0

We assume that $X \sim N(0,1)$ is a Gaussian RV, I wish to evaluate the probability

$$\Pr(X^2 \leq 4, X \geq 5)$$

Can I proceed as follows:

$\Pr(X^2 \leq 4, X \geq 5) = \Pr(X \geq -2, X \leq 2, X \geq 5) = \int\limits_{-2}^2 f_X(x) dx + \int\limits_{5}^\infty f_X(x) dx? $

Or is there a simpler way?

Olórin
  • 5,415
  • 3
    $X^2\le4$ and $X\ge 5$ cant happen simultaneously. – QED Jan 13 '18 at 03:41
  • 1
    For any two events $P(A\cap B) = P(A) + P(B) - P(A\cup B)$ or $P(X^{2}\leq 4, X\geq 5) = \int_{-2}^{2}\varphi(x)dx + \int_{5}^{\infty} \varphi(x)dx - \left( \int_{-2}^{2}\varphi(x)dx + \int_{5}^{\infty} \varphi(x)dx\right) = 0$. – JessicaK Jan 13 '18 at 03:58

1 Answers1

1

The expression you wrote $(\int_{-2}^2 +\int_5^\infty)$ is the probability that $X^2 \le 4$ or $X \ge 5.$ You want the probability $X^2\le 4$ and $X\ge 5.$ Since it is impossible for a number to have this property regardless of its distribution, the probability must be zero.

Another way to look at it is in terms of events. We have $$ \{X^2\le 4\}\cap\{X \ge 5\} = \emptyset$$ and it is always true that $P(\emptyset) = 0.$

spaceisdarkgreen
  • 58,508