# Probability question with Normal distribution and finding density function.

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.
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 $$Y$$.
• thanks for the answer, may I ask about why is it the sum? my only reason will be that when $P(X^2 \ge 1) = P(X \ge 1 or X \le 1)$ and then we sum the probabilities since they're disjoint, is my understanding correct? Jun 11 at 9:44
• @Pwaol Yes, you are splitting the event $(Y \leq \frac 1 2 )$ into two disjoint parts, one on which $X>0$ and the other on which $X<0$, so the probabilties add up. Jun 11 at 9:47
• Sorry for bothering again, I'm having a little trouble with differentiating the integral, I got $2\frac{1}{\sqrt{2\pi}}*e^{\frac{-(\frac{1}{y}-1)}{2}}*\frac{1}{2\sqrt{\frac{1}{y}-1}}*-\frac{1}{y^2}$ , but I'm stuck on the minus infinity bound, how do I approach that? do I just substitute it in the function? Jun 11 at 10:43
• @Pwaol There was a typo in the integral. The derivative of $\int_{g(y)}^{\infty} h(t)dt$ is $-h(g(y)) g'(y)$. (It doesn't matter whether the upper limit of the integral is a finite number or $\infty$) Jun 11 at 11:31