# Problems on expected value

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 $\mathbb{E}(Y_n)\rightarrow\mathbb{E}(Y)$.

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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

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.

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Excellent answer! Very useful and inspiring!:-) –  Stupid_Guy Nov 16 at 3:39