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Hey, what is the probability density function of the following random variable $\theta$ : $\theta=tan^{-1}\frac{Y_1-Y_2}{X_1-X_2}$ for $X_1-X_2>0$ and $\theta=tan^{-1}\frac{Y_1-Y_2}{X_1-X_2}+\pi$ for $X_1-X_2<0$. $X_1,X_2,Y_1,Y_2$ are all uniformly distributed random variables in the range 0 to 1. I arrived to this question when I was trying to find out the pdf of the angle between two randomly positioned nodes $(X_1,Y_1)$ and $(Y_1,Y_2)$. So basically I've 2 questions:(1). $X_1-X_2$ and $Y_1-Y_2$ have triangular distribution between -1 to +1. But their ratio distribution is difficult to find because the denominator takes 0 value outside the range (-1,+1). So, I attempted to approximate this triangular by a Gaussian of variance $\frac{1}{9}.$ Now, ratio distribution of 2 Gaussian is Cauchy and then by taking $tan^{-1}$ of that I got a uniform density function $\frac{1}{\pi}$ between $-\frac{\pi}{2}$ to $\frac{\pi}{2}$. Am I doing anything wrong in this? And, my 2nd ques (2). $tan^{-1}\frac{Y}{X}$ takes values only between $-\frac{\pi}{2}$ to $\frac{\pi}{2}$. So, for -ve $X$, I was adding a $+\pi$(See the 2$^{nd}$ line in the question). How does that reflect in the final pdf? The final pdf that I've obtained is only between $-\frac{\pi}{2}$ to $\frac{\pi}{2}$. Pls help.

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I don't think you should approximate anything like this by a Gaussian. There is no CLT involved. –  Leonid Petrov Mar 24 '13 at 16:40
    
I'm definitely not using CLT. I'm approximating a triangular distribution by a Gaussian by adjusting its variance such that 99% of its mass lie in the (-1,+1) range. How else do you suggest me to find the pdf of $tan^{-1}\frac{Y_1-Y_2}{X_1-X_2}$? –  Prakash Mar 24 '13 at 18:10
    
Okay, you're not using CLT, but the criticism remains valid: "approximating" the triangular distribution by a Gaussian changes the problem more or less completely. –  Greg Martin Mar 24 '13 at 20:12
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2 Answers

The way I was calculating the pdf of $Z=\frac{Y}{X}$ where $X$ and $Y$ are independent r.v. with triangular distribution in the range $(-1,1)$ is as follows :

\begin{align*} F_Z(z)=Pr(\frac{Y}{X}\leq z)=\displaystyle\int_{-1}^1 Pr(Y\leq xz)f_X(x)dx=\displaystyle\int_{-1}^1\int_{-1}^{min(xz,1)}f_Y(y)f_X(x)dydx \end{align*} Now, I calculate $Pr(Y\leq xz)$ as follows :\\ $\underline{z>0}$: [ Pr(Y\leq xz) = \begin{cases} \displaystyle\int_{-1}^{xz}(1+y)dy & \text{if } -\frac{1}{z} < x < 0 \\ \displaystyle\int_{-1}^0 (1+y)dy + \displaystyle\int_0^{xz} (1-y)dy & \text{if } 0 < x < \frac{1}{z}\\ 1 & \frac{1}{z} < x < 1 \end{cases} ]

$\underline{z\leq 0}$: [ Pr(Y\leq xz) = \begin{cases} \displaystyle\int_{-1}^{xz}(1+y)dy & \text{if } -\frac{1}{z} \geq x \geq 0 \\ \displaystyle\int_{-1}^0 (1+y)dy + \displaystyle\int_0^{xz} (1-y)dy & \text{if } 0 > x \geq \frac{1}{z}\\ 1 & -1 \leq x \leq \frac{1}{z} \end{cases} ]

