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.


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 $Y$.

  • $\begingroup$ 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? $\endgroup$
    – Pwaol
    Jun 11 at 9:44
  • 1
    $\begingroup$ @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. $\endgroup$ Jun 11 at 9:47
  • $\begingroup$ 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? $\endgroup$
    – Pwaol
    Jun 11 at 10:43
  • $\begingroup$ @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 $\infty$) $\endgroup$ Jun 11 at 11:31
  • $\begingroup$ thanks, to be honest I'm a little lost of why there was a typo, I would appreciate it if you could add another step or two where you said by symmetery, will help me alot to understand $\endgroup$
    – Pwaol
    Jun 11 at 12:59

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