# I've taken an alternative approach to a convergence in probability problem. Is there any mistake and/or is my conclusion correct?

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:

1. Was there any mistake made in the alternative solution?
2. Was the conclusion correct?

## 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
$$\begin{eqnarray} \mathbb{P}\left ( |X_n| \geq \varepsilon \right ) \leq \mathbb{P}\left ( |X_n - \mathbb{E}[X_n]| \geq \varepsilon - \mathbb{E}[X_n] \right ) \end{eqnarray}$$
$$\begin{eqnarray} \mathbb{P}\left ( |X_n| \geq \varepsilon \right ) \leq \frac{\mathbb{V}(X_n)}{{(\varepsilon - \mathbb{E}[X_n])}^2} \end{eqnarray}$$

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.

Your alternative proof does not appear to have any mistakes. And the conclusion is simply that based on the alternative proof alone, the result is indeterminate since you showed that an upper bound is larger than the desired limit.

If your goal is to prove this result with a chebyshev like approach you could use the markov inequality (of which chebyshev's inequality is a special case).

Markov's inequality:

$$P(Z \geq c) \leq \frac{E[g(Z)1\{Z \geq c\}]}{g(c)} \leq \frac{E[g(Z)]}{g(c)}$$

Where $$g()$$ is borel measurable and non-decreasing. (Williams, Probability with Martingales).

So when $$Z = |X - E[X]|$$ and $$g(z) = z^2$$ we get chebyshev's using the outside inequality. But as you showed, the desired limit is not achieved.

However, using markov and $$g(z) = z$$ we can prove convergence.

\begin{align} P(|X_n| \geq \epsilon) &\leq \frac{E[|X_n|]}{\epsilon}\\ &= \frac{2n^2 - 1}{\epsilon n^3} \overset{n\to\infty}{\to} 0 \end{align}