Thus, $Pr(Y\leq xz)$ becomes :\ [ Pr(Y\leq xz) = \begin{cases} xz+\frac{x^2z^2}{2}+\frac{1}{2} & \text{if } -\frac{1}{z} \geq x \geq 0, z > 0 \ xz-\frac{x^2z^2}{2}+\frac{1}{2} & \text{if } 0 < x < \frac{1}{z}, z > 0 \ 1 & \text{if } \frac{1}{z} < x < 1, z>0 \ xz+\frac{x^2z^2}{2}+\frac{1}{2} & \text{if } 0 \geq x \geq -\frac{1}{z}, z < 0 \ xz-\frac{x^2z^2}{2}+\frac{1}{2} & \text{if } \frac{1}{z} < x < 0, z < 0 \ 1 & \text{if } -1 < x < \frac{1}{z}, z<0 \

\end{cases} ]

Then, plugging in these values of $Pr(Y\leq xz)$ in the expression of $F_Z(z)$, I obtain :\ [ F_Z(z) = \begin{cases} \displaystyle\int_{min(-1,-\frac{1}{z})}^0(xz+\frac{x^2z^2}{2}+\frac{1}{2})(1+x)dx + \displaystyle\int_0^\frac{1}{z}(xz-\frac{x^2z^2}{2}+\frac{1}{2})(1-x)dx + \displaystyle\int_{\frac{1}{z}}^1 1.(1-x)dx & \text{if } z>0 \\ \displaystyle\int_0^{-\frac{1}{z}}(xz+\frac{x^2z^2}{2}+\frac{1}{2})(1+x)dx + \displaystyle\int_{\frac{1}{z}}^0(xz-\frac{x^2z^2}{2}+\frac{1}{2})(1-x)dx + \displaystyle\int_{-1}^{\frac{1}{z}} 1.(1-x)dx & \text{if } z \leq 0 \end{cases} ] So, now there are 3 cases for z: $0<z<1$, $1<z<\infty$ and $z\leq 0$\\ $\therefore$ Value of $F_Z(z)$ for each of these cases is as follows :\\ $\textbf{Case 1}$: $0<z<1$\ $\displaystyle\int_{-\frac{1}{z}}^0(xz+\frac{x^2z^2}{2}+\frac{1}{2})(1+x)dx + \displaystyle\int_0^\frac{1}{z}(xz-\frac{x^2z^2}{2}+\frac{1}{2})(1-x)dx + \displaystyle\int_{\frac{1}{z}}^1 1.(1-x)dx$\\ $\textbf{Case 2}$: $z>1$\ $\displaystyle\int_{-1}^0(xz+\frac{x^2z^2}{2}+\frac{1}{2})(1+x)dx + \displaystyle\int_0^\frac{1}{z}(xz-\frac{x^2z^2}{2}+\frac{1}{2})(1-x)dx + \displaystyle\int_{\frac{1}{z}}^1 1.(1-x)dx$\\ $\textbf{Case 3}$: $z\leq 0$\ $\displaystyle\int_0^{-\frac{1}{z}}(xz+\frac{x^2z^2}{2}+\frac{1}{2})(1+x)dx + \displaystyle\int_{\frac{1}{z}}^0(xz-\frac{x^2z^2}{2}+\frac{1}{2})(1-x)dx + \displaystyle\int_{-1}^{\frac{1}{z}} 1.(1-x)dx$

I don't know where exactly I'm going wrong ! But I'm calculating this in the same line as the ratio of 2 uniform r.v. which is coming same as given in Wiki.

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First, you should probably solve the problem first without the two cases and the $+\pi$; the symmetry of switching $(X_1,Y_1)$ with $(X_2,Y_2)$ will allow you to convert your answer to the simpler problem - some pdf $f(x)$ supported on $(-\frac\pi2,\frac\pi2)$ - to the answer you're really looking for - which will be $\frac12f(x) + \frac12f(x-\pi)$.

Second, I recommend trying to find the probability density function $g(x)$ for $\frac YX$, given that $X$ and $Y$ are independent random variables both with the triangle distribution on $(-1,1)$ - this is a ratio distribution. Afterwards you can adjust your answer to accommodate the $\tan^{-1}$ function: the resulting probability density function will be $g(\tan^{-1} x)/(1+x^2)$.

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