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I'm learning about functions of random variables and am trying to work out an example I made up. If $y = \sin(x)$ and $x$ has domain $[0, 4\pi]$, is the following the correct expression for the pdf of $y$: $$\begin{align*} f_Y(y) &= \frac{d}{dy}F_Y(y)\\ &= \frac{d}{dy}[F_X(2\pi) - F_X(\pi) + F_X(4\pi) - F_X(3\pi)]\\ &= \frac{d}{dx}F_X(2\pi)\left|\frac{dg^{-1}(y)}{dy}\right| - \frac{d}{dx}F_X(\pi)\left|\frac{dg^{-1}(y)}{dy}\right| + \frac{d}{dx}F_X(4\pi)\left|\frac{dg^{-1}(y)}{dy}\right| - \frac{d}{dx}F_X(3\pi)\left|\frac{dg^{-1}(y)}{dy}\right|\\ &=f_X(2\pi)\left|\frac{1}{\sqrt{1-0}}\right| - f_X(\pi)\left|\frac{1}{\sqrt{1-0}}\right| + f_X(4\pi)\left|\frac{1}{\sqrt{1-0}}\right| - f_X(3\pi)\left|\frac{1}{\sqrt{1-0}}\right|\\ &= f_X(2\pi) - f_X(\pi) + f_X(4\pi) - f_X(3\pi)\\ \end{align*}$$

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    $\begingroup$ This is hopelessly and irretrievably incorrect from the second equation onwards. Have you given any thought to the fact that the quantity in square brackets on line 2 is not a function of $y$ at all, and so the derivative with respect to $y$ must be $0$ and not the gobbledygook that you have conjured up out of thin air? $\endgroup$ – Dilip Sarwate Feb 22 '15 at 17:18
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    $\begingroup$ Thanks for your comment, @Dilip. I see $x = g^{-1}(y) = sin^{-1}(y)$. Although the question asks for $f_Y(y)$, I was looking at $f_Y(Y = 0)$ to simplify the intervals over which to write an expression for the CDF. I realize that this doesn't make sense either since the PDF at any one point is $0$. Drawing the picture, I see that for a particular $y$, we will have 0 (at $y = -1$), 1 (at $y = 1$), 2 (at $y = 0$), or 3 (other values of $y$) intervals over which we have to write an expression for the CDF. We can then differentiate this to get an expression for the PDF. Is this the right idea? $\endgroup$ – Vivek Subramanian Feb 22 '15 at 18:16
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Draw a picture.

Although you can apply a standard formula for changes of variable, this one is tricky because the transformation $X\to Y$ is not one-to-one. Often the most convenient and reliable method is to compute the distribution function (CDF) and then differentiate it.

The distribution function of $Y=\sin(X)$ is, by definition,

$$F_Y(y) = \Pr(Y \le y) = \Pr(\sin(X) \le y) = \Pr(\{X\,|\, \sin(X) \le y\}).$$

The latter probability is with respect to $X$. The graph of $\sin$ has been emphasized where its height is less than or equal to $y$. The values of $X$ where this occurs, shown in thick blue along the axis, show the set $\{x\in[0,4\pi]\,|\, \sin(x) \le y\}$.

Figure 1

When $0 \lt y$, this set consists of three disjoint intervals $\newcommand{s}{\sin^{-1}y}[0, \s]$, $[\pi -\s, 2\pi + \s]$, and $[3\pi - \s, 4\pi]$. Because they are disjoint, the chance that $X$ lies within this union is the sum of the chances of each interval:

$$\eqalign{ \Pr(\sin(X) \le y) &= F_X(\s) - F_X(0) \\ &+ F_X(2\pi + \s) - F_X(\pi - \s)\\ &+1 - F(3\pi - \s). }$$

The value $1 = F_X(4\pi)$ appeared because the range of $X$ is $[0, 4\pi]$. However, we may not replace $F_X(0)$ by $0$ because possibly $X$ has nonzero probability there.

When $y \lt 0$, the set $\{X \in[0,4\pi]\,|\, \sin(X) \le y\}$ is the union of just two disjoint intervals:

Figure 2

By emulating the preceding argument, you should have no trouble writing down an expression for their probability in terms of $F_X$.

The PDF, when it exists, is the derivative of $F_Y$. In the first case

$$f_Y(y) = \frac{d}{dy} F_Y(y) = \frac{d}{dy}\left(F_X(\s) - F_X(0) + F_X(2\pi + \s) \cdots - F(3\pi - \s)\right).$$

Apply the Chain Rule, recognizing that $\frac{d}{dy}\s = 1/\sqrt{1-y^2}$:

$$f_Y(y) = \frac{1}{\sqrt{1-y^2}}\left(f_X(\s) + f_X(\pi-\s) + f_X(2\pi+\s) + f_X(3\pi-\s)\right)$$

This formula can be understood for all $y$ provided we replace $f_X(\s)$ by $f_X(4\pi + \s)$ whenever $y \lt 0$. Equivalently, and much more generally (with no restrictions on the range of $X$),

$$f_Y(y) = \frac{1}{\sqrt{1-y^2}}\sum_{i=-\infty}^\infty f_X(2i \pi + \s) + f_X((2i+1)\pi - \s).$$

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    $\begingroup$ Fantastic, @whuber! Makes completely sense and is what I was hinting at in my comment to @DilipSarwate. Thank you! $\endgroup$ – Vivek Subramanian Feb 23 '15 at 23:41

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