Probability question with Normal distribution and finding density function.

Question

Let $X$ be a normal random variable, such that $X$~$N(0,4)$.
We define $Y=\frac{1}{X^2+1}$.
Find $P(Y\le\frac{1}{2}), P(Y\le-1)$.
Find the density function of $Y$.

My Work:
$P(Y\le\frac{1}{2})=P(\frac{1}{X^2+1}\le\frac{1}{2})=P(1\le \frac{X^2}{2}+\frac{1}{2})=P(1 \le X^2)$ And now, when $X>0 \Longrightarrow P(1\le X) \Longrightarrow Z=\frac{X}{2}$~$N(0,1)$$ \Longrightarrow P(\frac{1}{2}\le Z)=1-P(\frac{1}{2}>Z)=1-\phi(\frac{1}{2})=1-0.6915=0.3085$

And when $X<0\Longrightarrow P(1\le-X)=P(X\le -1)=P(Z\le\frac{-1}{2})=1-P(Z\le\frac{1}{2})=0.3085$.

Note: I defined $Z=\frac{X}{2}$ to normalize $X$ to $N(0,1)$.

It's my first time solving a question with normal distribution and I feel like I've made some mistakes calculating this part, and I would appreciate any hints on how to find the density function of $Y$, because I have no idea how to start.
Thanks in advance.

Answer 1

You have to give a single number as the answer for $P(Y \leq \frac 1 2 )$. The correct value of this probability is the sum of the two numbers you got. Of course, $P(Y \leq -1)=0$ since $Y$ is a positive random variable.

For $y >0$ we have $P(Y \leq y)=P(X^{2} \geq \frac 1 y -1)$. By symmetry we can write this as $2\frac 1 {\sqrt {2 \pi}} \int_{\sqrt {\frac 1 y -1}}^{\infty} e^{-t^{2}/2} dt$. Differentiate this (using Chain Rule) to get the density function of $Y$.