Calculating the probability of a transformed normal random variable

Question

I'm trying to find the probability of the following:

$\ P(X^2-2X+1>0) $

where X is $\ N ∼ (1,2.25) $

Now I've used the fact that we can express X in the form:

$ \ X = 1.5Z+1 $

where Z is a standard normal random variable.

We can express the first equation as:

$\ P((X-1)^2>0) $

Then we can express this in the form of Z:

$\ P((1.5Z)^2>0) $

How do I solve this? Intuitively I see that $\ Z^2 $ cannot take negative values and is always greater than 0 giving the answer 1. But this is not correct. How do I solve the last part. I believe my intuition is wrong.

Answer 1

There's no need to make things too complicated. Let's ask ourselves: $$ X^2 - 2X +1>0 $$ When is this true? That's an easy question, it's true almost always, only $X=1$ doesn't work. So $$ X^2 - 2X +1>0 \qquad \Leftrightarrow \qquad X \neq 1 $$ Therefore, the probability is the same as $$ P(X\neq 1) = 1 - P(X=1) $$ And if $X$ is a continous random variable, the probability is equal to $1$.

Answer 2

Why do you think it's not 1? I've made monte-carlo and got 1. Here is python code:

import numpy as np

x = np.random.normal(loc=1, scale=1.5, size=100000)
(x**2 - 2*x + 1 > 0).mean()