I am currently studying convergence on my own, which means that I don't have many alternatives for discussing problems in order to improve my understanding. This post was an alternative to get around this situation.
The problem will be introduced, as well as the solution to it. After solving the problem, I'll introduce an alternative solution and my conclusion on why it could not solve the original problem. This is what I'm looking for:
- Was there any mistake made in the alternative solution?
- Was the conclusion correct?
Thank you in advance!
Problem
Let $X_1, X_2, ..., X_n$ be a sequence of random variables such that
\begin{eqnarray} \mathbb{P}\left ( X_n = \frac{1}{n} \right ) = 1-\frac{1}{n^2} \ \ \ \ \ and\ \ \ \ \ \ \mathbb{P}\left ( X_n = n \right ) = \frac{1}{n^2} \end{eqnarray}
Does $X_n$ converge in probability?
Solution
Convergence in probability is defined as \begin{eqnarray} \lim_{n \to \infty} \mathbb{P}\left ( |X_n - X| \geq \varepsilon \right ) = 0 \end{eqnarray}
which is equivalent to
\begin{eqnarray} \lim_{n \to \infty} \mathbb{P}\left ( |X_n - X| \leq \varepsilon \right ) = 1 \end{eqnarray}
Assuming convergence to 0, then
\begin{eqnarray} \lim_{n \to \infty} \mathbb{P}\left ( |X_n| \leq \varepsilon \right ) = 1 \end{eqnarray}
Which can be proven to be true, as $n\rightarrow \infty $, for
\begin{eqnarray} \mathbb{P}\left ( |X_n| \leq \varepsilon \right ) = 1-\frac{1}{n^2} \ \ \ \ \ \ \ \ \ \ if\ \ \ n > \frac{1}{\varepsilon} \end{eqnarray}
Alternative Approach
Assuming convergence to 0, the convergence in probability will be given by
\begin{eqnarray}
\lim_{n \to \infty} \mathbb{P}\left ( |X_n| \geq \varepsilon \right ) = 0
\end{eqnarray}
Subtracting $\mathbb{E}[X_n]$ on both sides of the inequality, we can bound the $\mathbb{P}\left ( |X_n| \geq \varepsilon \right )$ by the Chebyshev's inequality
\begin{eqnarray}
\mathbb{P}\left ( |X_n| \geq \varepsilon \right ) \leq \mathbb{P}\left ( |X_n - \mathbb{E}[X_n]| \geq \varepsilon - \mathbb{E}[X_n] \right )
\end{eqnarray}
Which leads to
\begin{eqnarray}
\mathbb{P}\left ( |X_n| \geq \varepsilon \right ) \leq \frac{\mathbb{V}(X_n)}{{(\varepsilon - \mathbb{E}[X_n])}^2}
\end{eqnarray}
In that way, if Chebyshev's inequality converges to $0$ as $n \rightarrow \infty$, we can prove the convergence in probability.
\begin{eqnarray} \lim_{n \to \infty} \frac{\mathbb{V}(X_n)}{{(\varepsilon - \mathbb{E}[X_n])}^2} = 0 \ \ (?) \end{eqnarray}
$\mathbb{E}[X_n]$ is given by
\begin{eqnarray}
\mathbb{E}[X_n] = \frac{1}{n}{\left (1-\frac{1}{n^2} \right )} + n{\left (\frac{1}{n^2} \right)} = \frac{2n^2-1}{n^3}
\end{eqnarray}
and $\mathbb{V}(X_n)$ is given by
\begin{eqnarray}
\mathbb{V}(X_n) = {\left (\frac{1}{n}-\frac{2n^2-1}{n^3}\right )}^2{\left (1-\frac{1}{n^2} \right )} + {\left (n-\frac{2n^2-1}{n^3}\right )}^2{\left (\frac{1}{n^2} \right)}
\end{eqnarray}
\begin{eqnarray} \mathbb{V}(X_n) = \frac{{(1-n^2)}^2(n^2-1) + {(n^4 - 2n^2 + 1)}^2}{n^8} \end{eqnarray}
As $n \rightarrow \infty$
\begin{eqnarray} \mathbb{E}[X_n] \rightarrow 0\\ \end{eqnarray}
\begin{eqnarray}
\mathbb{V}(X_n) \rightarrow 1\\
\end{eqnarray}
Leading to
\begin{eqnarray}
\lim_{n \to \infty} \mathbb{P}\left ( |X_n| \geq \varepsilon \right ) \leq \lim_{n \to \infty} \frac{\mathbb{V}(X_n)}{{(\varepsilon - \mathbb{E}[X_n])}^2} = \frac{1}{\varepsilon^2}
\end{eqnarray}
Conclusion
Since Chebyshev did not converge to $0$, the convergence in probability can not be proven through Chebyshev's Inequality.