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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!

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    $\begingroup$ A joint PDF does not exist. The variable $(X,Y)$ has a singular distribution: all its probability is concentrated on a lower-dimensional space (the unit circle). $\endgroup$
    – whuber
    Commented Apr 22, 2020 at 19:12
  • $\begingroup$ Refer to math.meta.stackexchange.com/questions/5020/… for typesetting math. $\endgroup$ Commented Apr 23, 2020 at 15:13

2 Answers 2

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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

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

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  • $\begingroup$ Are you suggesting it's possible to compute a PDF for $(X,Y)$?? $\endgroup$
    – whuber
    Commented Apr 22, 2020 at 19:13
  • $\begingroup$ Which step is incorrect? $\endgroup$
    – dlnB
    Commented Apr 22, 2020 at 19:35
  • $\begingroup$ You haven't answered the question: it asks for $f,$ not $F.$ By offering a calculation of $F$ you are implying its mixed second partial derivative will yield $f,$ as stated in the question--but that derivative does not exist everywhere and does not produce a valid PDF. $\endgroup$
    – whuber
    Commented Apr 22, 2020 at 20:01
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    $\begingroup$ Better, yes--but not quite right. There are plenty of CDFs that are not everywhere differentiable but still yield PDFs. For instance, writing $u^{*}=\max(\min(u,1),0),$ consider $F(x,y)=x^{*}y^{*}.$ This is not second differentiable on the boundary of the unit square $[0,1]\times[0,1]$ but it has a PDF (it's the uniform distribution on the unit square). Thus, failing to be differentiable is not the reason the PDF doesn't exist. You need to invoke the second clause of my comment: computing $f$ from your $F$ and integrating $f$ does not reproduce $F.$ $\endgroup$
    – whuber
    Commented Apr 22, 2020 at 21:15
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    $\begingroup$ +1. The idea is clear -- but the use of $\arcsin$ is not correct, because by definition its values will always lie between $-\pi$ and $\pi$ and its domain is $[-1,1]$ rather than all the real numbers. $\endgroup$
    – whuber
    Commented Apr 23, 2020 at 13:46
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  1. $\Theta \sim U(-\pi, \pi)$ so the density of $\Theta$ is given by $\frac{1}{2 \pi}$ in $-\pi, \pi$.

  2. $F(x, y) = P(X \le x, Y \le y) = P(\Theta \le \arcsin(x) \wedge \arccos(y))$.

  3. $ F(x, y) = \frac{1}{2\pi}\int_{-\pi}^{\arcsin(x) \wedge \arccos(y)} 1 \cdot d\theta = \frac{1}{2\pi}\arcsin(x) \wedge \arccos(y) + \frac{1}{2}. $

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  • $\begingroup$ One problem with this analysis is that neither the arcsine nor the arccosine are uniquely defined, requiring care in explaining what they mean. Another problem is that it begs the question: how can one describe some kind of a density related to $F$? $\endgroup$
    – whuber
    Commented Jan 20, 2021 at 19:28

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