Probability density function of $Z = X \sin Y$

Question

The probability density function of a random variable $X$ is a uniform distribution $U(a,b)$. Likewise, the probability density function of a random variable $Y$ is a uniform distribution $U(0, 2\pi)$. $X$ and $Y$ are independent of each other.

What is the probability density function of $Z = X \: \sin{Y}$ ?

Answer 1

Recall that a general procedure to compute the distribution of $Z=X\sin Y$ when $(X,Y)$ are independent with densities $f_X$ and $f_Y$ is to compute $E[u(Z)]$ for every bounded measurable function $u$. If it happens that $$ E[u(Z)]=\int_\mathbb Ru(z)g(z)\mathrm dz, $$ then $g$ is the density of $Z$. Here, $$ E[u(Z)]=\iint u(x\sin y)f_X(x)f_Y(y)\mathrm dx\mathrm dy, $$ hence the job is to transform this double integral into the simple integral above. The way to do that is rather clear: change of variables.

Consider $(z,t)=(x\sin y,x\cos y)$ and assume for example that $X$ is uniform on $(0,1)$ and $Y$ uniform on $(0,2\pi)$, then $\mathrm dz\mathrm dt=x\mathrm dx\mathrm dy$ and $x^2=z^2+t^2$ hence $$ E[u(Z)]=\iint_{z^2+t^2\leqslant1} u(z)\frac{\mathrm dz\mathrm dt}{2\pi\sqrt{t^2+z^2}}, $$ and one sees that $$ g(z)=\mathbf 1_{|z|\leqslant1}\int_{-\sqrt{1-z^2}}^{\sqrt{1-z^2}}\frac{\mathrm dt}{2\pi\sqrt{t^2+z^2}}=\frac1\pi\mathbf 1_{|z|\leqslant1}\int_0^{\sqrt{1-z^2}/z}\frac{\mathrm dt}{\sqrt{t^2+1}}, $$ that is, $$ g(z)=\frac1\pi\cosh^{-1}\left(\frac1z\right)\mathbf 1_{|z|\leqslant1}. $$

Answer 2

Case 1: $(0 < a < b)$ To proceed, let $X$ have pdf $f(x)$:

where I am initially assuming parameter $a$ is positive. I will consider the alternative cases later. Second, let $W = sin(Y)$ with pdf $\phi(w)$:

The latter is relatively straightforwards to derive.

Next, we seek the pdf of the product of two random variables, namely $Z = X * W$. This can be cumbersome to do by hand, but can be derived easily using the TransformProduct function in the mathStatica suite. In particular, the pdf of $Z$, say $g(z)$ is:

All done. (I should add I am one of the authors of the software.)

Here is a plot of the theoretical pdf derived $g(z)$ (in red) when $a = 2$ and $b = 5$, and superimposed on top a Monte Carlo check (in blue).

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The other 2 cases (with parameter $a< 0$) are derived in identical fashion ... just change the assumptions on parameters $a$ and $b$ in the first input. Doing so yields:

Case 2: $(a < 0, b > 0)$

The solution pdf is $g(z)$:

Here is a plot of the solution pdf $g(z)$ when $a = -3, b = 2$:

Case 3: $(a < b < 0)$

The solution pdf is $g(z)$:

Here is a plot of the solution pdf $g(z)$ when $a = -4, b = -2$:

Answer 3

Remember that c. d. f. of $Z$ is $$P\{Z<z\}=P\{X\sin Y<z\}=\iint\limits_{x\sin y<z} p_{XY}(x,y)dxdy=\int\limits_{-\infty}^{+\infty}dx\int\limits_{-\infty}^{z/x}p_{XY}(x,y)dy=\\=\int\limits_a^{b}dx\int\limits_{0}^{z/x}p_{XY}(x,y)dy$$

The probability density function will then be $$\frac{d}{dz}\left(\int\limits_a^{b}dx\int\limits_{0}^{z/x}p_{XY}(x,y)dy\right)$$