Let $_1,_2, … , _$ be a sequence of random variables, with $(_) = 7$ and $(_) = 1/sqrt(^)$ for each . Prove or disprove that ${_}$ must converge to 7 with probability 1 if $ > 2$?
I'm thinking of a counter example here:
Let $Xn = 7 + Z(n^{-\frac{\alpha}{4}})$ where $Z$ is a standard normal random variable with mean 0 and variance 1
$E(X_n) = E(7 + Z(n^{-\frac{\alpha}{4}})) = 7 + E(Z)n^{-\frac{\alpha}{4}}= 7$
$Var(X_n) = Var(Z(n^{-\frac{\alpha}{4}})) = n^{-\frac{\alpha}{2}}Var(Z) = n^{-\frac{\alpha}{2}}$
The variance of Xn decreases at a rate of $n^{-\frac{\alpha}{2}}$, which is slower than $\frac{1}{n}$ for $a > 2$. This slower rate of convergence allows for the possibility of $X_n$ taking on values far from 7 with a non-zero probability, even as $n$ grows infinitely large.
From here, although it satisfies the condition for convergence in probability, the set of all possible outcomes where Xn remains close to 7 for all n has a probability of less than 1, violating the strict requirement of almost sure convergence.
Does this disprove correct? Is there any other ways to do this?
Well, the counterexample is probably wrong since the statement is true.
In particular, Chebyshev's inequality gives that for any $\epsilon > 0$, $$ P(|X_n - 7| \geq \epsilon) \leq \frac{\operatorname{Var}(X_n)}{\epsilon^2} = \frac{n^{-a/2}}{\epsilon^2}. $$ Then, when $a > 2 \iff a/2 > 1$, $$ \sum_{n=1}^\infty P(|X_n - 7| \geq \epsilon) \leq \epsilon^{-2}\sum_{n=1}^\infty \frac{1}{n^{a/2}} < \infty $$ so Borel-Cantelli gives that, with probability 1, $|X_n - 7| \geq \epsilon$ only finitely often; since this holds for any positive $\epsilon$, $X_n \to 7$ almost surely.