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Convergence of this set of random variables $Y_n$ where: $X_n: \mathcal U(0,1)$ independent random variables, $Y_n=\frac{1}{nX_n}\ \ n=1,2,...$

I can easily prove that this does not converge almost surely but what I am having troubles with are the other convergences. They are given, I just do not understand them.

$1.$Could someone explain why $E(\frac{1}{n X_n})^2$ does not exist?

and convergence in probability is explained like so: $$P\{Y_n<y\}=P\{\frac{1}{nX_n}<y\}=P\{X_n > \frac{1}{ny}\}=\begin{cases} 0, y \leq \frac{1}{n} \\ 1-\frac{1}{ny}, y>\frac{1}{n}, \end{cases}=\begin{cases} 0, y \leq 0 \\ 1, y>0, \end{cases}$$ this is the distribution function of $Y \equiv 0$, and this is very clear to me, all of these steps.

2. But then it says that because of this the random variable $Y_n$ converges in probability to $0$. I don't understand this, does this mean asymptotic convergence in probability implies convergence in probability??

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Let $\epsilon>0$ and observe that $$\lim_{n\to\infty}P(|Y_n|\geq\epsilon)=\lim_{n\to\infty}P(|X_n|\leq\frac1{n\epsilon})=0$$

So we have convergence in probability towards constant random variable $Y\equiv0$.

Generally convergence in distribution does not imply convergence in probability, but in the special case of convergence in distribution to a constant random variable it does.

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  • $\begingroup$ I see, why does $E(\frac{1}{n X_n})^2$ not exist? E = expectation. $\endgroup$
    – Jerry West
    Commented Jan 3, 2016 at 14:01
  • $\begingroup$ Formally looking at this. $\endgroup$
    – Jerry West
    Commented Jan 3, 2016 at 14:01
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    $\begingroup$ Because $\int_0^1u^{-2}du=\infty$ $\endgroup$
    – drhab
    Commented Jan 3, 2016 at 14:03

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