Probability question with cumulative distribution function and density function calculation.

Question

Let $X$ be a continuous random variable with the density function: $$f_X(x)=\frac{1}{\pi}\frac{1}{1+x^2},$$ $-\infty \le x\le\infty$.
We define the random variable $Y=\frac{1}{X}$. Find the density function of $Y$, and find the connection between $X,Y$ distributions.

My Work:
My plan is to find the Cumulative distribution function $F_Y(y)=P(Y\le y)$, and then take derivative to get $f_Y(y)$.
$$F_Y(y)=P(Y\le y)=P(\frac{1}{X}\le y)=P(1 \le Xy)\Longrightarrow(\text{When $y> 0$ }) \Longrightarrow =P\left(X \ge \frac{1}{y}\right)=\int_{\frac{1}{y}}^\infty f_X(x) \, dx=\frac{1}{\pi} \int_{\frac{1}{y}}^\infty \frac{dx}{1+x^2}=\frac{1}{\pi} \left[\frac{\pi}{2}-\arctan\left(\frac{1}{y}\right)\right]$$ $$\text{For $y < 0$:} \Longrightarrow P\left(X \le \frac{1}{y}\right) =\frac{1}{\pi}\int_{-\infty}^\frac{1}{y}\frac{dx}{1+x^2}=\frac{1}{\pi} \left[\arctan\left(\frac{1}{y}\right)-\left(-\frac{\pi}{2}\right)\right]$$

Now I'm not sure if what I did was correct, and I got stuck trying to figure out what happens when $y=0$. And I couldn't figure out what connection there is between the distributions of $X,Y$ until now.

Would appreciate any help or feedback,

thanks in advance to everyone!

Answer 1

There is no need to address the case where $y=0.$ Probability density functions are not defined pointwise (unlike likelihood functions in statistics, for which the distinction between $\text{“}{<}\text{”}$ and $\text{“}{\le}\text{”}$ sometimes matters in ways in which it does not with density functions). Changing a density function at an isolated point does not change the value of its integral over any set. To say that $f$ is the density function of the probability distribution of a random variable $W$ means only that for every measurable set $A,$ $$ \Pr(W\in A) = \int_A f(x)\,dx. $$ Altering $f$ at just one point does not alter the integral, so the value of $f$ at an isolated point isn't really a part of the identity of $f$ at all.

In the case $y>0$ and in the case $y<0,$ the derivative of the c.d.f. with respect to $y$ is $$ \frac 1 \pi \frac 1 {1+y^2}, $$ and the conclusion is that the probability distribution of $1/X$ is the same as that of $X.$

Answer 2

Correct.   Of course, you don't need to explicitly find the CDF by integration since you would immediately differentiate that to find the pdf, which is all that you want.   So to save some work:

$$\begin{align}f_{1/X}(y)&=\dfrac{\mathrm d \mathsf P(1/y\leq X)}{\mathrm d y\hspace{10.5ex}}\\[1ex]&=\dfrac{\mathrm d ~~}{\mathrm d y}\int_{1/y}^\infty f_X(x)\,\mathrm d x\\[1ex] &= -\dfrac{\mathrm d (1/y)}{\mathrm d y}\cdot f_X(1/y)\\[1ex]&=\dfrac{1}{y^2}\cdot\dfrac 1\pi\cdot\dfrac{1}{1+y^{-2}}~\mathbf 1_{y\in\Bbb R}\\[2ex]\therefore\qquad f_Y(y)&=\dfrac 1\pi\cdot\dfrac 1{y^2+1}~\mathbf 1_{y\in\Bbb R}\end{align}$$

Answer 3

Here's another way. A trigonmetric identity says: $$ \begin{align} \text{If } y>0 \text{ then } & \arctan \frac 1 y = \frac\pi2-\arctan y. \\[8pt] \text{If } y<0 \text{ then } & \arctan\frac 1 y = -\frac\pi2 - \arctan y. \end{align} $$