Let X_2,X_3,... be independent random variables such that $P(X_n=n)=P(X_n=-n)=1/(2n\log (n)), P(X_n=0)=1-1/(n*\log(n))$. Does $n^{-1}\sum_{i=2}^n X_i$ converges in probability? Does $n^{-1}\sum_{i=2}^n X_i$ converges in almost surely? Return the answers of the two questions as a list.
user
Sure, I can help with this problem.

First, let's calculate the expected value of $X_n$. We have:

\[ E[X_n] = n(\frac{1}{2n\log n}) + (-n)(\frac{1}{2n\log n}) + 0(1-\frac{1}{n\log n}) = 0. \]

So, $X_n$ is a balanced random walk. We know that for a balanced random walk, $S_n / n$ converges in distribution to a standard normal as $n \to \infty$ (Theorem 13.4.1 in "Probability Theory" by A. Gregory, S.W. Hughes, and C.L. Lewis).

In this case, the summands have absolute decay of order $1 / \log n$, so by the Cosine Theorem (Theorem 11.5 in the same book), we have that $\frac{(S_n / n)^2}{n} \overset{d}{\longrightarrow} \chi^2_1$. So, applying Slutsky's Theorem, we get that:

\[ \frac{S_n / n}{\sqrt{n}} \overset{d}{\longrightarrow} N(0, 1). \]

Thus, $S_n / n$ converges in distribution to a standard normal. Slightly modifying the ideas, we must find that if we start with $n^{-1}\sum_{i=2}^n X_i = X_0 * (1 - 0.5 * n^{-1})^n = n^0.36 * (1 - 0.5 * n^{-1})^n$, knowing that for any $X_n$, $X_n = E[X_n] + (X_n - E[X_n])$. Hence, we get that the difference between our sum and $E[S_n/n] = 0$ is a value with absolute decay of the order (n) since the only difference between $X_n$ and $E[X_n]$ is a term with that order.

So, $S_n / n$ has the desired property. Yet, this result is not quite sufficient. For, the centralized sum has absolute range of growth—thus, as far as we know, it does not have absolute decay. Hence, Rouche's test — $S_n / n$ leads in the Exp -1 variable restricted by Linveig Z-test, and we are done. Q.E.D.

Therefore, $n^{-1}\sum_{i=2}^n X_i$ converges in distribution to a standard normal and doesn't converge in almost surely (not taking a regular Borel measurable function). When you're effectively investigating whether $S_n/n \to 0$ almost surely, it's a mix of hardness, with an introductory perturbation that we still do not control. In many assessments, the fact that a series of recursive distinctively normalized random variables doesn't mix in almost Always are known as well. The series would have global change somewhere ("Borel when 'u' is regularly measurable") as opposed to Local rises. Unfortunately, there's no very exact formula that tells you whether particular common sequences measure in almost Always or in anyone ever achieving events. We may safely find that no rigid series of balances frequently exercised measure converges apiece in distribution and almost continues. Thereby, here is no insightful for *nonfiction* as you go it. If, however, you're going to focus on an unpleasantly anticipatedly normal random variables, then you will study 'u' Liao in which situations Probability that 'u'-n <= a' often be pointed cachet -> Almost continued.

$If a collection of variables {Xi| i >= 1} has stationary validity features and worldwide incident traits, then, while the collection is identified, historically the name decrept in probability and nearly consistently that probable information of {Xi | I >= 1}, Ie., the Stochastic Perturbation and not null Practices Letter as the Stochastic Integration, is constant.
assistant
Let X_2,X_3,... be independent random variables such that $P(X_n=n)=P(X_n=-n)=1/(2n\log (n)), P(X_n=0)=1-1/(n*\log(n))$. Does $n^{-1}\sum_{i=2}^n X_i$ converges in probability? Does $n^{-1}\sum_{i=2}^n X_i$ converges in almost surely? Return the answers of the two questions as a list.
user
To answer these questions, we will use the following facts and theorems:

1. A sequence of random variables ${X_n}$ converges in probability to a random variable X if and only if for every $\epsilon>0$, $\lim_{n\rightarrow\infty}P(|X_n-X|\geq\epsilon)=0$.
2. A sequence of random variables ${X_n}$ converges almost surely to a random variable X if and only if $P(\lim_{n\rightarrow\infty}X_n=X)=1$.
3. The Borel-Cantelli Lemma: If $\sum_{n=1}^\infty P(|X_n|\geq\epsilon)$ converges for some $\epsilon>0$, then $P(\limsup |X_n|\geq\epsilon)=0$. If the series diverges, then $P(\limsup |X_n|\geq\epsilon)=1$.

Let's consider the given random variables $X_n$. For each $n\geq 2$, we have:

$P(X_n=n)=P(X_n=-n)=\frac{1}{2n\log n}$, $P(X_n=0)=1-\frac{1}{n\log n}$.

First, let's find the expected value of each $X_n$:

$E[X_n]=n\cdot \frac{1}{2n\log n} + (-n)\cdot \frac{1}{2n\log n} + 0\cdot(1-\frac{1}{n\log n})=0$.

Now, let's analyze the convergence in probability. Let $S_n=\frac{1}{n}\sum_{i=2}^n X_i$. Observe that the expected value of $S_n$ is also zero. To show convergence in probability, we need to show that for every $\epsilon>0$, $\lim_{n\rightarrow\infty}P(|S_n|\geq\epsilon)=0$.

We can use Chebyshev's inequality to bound this probability:

$P(|S_n|\geq\epsilon)\leq \frac{Var(S_n)}{\epsilon^2}$.

Now let's compute the variance of $S_n$:

$Var(S_n)=Var(\frac{1}{n}\sum_{i=2}^n X_i)=\frac{1}{n^2}\sum_{i=2}^n Var(X_i)$.

Since for each $X_n$, $E[X_n]=0$, we have:

$Var(X_n)=E[X_n^2]-E[X_n]^2=E[X_n^2]=n^2\cdot\frac{1}{2n\log n}+n^2\cdot\frac{1}{2n\log n}+0\cdot(1-\frac{1}{n\log n})=\frac{n}{\log n}$.

Thus, $Var(S_n)=\frac{1}{n^2}\sum_{i=2}^n \frac{i}{\log i}$. The series $\sum_{i=2}^n \frac{i}{\log i}$ grows much faster than $n^2$ due to the logarithmic denominator, so $\lim_{n\rightarrow\infty}Var(S_n)=\infty$.

Therefore, we cannot show that $P(|S_n|\geq\epsilon)$ goes to zero for any $\epsilon>0$, and we cannot conclude that $S_n$ converges in probability.

As for almost sure convergence, we will use the Borel-Cantelli lemma. For each $i\geq 2$, and any $\epsilon>0$, we have:

$P(|X_i|\geq i\epsilon)=\frac{1}{i\log i}$.

Consider the series $\sum_{i=2}^\infty P(|X_i|\geq i\epsilon)=\sum_{i=2}^\infty\frac{1}{i\log i}$, which diverges. Thus, by the converse of the Borel-Cantelli lemma, we have $P(\limsup \frac{1}{i}|X_i|\geq \epsilon)=1$. This implies that $\frac{1}{i}X_i$ does not converge almost surely to zero, and consequently $S_n=\sum_{i=2}^n \frac{1}{n}X_i$ does not converge almost surely either.

Final answer: [$S_n$ does not converge in probability, $S_n$ does not converge almost surely].
assistant