# Computing the probability a sequence of random variable limit is equal to 0.

I am trying to read my probability book on the "Strong Law of Large Numbers", and came across this example that is really confusing me.

Let $X_i$ be a sequence of independent uniformly distributed random variables in $[0, 1]$ and $Y_n = \min(X_1, ..., X_n)$. Show that $Y_n$ converges to zero with probability 1.

The book says let $Y$ be the limit of the $Y_n$s which exists because nonincreasing and bounded below. Then for $1 > \epsilon > 0$, $$P(Y \ge \epsilon) = P(X_1\ge \epsilon \text{ & }\cdots\text{ & }X_n \ge \epsilon) = (1 - \epsilon)^n.$$ Here it means $Y_n$ I think, right?

Then it says $$P(Y \ge \epsilon) \le \lim_{n \to \infty} (1 - \epsilon)^n = 0.$$ So we can conclude $P(Y \ge \epsilon) = 0$, so $P(Y = 0) = 1$.

Why does it have a less than or equal sign for the last step? Shouldn't it be equality?

Thanks for any help.

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The $Y_n$ is a typo. And yes, it should be equality. –  André Nicolas Feb 21 '13 at 5:21

The reason is that one starts from the almost sure inequality $Y\leqslant Y_n$ for each fixed $n$, which yields $[Y\geqslant\varepsilon]\subseteq[Y_n\geqslant\varepsilon]$, hence $\mathbb P(Y\geqslant\varepsilon)\leqslant\mathbb P(Y_n\geqslant\varepsilon)$. Finally, $\mathbb P(Y\geqslant\varepsilon)\leqslant\lim\limits_{n\to\infty}\mathbb P(Y_n\geqslant\varepsilon)$.