Finding P($X^2$<1) , after verifying that $X^2$ is distributed $\chi^2_1$ The Problem
Suppose $X$ and $Y$ are independent n(0,1) random variables. Find $P(X^2 < 1)$, after verifying that $X^2$ is distributed $\chi^2_1$.
My Work
I understand that I need to find the probability that $X^2$ is less than 1, but I am not too sure what my upper and lower limits would be, if I integrate. I am new to the whole chi-squared distribution, and I have been trying to find resources that I will share below.
What I THINK my first step should be is to, square $X^2$ and find the corresponding chi - squared value.
I do not know for sure. I am brand new to this, so please bear with me.
My Question
How do I verify that $X^2$ is distributed $\chi^2_1$? Also, intuitively speaking, would the integration upper limit be $1$ and the lower limit $0$?
Resources
https://www.khanacademy.org/math/statistics-probability/inference-categorical-data-chi-square-tests/chi-square-goodness-of-fit-tests/v/chi-square-distribution-introduction
https://www.bmj.com/about-bmj/resources-readers/publications/statistics-square-one/8-chi-squared-tests
 A: a
Let $Y=X^2$ with $X$ being standard normal, and let $F$ be the cdf (cumulative distribution function) of $Y$. A standard procedure is to find the cdf and then take the derivative to get the pdf.
$F(y)=P(Y\le y)=P(X^2\le y)=P(-y^{1/2}\le X\le y^{1/2})=\Phi(y^{1/2})-\Phi(-y^{1/2})$.
Taking the derivative wrt $y$ I get: $f(y)=\phi(y^{1/2})\frac 12 y^{-1/2}-\phi(-y^{1/2})\cdot -\frac 12 y^{-1/2}$. Substituting in $\phi(y^{1/2})=\phi(-y^{1/2})=\frac 1 {\sqrt{2\pi}}e^{-\frac 12y}$ this becomes:
$$f(y)=\frac{1}{\sqrt{2\pi}}e^{-\frac 12 y}y^{-1/2}$$ for $y\ge 0$. Comparing this with the pdf of a chi-squared random variable with $m$ degrees of freedom: $f(x)=\frac 1{2^{m/2}\Gamma(m/2)}x^{m/2-1}e^{-x/2}$, if you substite $m=1$ you will get the pdf above. Hence $Y$ is chi-squared with 1 degree of freedom.
b
For finding $P(X^2\le 1)$ you can use Econbernardo's method (+1) or use a chi-squared distribution table. My table says $P(X^2\le 1.074)=.7$ and $P(X^2\le .7083)=.6$, so a reasonable estimate for $P(X^2\le 1)$ is somewhere in between, say $.68$.
Using econbernardo's method, $P(X^2\le 1)=P(-1\le X\le 1)=2\Phi(1)-1=2(.8413)-1=.6826$
A: $$\{\omega \in \Omega: X^2(\omega) \le 1 \} = \{\omega \in \Omega: X(\omega) \in [-1,1]\} \implies \mathbb{P}(X^2 \le 1) = \mathbb{P}(-1\le X \le 1) = \Phi(1)-\Phi(-1) = 2\Phi(1)-1$$
where $\Phi$ is the standard normal cdf.

How do I verify that $X^2$s distributed $\chi_1^2$

Are you looking for a proof on this? It is a simple application of the transformation of random variables theorem: $$Y = g(X) \implies f_Y(y) = f_x(g^{-1}(y))\frac{d}{dy}g^{-1}(y)$$
But the fact that the square of a standard normal variable is a chi-squared with 1 degree of freedom is a standard result, I am not sure what you mean by verify? Let me know if this answers your question.
