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The exercise states:

Show that $X_n{\buildrel 1 \over \to} X \implies E(X_n) \to E(X)$.

Does the converse implication hold?

$\underline{SOLUTION}$

For the first part i did the following:

$|E(X_n)-E(X)| \leq\{Jensen's\hspace{1mm} inequality\} \leq E(|X_n - X|) \to 0\text{(by assumption)}$.

I'm having trouble proving or disproving the second part (perhaps a counter-example).

Thanks in advance.

Davide Giraudo
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teo theo
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1 Answers1

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Hint for the counter-example : you could find a very simple sequence of random variables such that $\mathbf{E}[X_n] = 0$ for every $n$, and so $\mathbf{E}[X_n] \to \mathbf{E}[0]$ obviously, but such that $X_n$ does not converge to $0$ in $L^1$, i.e. such that $\mathbf{E}[|X_n|]$ does not goes to zero...

Tlön Uqbar Orbis Tertius
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  • I used the following example. Let$$X_n=\begin{cases} \hspace{3mm}1&, & \text{with $p=\frac{1}{2}$} \ -1&, & \text{with $p=\frac{1}{2}$} \end{cases} $$ Hence, $E(X_n)=1\cdot\frac{1}{2}-1\cdot\frac{1}{2} = 0 = E(0)$. So, $E(X_n) \to E(0)$.

    But, $ X_n{\buildrel 1 \over \nrightarrow}0$ because $$E|X_n-0|^1 = E|X_n| = |1|\cdot\frac{1}{2}+|-1|\cdot\frac{1}{2} = 1$$

     – teo theo Nov 11 '15 at 16:16
  • This is exactly the example I was thinking of. – Tlön Uqbar Orbis Tertius Nov 11 '15 at 16:42