do the same for $X^-$.

Once what is above is grasped, it is possible to derive the rest in the same way.

share|cite|improve this answer
Excellent answer! Very useful and inspiring!:-) – Stupid_Guy Nov 16 '14 at 3:39

Your Answer


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.