While reading "Introduction to probability, statistics, and random processes" by H. Pishro-Nik, I encountered example 7.11 in the 7.2.6 chapter,
The example defines a sequence of random variabls $X_1, X_2, X_3, ...$ such that:
$\begin{equation} \nonumber X_n = \left\{ \begin{array}{l l} n^2 & \qquad \textrm{with probability } \frac{1}{n} \\ & \qquad \\ 0 & \qquad \textrm{with probability } 1-\frac{1}{n} \end{array} \right. \end{equation}$
then it shows that:
- $X_n \ \xrightarrow{p}\ 0$
- $X_n$ does not converge in the $r$th mean for any $r≥1$
Then, the example concludes with an interesting fact, which is:
although $X_n \ \xrightarrow{p}\ 0$, the expected value of $X_n$ does not converge to $0$
The last fact stated by the example is confusing me a little bit.
If $X_n \ \xrightarrow{p}\ Y$, and $Y$ is r.v. that has a divergent expected value, then the last statment whould make since. However, if $Y$ does have a convergent expected value (in our case it is the degenerate case of the zeroth r.v.), I do not see why the expected value of $X_n$ does not converge. Why does the definition of convergence in probability fail to guarantee the convergence of the expected value of $X_n$ to $E[Y]$. At the end, the probability of $X_n$ being further from $Y$ by $\epsilon$ do reach $0$ as n grows by definition of convergnece in probability.
Can anyone explain what am I missing? Thanks,,
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In the counterexample, the probability of being $0$ increases towards $1$ and of not being $0$ decreases towards $0$. So you have convergence towards $0$ in probability.
But the mean takes account of the whole distribution and does not ignore the small probability of being non-$0$, and in this case the expected value has been constructed to be $E[X_n]=n$, which increases without limit with $n$.
A similar example could target any desired function of $n$ as the expectation: $P(\, X_n=n\,f(n)\,)=\frac1n$ and $P(X_n=0)=1−\frac1n$ while still having convergence towards $0$ in probability.