weak law of large numbers, exponential(3) distribution

249 Views Asked by At

I'm going through an example in my textbook but can't seem to follow every step.

Let $ W_1, W_2,... $ be i.i.d with distribution $ Exponential(3). $ Prove that for some $n$, we have $ P(W_1+W_2+...+W_n < \frac{n}{2}) > 0.999. $

The solution in the book says:

$ P(W_1+W_2+...+W_n < \frac{n}{2}) = 1-P(W_1+...+W_n \geq \frac{n}{2}) \geq (**)$ $ 1-P(|\frac{1}{n}(W_1+...+W_n)-1/3| \geq 1/6). $

What I don't understand is the inequality (**), I simply don't get why it's there, and I also don't understand why there is $ \frac{1}{6} $ in the last row (why have they chosen $\varepsilon$ as $\frac{1}{6}$ and not any other number larger than 0?). What I do understand is that $ P(|\frac{1}{n}(W_1+...+W_n)-\frac{1}{3}| > \varepsilon) = 0 $ and hence, it has been shown that $ P(W_1+...+W_n < \frac{n}{2}) \geq 1 $ as well as larger than 0.999.