# Is this equality generally true? | Probability

1. Oct 27, 2010

### michonamona

Suppose Xn is a random variable. Let b and c be a constant.

Is the following generally true?

$$P(|X_{n}-b| \geq \epsilon) = P(|X_{n}-b|^{2} \geq \epsilon^{2})$$

This says that the probability that Xn minus b is greater than or equal to epsilon is equal to the probability that Xn minus b squared is greater than epsilon squared.

My prof keeps saying that they are the same event, therefore, they have the same probability. But I still don't understand. Any insight?

Thanks,
M

2. Oct 27, 2010

### jbunniii

Yes, it's generally true. It is because for any positive numbers $a$ and $b$,

$$a \geq b$$

if and only if

$$a^2 \geq b^2$$