How can $ P\big(|X_n| \geq \varepsilon\big) \leq P\big(\big|X_n-\frac{1}{n}\big| \geq \varepsilon+\frac{1}{n}\big) $?

Question

I am following a probability course on MIT OCW, and on a tutorial on probability convergence, in an attempt to use the Chebyshev's Inequality to prove it, the TA claimed that $$ P\big(|X_n| \geq \varepsilon\big) \leq P\bigg(\bigg|X_n-\frac{1}{n}\bigg| \geq \varepsilon+\frac{1}{n}\bigg). $$ To prove it, he said "The right hand side will cover less of $X_n$ hence the inequality holds".

Shouldn't "covering less" of the random variable result in less probability, and not more?
Am I missing something or did he made a mistake?

[Link for the video, discussed around 8:10.]

Answer 1

The TA made a sign error. It should have read

$$\mathbb{P}(|X_n| \geq \varepsilon) \leq \mathbb{P}\left(\left|X_n - \frac1n\right| \geq \varepsilon - \frac1n\right)$$

The rest of the argument follows as expected for any $n > \frac1\varepsilon$, hence it works in the limit as $n\to\infty$ as well.

Answer 2

Edit Note: I refer to the RHS of the equation in question as the second event, while the LHS is referred to as the first event.

Hmmm... Let me see if I can give it a shot(no guarantee if this is correct)

The inequality is incorrect and should actually face the other way. That is, $$P(|X_n| \geq \epsilon) \geq P(|X_n - \frac{1}{n}| \geq \epsilon + \frac{1}{n})$$

If we take the fist event and decompose it,

$$P(|X_n| \geq \epsilon) = P(X_n \geq \epsilon \cup -X_n \leq -\epsilon) \\ =P(X_n \geq \epsilon) + P(-X_n \leq -\epsilon)$$

Does that make sense? The last equality comes from the fact that the events are disjoint.

Now on the other hand, $$P(|X_n - \frac{1}{n}| \geq \epsilon + \frac{1}{n}) = P(X_n - \frac{1}{n} \geq \epsilon + \frac{1}{n}) + P(-X_n + \frac{1}{n} \leq -\epsilon - \frac{1}{n})$$ Simplifying the algebra inside the parenthesis now $$P(|X_n - \frac{1}{n}| \geq \epsilon + \frac{1}{n}) = P(X_n \geq \epsilon + \frac{2}{n})+P(-X_n \leq -\epsilon - \frac{2}{n})$$

So now if we compare the two probabilities -

For first probability/event $X_n$ just has to be bigger than $\epsilon$ or $-X_n$ has to be smaller than$-\epsilon$. On the other hand, for the second probability/event $X_n$ has to be bigger than $\epsilon$ PLUS $\frac{2}{n}$ or $X_n$ has to be smaller than $-\epsilon$ MINUS $\frac{2}{n}$

So the second probability is less likely of happening because we have more conditions(on X_n) needed to satisfy the event happening. Said differently, the probability space for the second event happening is smaller than the probability space of the first event happening.

Answer 3

The inequality is obviously wrong. $X_n=\frac 1 n$ with $0 <\epsilon <\frac 1 n$ gives a counter-example.