# Convergence in probability does not imply convergence in $r^{th}$ mean

I am confused regarding convergence in probability and convergence in $$r^{th}$$ mean.

I am able to prove that convergence in $$r^{th}$$ mean implies convergence in probability, which is not true. Let me know where I went wrong in my proof.

Before I start the proof, let me state two results that will be used:

1. $$X_n\xrightarrow{p} X \ \ \implies X_n -X\xrightarrow{p} 0 \ \$$

2. $$X_n\xrightarrow{p} X \ \ \implies f(X_n)\xrightarrow{p} f(X)$$, where $$f$$ is a continuous function

Proof:

Let $$X_n$$ be a sequence of random variables that converges in Probability to $$X$$, then using 1:

$$X_n-X\xrightarrow{p}0 \\\implies|X_n-X|\xrightarrow{p}0\ \ \\\\\implies |X_n-X|^r\xrightarrow{p}0\ \ \\$$

Now using 2: $$\\\implies E|X_n-X|^r\xrightarrow{}0\ \ \\$$

Hence $$X_n$$ converges in $$r^{th}$$ mean to X.

• See these notes Exercise 2 (proof). – BruceET Jun 21 '19 at 6:05

Let $$\Pr(z_n=0)=(n-1)/n$$ and $$\Pr(z_n=n)=1/n$$. Then, for any $$\epsilon$$, $$\Pr[|z_{n}-0|<\epsilon]=\Pr(z_n=0)=(n-1)/n\rightarrow 1,$$ i.e. $$z_{n}\to_p0$$. However, $$E(z_n-0)^2=n^2\cdot 1/n\rightarrow\infty,$$ so that $$z_{n}\nrightarrow_{\mathrm{m.s.}}0$$.
The reason you cannot just apply the expected value operator in your final step is that, as my counterexample illustrates, convergence in probability essentially only looks at the probability of events, while expected values also weigh by the event. If, as in this example, the event is "extreme" in that it takes large values as $$n$$ gets large, we do not have convergence in mean.
After you get $$2$$, you have a convergence in probability of the $$r$$-power of the absolute difference. When you take your expectation, that's again a convergence in probability. Therefore, you conclude that in the limit, the probability that the expected value of de rth power absolute difference is greater than $$\epsilon$$, is $$0$$. That's not the definition of convergence in $$r$$-mean.