Solved – If $X=\sin\Theta$ and $Y=\cos\Theta$ with $\Theta$ uniformly distributed, how can I compute the joint pdf of $(X,Y)$

Question

I have a random variable $\Theta$ uniformly distributed between $[-\pi ,\pi]$, two functions $X=\sin\Theta$ and $Y=\cos\Theta$. I know that $X$ and $Y$ are uncorrelated but not independent. I want to find the joint pdf $f(x,y)$ of $X$ and $Y$. How can I compute this?

I tried by using cdf $F(x,y)$, defined as:

\begin{align}
F(x,y)&=P(X<x,Y<y)
\\&=P(\sin\Theta<x,\cos\Theta<y)
\\&=P(\Theta<\arcsin x,\Theta<\arccos y)
\\&=P(\Theta<\max(\arcsin x,\arccos y))
\end{align}

(Is the last equality right?)

So:

\begin{align}
F(x,y) = \begin{cases}c_1\arccos x+c_2&,\text{ if }x,y\le \frac{\sqrt 2}{2} \\ c_3\arcsin y+c_4 &,\text{ if }\frac{\sqrt 2}{2}\le x,y\le 1\end{cases}
\end{align}

By imposing the property of cdf (i.e. $F(-1,-1)=0$ and $F(1,1)=1$):

\begin{align}
F(x,y) = \begin{cases}-\frac{2}{3\pi}\arccos x+ \frac23 &,\text{ if }x,y\le \frac{\sqrt 2}{2}
\\ \frac2{\pi}\arcsin y &,\text{ if }\frac{\sqrt 2}{2}\le x,y\le 1\end{cases}
\end{align}

Now, I want to find the joint pdf $f(x,y)$ as:

$$f(x,y) = \frac{\partial ^2 F(x,y)}{\partial x\partial y}$$

How can I proceed (if the procedure that I used is correct)?

If the my procedure isn't right, how do I calculate the joint PDF of $(X,Y)$?

Thank you in advance!

Answer 1

We can derive a CDF, but not a valid pdf, as pointed out by @whuber. I will demonstrate how to derive the CDF.

You are correct up until here: $$\eqalign{ F(x,y) &= P(X \leq x, Y \leq y) = P (\sin(\theta) \lt x, \cos(\theta)\lt y) \\ &= P(\theta \leq \arcsin(x), \theta \leq \arcsin(y)).}$$

However, in your next step, you write $\max$ where you should have $\min$ (since $\theta$ must be less than both, it must be less than the smaller of the two). Therefore, we have

$$F(x,y) = P\left(\theta \leq \min\{\arcsin(x), \arcsin(y)\}\right).$$

Since $\theta \sim U(-\pi, \pi)$, it follows that

$$F(x,y) = \begin{cases} 0, & \min\{\arcsin(x),\, \arcsin(y)\} \leq -\pi \\ \frac{\min\{\arcsin(x),\ \arcsin(y)\} + \pi}{2\pi}, & -\pi \leq \min\{\arcsin(x),\, \arcsin(y)\}\leq \pi \\ 1, & \pi \leq \min\{\arcsin(x),\, \arcsin(y)\} \end{cases} $$