# Transform bivariate uniform variable

Let $$X_1 = U(0,1)$$ and $$X_2 = U(0,1)$$. $$X_1$$ and $$X_2$$ are independent. Then $$f(x_{1}, x_{2})=1, {0}\le{x_1}\le{1}, {0}\le{x_2}\le{1}$$.

Let $$Y_1 = \arctan(X_{2}/X_{1})$$, $$Y_2 = X_2$$. I need to find the density function $$g(y_1)$$.

Here's what I've done so far.

$$X_{2}/X_{1} = \tan(Y_{1})$$ $$X_{1} = X_{2} / \tan(Y_{1}) = X_{2}\cot(Y_{1})$$

Since $$X_{2}=Y_{2}$$, $$X_{1}=Y_{2}\cot(Y_{1})$$.

The Jacobian of the transformation is give by the matrix $$J = \begin{bmatrix} \frac{\partial{y_2\cot(y_1)}}{\partial{y_1}} & \frac{\partial{y_2\cot(y_1)}}{\partial{y_2}} \\ \frac{\partial{y_2}}{y_1} & \frac{y_2}{y_2} \end{bmatrix} = \begin{bmatrix} -y_{2}\csc^{2}(y_{1}) & \cot(y_1) \\ 0 & 1 \end{bmatrix}$$. So, $$\det(J)= -y_{2}\csc^{2}(y_{1})$$.

At this point, I am kinda lost. I know that I need to compute $$g(y_1, y_2)$$, find its domain, and then integrate with respect to $$y_2$$. I do not understand how to find the joint density $$g(y_1, y_2)$$, could you please help?

• The change of variable formula indicates how to get from $f(x_1,x_2)$ to $g(y_1,y_2)$:$$g(y) = f\big(h^{-1}(y)\big) \left| \frac{d}{dy} \big(h^{-1}(y)\big) \right|$$ Commented Apr 4, 2023 at 6:54

Find the distribution function of $$Y_1$$ with a picture, then differentiate it.

Observe that because $$(X_1,X_2)$$ lies in the first quadrant and $$Y_1$$ is the angle subtended by this point, $$0\le Y_1 \le \pi/2.$$ Moreover, for any possible value $$\theta$$ in this interval, the event $$Y_1 \le \theta$$ is the region in the unit square bounded above by the line at angle $$\theta,$$ highlighted in this diagram:

Because $$(X_1,X_2)$$ is uniformly distributed on the unit square, the probability of any event equals its area.

Clearly when $$0\le \theta\le \pi/4,$$ where this event is a triangle, its area is half its height:

$$\Pr(Y_1 \le \theta) = \frac{1}{2}\tan\theta, \quad \theta \le \pi/4.$$

Its derivative is $$\sec^2(\theta)/2.$$

When $$\pi/4\le\theta\le \pi/2,$$ the event is the complement of a triangle, immediately giving

$$\Pr(Y_1 \le \theta) = 1 - \frac{1}{2}\cot\theta, \quad \pi/4 \le \theta \le \pi/2.$$

Its derivative is $$\csc^2(\theta)/2 = \sec^2(\pi/2-\theta)/2.$$

Combining these results into a common formula gives

$$g(\theta) = \frac{1}{2}\sec\left(\min\left(\theta, \frac{\pi}{2}-\theta\right)\right)^2,\quad 0\le\theta\le\frac{\pi}{2}.$$

Of course $$g\equiv 0$$ for all other arguments.

To illustrate $$g,$$ here is a histogram of a million realizations of $$Y_1$$ (computed with the R code n <- 1e6; y <- atan2(x2 <- runif(n), x1 <- runif(n))) over which the graph of $$g$$ is plotted in red.

The transformation from $$(x_1,x_2)$$ to $$(y_1,y_2)$$ is given by $$h(x_1,x_2) = (h_1(x_1,x_2),h_2(x_1,x_2))^t = \left(\arctan(\frac{x_2}{x_1}), x_2\right)^t$$. The jacobian matrix $$J(x_1,x_2)$$ is given by:

$$J(x_1,x_2) = \begin{bmatrix} \frac{\partial h_1}{\partial x_1} & \frac{\partial h_1}{\partial x_2} \\ \frac{\partial h_2}{\partial x_1} & \frac{\partial h_1}{\partial x_2} \end{bmatrix} = \begin{bmatrix} \frac{-x_2}{x_1^2 + x_2^2} & \frac{x_1}{x_1^2 + x_2^2} \\ 0 & 1 \end{bmatrix},$$

with the absolute value of the determinant $$|J(x_1,x_2)| = \frac{x_2}{x_1^2 + x_2^2}$$.

As mentioned by Xi'an in the comments, you can find the density probability function $$g(y_1,y_2)$$ of $$(Y_1,Y_2)$$ with the following formula: $$g(y_1,y_2) = \frac{f(x_1,x_2)}{|J(x_1,x_2)|}|_{(x_1,x_2) = h^{-1}(y_1,y_2)} = \frac{x_1^2 + x_2^2}{x_2}|_{(x_1,x_2) = h^{-1}(y_1,y_2)},$$

$$= \frac{y_2^2\cot(y_1)^2 + y_2^2}{y_2} = y_2(\cot(y_1)^2 + 1),$$

where $$f(x_1,x_2)$$ is the probability density function of $$(X_1,X_2)$$. Note that this works only for bijective transformations (mapping) $$h(x_1,x_2)$$.

• To find the marginal of $Y_1,$ then, you would integrate out $y_2$ in $g,$ giving $(\cot(y_1)^2+1)/2.$ Clearly that's not correct for $y_1 \le \pi/4:$ it blows up near $y_1=0.$ For $\pi/2\le y_1\le \pi/2$ it can also be expressed as $\sec(\pi/2-y_1)^2/2,$ which is correct.
– whuber
Commented Apr 4, 2023 at 22:35