Probability question with cumulative distribution function and density function calculation.

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,

• Jun 10 '21 at 19:16
• I would have written $-\infty<y<+\infty$ rather than $\text{“}{\le}\text{”}. \qquad$ Jun 10 '21 at 19:40
• An explanation for the fact that $1/X$ has the same distribution as $X$ is that this distribution is that of the ratio $Z_1/Z_2$ of independent standard normal random variables. Jun 11 '21 at 18:48

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 $$X.$$

• Thanks for the great explanation, just a question, The derivative in case of $y<0$, is coming out with a negative sign, did I make a mistake calculating the integral or derivative? or there's something I'm missing about the signs of the function? Jun 10 '21 at 20:06
• @Pwaol : My first suspicion without seeing the details of what you did is that a minus sign was missed whey you applied the chain rule. Jun 10 '21 at 20:07
• The thing is the $arctan(\frac{1}{y})$ when $y>0$ we have a minus before it, so when I take derivative it cancels with the minus from the chain rule giving the desired result, but when $y<0$ the minus comes from the chain rule, and there's no minus before the $arctan$ there so it doesn't cancel and stays with the result. Jun 10 '21 at 20:11
• $$\frac d{dy} \left( - \arctan\frac 1 y \right) = -\frac 1 {1+\left( 1/y \right)^2} \cdot \frac{-1}{y^2} = \frac 1 {y^2+1}.$$ Jun 10 '21 at 20:21
• @Pwaol : I should actually have written that if $y<0$ then $$\Pr\left( \frac 1 X \le y\right) = \Pr( 1 \ge Xy) = \Pr\left( \frac 1 y \le X \right) = \int_{1/y}^0 \cdots.$$ Jun 11 '21 at 19:11

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}

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}