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(from The Probability Tutoring Book, C. Ash, p. 157)

Find the density of $Y$ if $Y = 1/X$ and $X$ is uniform on $[-1,1]$.

The distribution function given in the answer key is

$$ F(y) = \begin{cases} \frac{-1}{2y} & \text{if }y \leq -1\\ \frac{1}{2} & \text{ if }-1 \leq y \leq 1\\ 1-\frac{1}{2y} & \text{if } y \geq 1 \end{cases} $$

My question is why the first and last cases aren't $0$ and $1$. Doesn't $Y$ only range over $[-1, 0)$ and $(0, 1]$?

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  • $\begingroup$ No, when $X \approx 0,$ you have extreme positive and negative values of $Y.$ $\endgroup$
    – BruceET
    Apr 18, 2020 at 23:17
  • $\begingroup$ Yes, I will adjust my question. I'm really trying to understand the equations for y <= -1 and y >= 1 $\endgroup$
    – planarian
    Apr 18, 2020 at 23:23
  • $\begingroup$ Maybe start with $F_Y(y) = P(Y \le y) = P(1/X \le y) = \cdots,$ for $y$ in each of the three intervals. $\endgroup$
    – BruceET
    Apr 18, 2020 at 23:31
  • $\begingroup$ Similar to stats.stackexchange.com/q/450921/119261 since the distribution of $X$ is symmetric about zero. To derive the pdf from the cdf, it suffices to find the cdf for $y>0$ as the distribution of $1/X$ will also be symmetric about zero. $\endgroup$ Apr 19, 2020 at 18:22

2 Answers 2

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I hope a quick simulation in R will help you visualize this transformation. For readable graphs, I use only transformed values in $(-10,10).$

set.seed(2020)
x = runif(10^5, -1,1)
y = 1/x
Y = y[abs(y) < 10]  # plotted values
length(y)
[1] 100000
length(Y)
[1] 89815
par(mfrow=c(1,2))
 hist(Y, prob=T, col="skyblue2")
 plot(ecdf(Y))
par(mfrow=c(1,1))

enter image description here

Note: The ECDF of a sample plots values of the sample from smallest to largest. It is a stairstep function (with jumps too small to see here). At each value of a sample of size $n$ the ECDF jumps up by $1/n.$ If there are $K$ values tied at a point, then the function jumps by $k/n$ at that point. (No ties here) For a sufficiently large sample the ECDF closely imitates the population CDF.

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Observe that $-1\le x\le 1\implies |x|\le 1\implies \frac1{|x|}\ge 1$, so support of $|Y|$ is $[1,\infty)$.

In other words, support of $Y$ is $(-\infty,-1]\cup [1,\infty)$.

For any continuous random variable $X$, distribution function of $Y=\frac1X$ is

\begin{align} P(Y\le y)&=P\left(\frac1X\le y,X<0\right)+P\left(\frac1X\le y,X>0\right) \\&=\begin{cases} P\left(\frac1y\le X<0\right) &,\text{ if }y<0 \\ P(X<0)+ P\left(X\ge \frac1y\right) &,\text{ if }y>0\end{cases} \end{align}

That is,

$$P(Y\le y)=\begin{cases} P(X<0)-P\left(X\le \frac1y\right) &,\text{ if }y<0 \\ 1 + P(X<0)-P\left(X < \frac1y\right) &,\text{ if }y>0\end{cases}$$

As $P\left(\frac1X\le y\right)=P(X\in A)$ where $A=\left\{x: \frac1x \le y\right\}$, drawing a picture of the region $A$ might help in verifying the calculation.

Now if you carefully find $P\left(X \le \frac1y\right)$ from the distribution function of $X$, you will end up with the answer in your post.

Alternatively, you can find the density of $Y$ directly from the change of variables $y=\frac1x$:

$$f_Y(y)=f_X\left(\frac1y\right)\left|\frac{\mathrm dx}{\mathrm dy}\right|$$

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