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Let $R$ and $X$ be independent non-negative random variables such that $R^2\sim \chi^2_2$ and $X\sim U(0, 2\pi)$. Fix $a$ belonging to $(0, 2\pi)$. Find the distribution of $R\sin(X+a)$.

Michael Hardy
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kris91
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    I wonder if you meant $R\sin(X+a)$ rather than $R\sin(x+a)$? – Michael Hardy Apr 26 '13 at 16:38
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    When you say "Find the distribution of . . .", you're phrasing this in a way appropriate for a homework assignment. That makes it look as if you're more-or-less stenographically passing along to us a question written by someone other than your self, when you haven't demonstrated that you've understood the question. If you can say something about your own thoughts on this, or what questions you have about it, you'll get a better response here. – Michael Hardy Apr 26 '13 at 16:47
  • yes i meant Rsin(X+a) – kris91 Apr 26 '13 at 17:21

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The chi-square distribution with $2$ degrees of freedom is the distribution of the sum of squares of two independent $N(0,1)$ random variables, which let us call $A$ and $B$. Thus $R$ is the distance from the origin to $(A,B)$. Because of the circular symmetry of the distribution of the pair $(A,B)$, the distribution of the angle from the $x$-axis in a counterclockwise direction to the ray from the origin through $(A,B)$ is uniform on the interval $(0,2\pi)$. Since that latter distribution remains the same regardless of how far $(A,B)$ is from the origin, the two random variables are independent. When you take the sine of the angle, you're looking at the component in the direction of one line through the origin. Which line it is depends on (lower-case) $a$ (not to be confused with the random variable (capital) $A$). But again, because of the circular symmetry, the distribution doesn't depend on which line you pick, so it may as well be the $x$-axis, and then the projection of $(A,B)$ onto that line is just the random variable $A$.

Bottom line: It's the distribution of $A$, so it's a standard normal distribution.

Michael Hardy
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  • I tried using transformation but got stuck at finding the marginal distribution.i took X1=Rsin(X+a), X2=Rcos(X+a) – kris91 Apr 26 '13 at 17:20
  • The distribution of $\sin(X+a)$ does not depend on $a$. Notice that $X+a$ is uniformly distributed on $(a,a+2\pi)$ $=(a,2\pi)\cup[2\pi,a+2\pi)$, and on $[2\pi,a+2\pi)$, the sine function behaves the same way it does on $[0,a)$. So the distribution you're looking for is the same as the distribution you'd get if $a$ were not there at all. The distribution of $\sin(X+a)$ is therefore the same as the distribution of $\sin X$. ${}\qquad{}$ – Michael Hardy Apr 26 '13 at 17:57
  • Nicely done ...! – wolfies Apr 27 '13 at 14:33
  • Thank you. I just didnt get one part of the answer-which line it is depends on a. – kris91 Apr 28 '13 at 06:13
  • @kris91 : $R\sin X$ is the projection of $(A,B)$ onto the $y$-axis. $R\sin X$ is the same as $B$. And $R\sin(X+a)$ is the projection of $(A,B)$ onto the line that results from rotating the $y$-axis through the angle $a$. ${}\qquad{}$ – Michael Hardy Apr 28 '13 at 16:56