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Let $X_n$ a sequence such that $P(X_n=1)=P(X_n=n)=\frac{1}{n^3+1}$ and $P(X_n=2)=\frac{n^3-1}{n^3+1}$

a) Check if $X_n \xrightarrow{p} X$, identifying the random variable X.

b) Check if $X_n \xrightarrow{L^2} X$, identifying the random variable X.

c) Check if the Law of larger numbers applies in this case.


How do I find where a random variable is converging to ?

I tried to solve the first item using the definition of convergence of probability and markov...

I can see that $X_n$ does not converge when $X = 0$

$P(|X_n-X|\ge \epsilon) = P(X_n\ge \epsilon) \le E(X_n)/\epsilon$

And $E(X_n)/\epsilon$ goes to 2 when $n \rightarrow \infty$

I found, by guessing, that $X_n \xrightarrow{p} X$ when $X=n$, but is that correct ? And $n$ surely is not the only value, right ? How can i make it formal ?

Thank you

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  • $\begingroup$ The limit is $X=2$. a) and b) are true. I don't have an answer to c) yet. $\endgroup$ – Kavi Rama Murthy Dec 5 '17 at 5:51
  • $\begingroup$ You're right. Limite is equal to 2 when X=0. I will edit that! Thanks $\endgroup$ – lenosane Dec 5 '17 at 6:53
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Maybe first let's take it from a very explicit angle. Let's say $n=100.$ Then we have that $$P(X_{100} = 1) = \frac{1}{1000001}\\ P(X_{100} = 2) = \frac{999999}{1000001}\\ P(X_{100} =100) = \frac{1}{1000001}.$$

It looks like it's overwhelmingly probable that $X_{100} =2.$ If $X_n$ is converging to anything in any sense, it's probably $2.$

In fact, $X_n\to_P 2,$ since we have, for any $\epsilon > 0,$ $$ P(|X_n-2| \ge \epsilon) \le \frac{2}{n^3+1} \to 0.$$ The $\frac{2}{n^3+1}$ is just the probability that $X_n$ is not exactly equal to $2,$ which will definitely be the case if $|X_n-2|\ge \epsilon.$

For $L^2,$ you need to compute $E(|X_n-2|^2) $ and see if it goes to zero or not. You have the exact distribution of $X_n$ (it can only take three values for each $n$, thus $|X_n-2|^2$ also can only take three values... so this expectation is easy to compute).

For the law of large numbers consider proving also that $X_n\to 2$ almost surely (it does... moreover $X_n$ is eventually equal to two with probability one). Then what can you say (almost surely) about $ \lim_{n\to\infty}\frac{1}{n}(X_1+\ldots+X_n)?$ (Does the name Cesaro ring a bell?)

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  • $\begingroup$ Thank you very much for your answer, really helped me a lot! Using markov would fail anyway... I’m having trouble to understand and visualize why the probability that Xn is not exactly equal to 2 is bigger than P(|Xn-2|>=epsilon) $\endgroup$ – lenosane Dec 5 '17 at 8:47
  • $\begingroup$ $|X_n-2|>\epsilon$ implies $X_n\ne 2.$ so the former event is a subset of the latter and thus its probability is thus less than or equal. $\endgroup$ – spaceisdarkgreen Dec 5 '17 at 14:33

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