# convergence of random variables

This is related to this question. How exactly does one go about computing the limits in the answer to the linked question.

Thanks.

P.S: I would have commented on the linked question, but I don't have enough points.

So we have the sequence of constant variables $X_n = 1+1/n$ ($X_n (\omega) = 1+1/n$ for any $\omega \in \Omega$). how does one go about showing the following:
a) $$\mathop {\lim }\limits_{n \to \infty } P(1 + 1/n \le x) = P(1 \le x),$$ for any $x \neq 1$, showing that the sequence converges in distribution.
b) $$\mathop {\lim }\limits_{n \to \infty } P(|(1 + 1/n) - 1| > \varepsilon ) = 0,$$ for any $\varepsilon > 0$. This shows that the sequence converges in probability.
c) $$P(\lim _{n \to \infty } (1 + 1/n) = 1) = 1.$$ which shows that the sequences converges almost surely.
d) $$\mathop {\lim }\limits_{n \to \infty } {\rm E}|(1 + 1/n) - 1|^p = 0.$$ showing that the sequence converges in the $p$-th moment.

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Please make this question standalone readable. – Jonas Teuwen Jul 17 '11 at 22:29
should I copy the entire question? – John1 Jul 17 '11 at 22:36
No. Just make sure we can understand what your question is without following that link. Even when I do follow the link, I'm not sure what your question is. – Jonas Teuwen Jul 17 '11 at 22:41
Ok. I have added to my question. I hope its better. – John1 Jul 17 '11 at 22:45

For a), fix any $x \neq 1$. If $x < 1$, then obviously $$\mathop {\lim }\limits_{n \to \infty } P(1 + 1/n \le x) = \mathop {\lim }\limits_{n \to \infty } 0 = 0 = P(1 \leq x).$$ If, on the other hand, $x > 1$, then there exists $N \in \mathbb{N}$ such that $1+1/n < x$ for all $n > N$. Hence, $P(1 + 1/n \le x) = 1$ for all $n > N$, and so obviously $$\mathop {\lim }\limits_{n \to \infty } P(1 + 1/n \le x) = 1 = P(1 \leq x).$$

For b), first fix $\varepsilon > 0$. Note that $$\mathop {\lim }\limits_{n \to \infty } P(|(1 + 1/n) - 1| > \varepsilon ) = \mathop {\lim }\limits_{n \to \infty } P(1/n > \varepsilon ).$$ Since there exists $N \in \mathbb{N}$ such that $1/n < \varepsilon$ for all $n > N$, it holds $P(1/n > \varepsilon )=0$ for all $n > N$, and hence obviously $$\mathop {\lim }\limits_{n \to \infty } P(1/n > \varepsilon ) = 0.$$

For c), $$P(\mathop {\lim }\limits_{n \to \infty } (1 + 1/n) = 1) = P(1 = 1) = 1.$$

For d), $$\mathop {\lim }\limits_{n \to \infty } {\rm E}|(1 + 1/n) - 1|^p = \mathop {\lim }\limits_{n \to \infty } {\rm E}|1/n|^p = \mathop {\lim }\limits_{n \to \infty } \frac{1}{{n^p }} = 0.$$

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Thanks Shai. Can you please elaborate on why $P(1/n >\epsilon )=0$. Also I don't understand the 2nd to last equlity in (d). – John1 Jul 18 '11 at 0:29
A) If $1/n < \varepsilon$ $\forall n > N$, then $P(1/n \leq \varepsilon) = 1$ $\forall n > N$, and hence $P(1/n > \varepsilon) = 0$ $\forall n > N$; B) Since the $|1/n|^p$ are constants, ${\rm E}(|1/n|^p) = |1/n|^p = (1/n)^p = 1/n^p$. – Shai Covo Jul 18 '11 at 0:39
Thanks. I understand b) now. what I don't get about (A) is why $P(1/n \leq \epsilon )=1$. – John1 Jul 18 '11 at 0:51
If $1/n < \varepsilon$, then in particular $1/n \leq \varepsilon$, and hence $P(1/n \leq \varepsilon) = 1$. – Shai Covo Jul 18 '11 at 0:57
Thanks Shai; you've been immense! – John1 Jul 18 '11 at 1:06

Let $\varepsilon >0$ and choose $n$ large enough so that $(1+1/n)-1<\varepsilon$. Then, $$\left| \Pr [1\leq x]-\Pr [1+1/n\leq x]\right| \leq \Pr [1+1/n-\varepsilon <x]-\Pr [1+1/n\leq x]=\Pr [x<1+1/n<x+\varepsilon ]<\varepsilon .$$

For the second one, you can use the fact that pointwise (almost everywhere) convergence implies convergence in probability.

The third one is the easiest. It just follows from the fact that $\left\{ \omega \in \Omega |\, \lim _{n\to \infty}[1+1/n]=1\right\} =\Omega$.

The fourth one is pretty easy too. Just write down the integral: $$\mathrm{EV}\left| (1+1/n)-1\right| ^p=\int _\Omega 1/n^pd\Pr =1/n^p.$$

Hope that helps!

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Thanks, GleasSpty. – John1 Jul 18 '11 at 1:19