I have a two dimensional constant vector $\mathbf{A} = \left < 2,1 \right>$. Also, I have a vector $\mathbf{e} = \left < \epsilon_x, \epsilon_y\right >$. Both $\epsilon_x$ and $\epsilon_y$ are drawn from a uniform distribution $U(0,1)$.

There the resulting vector $\mathbf{A} + \mathbf{e}$ would be uniform too, with the $x$ component be uniform in $[2,3]$ and the $y$ component be uniform in $[1,2]$.

I want to make this vector unit. So what will happen to the distribution?

Solving for the magnitude of the vector $\left< A_x, A_y\right > + \left <e_x, e_y \right> = \left < A_x +e_x, A_y + e_y \right >$.

$M^2 = |\mathbf{A}|^2 + |\mathbf{e}|^2 + 2\mathbf{A}\cdot\mathbf{e}$

What will happen to unit vector after this?

Thanks for any help.


Call your new RVs as $X \sim U[2,3], Y \sim [1,2]$. The new vector will have elements $\frac{X}{\sqrt{X^2+Y^2}}$ and $\frac{Y}{\sqrt{X^2+Y^2}}$; call these as $Z, W$. We might be interested in marginals, but also in joint PDF, which can also be reduced down to marginals if necessary. For this, first we write the joint CDF: $$\begin{align}F_{Z,W}(z,w)&=P(Z\leq z, W \leq w)=P(\sqrt{\frac{1-z^2}{z^2}} X\leq Y, Y \leq \sqrt{\frac{w^2}{1-w^2}} X)\\ &= P(\alpha \leq \frac{Y}{X} \leq \beta) = P(\alpha\leq T\leq\beta)=F_T(\beta)-F_T(\alpha)\end{align}$$ It'll be easier if we could find CDF of $T=\frac{Y}{X}$, which is distributed in $[\frac{1}{3},1]$. Here, we need to draw a unit area square in $[2,3]$ and $[1,2]$ and examine different cases for $\frac{Y}{X}$, (It's better that you a draw while reading here).

We're going to sweep a line $Y=tX$ from x-axis to y-axis. Every time it touches a tip of the square, we have a new region to think of. There are three important cases to consider in out range: At $t=\frac{1}{3}$, line touches lower right corner of the square; at $t = \frac{1}{2}$, line touches lower left corner; at $t = \frac{2}{3}$, line touches upper right corner; and at $t = 1$, it gets out of the square. So, we have three cases: $[\frac{1}{3},\frac{1}{2}], [\frac{1}{2},\frac{2}{3}], [\frac{2}{3},1]$.

The integration of joint PDF is easy, since it is constant and $1$. We just have to find the area of the intersection of the region under the line and the square. A careful (hope so:)) analysis would lead us to:

$$F_T(t)= \begin{cases} 0 & t < \frac{1}{3} \\ \frac{(3t-1)^2}{2t} & \frac{1}{3} \leq t < \frac{1}{2} \\ \frac{5t-2}{2} & \frac{1}{2}\leq t < \frac{2}{3} \\ 1-\frac{2(1-t)^2}{t} & \frac{2}{3} < t \leq 1 \\ 1 & t > 1 \end{cases} $$

Recall that we want to find $F_T(\beta)-F_T(\alpha)$. Of course, we need $\alpha \leq \beta$, o/w the probability was 0. So, there are $15$ legitimate cases here. It's upto you to expand this formulation into those cases, and put $\alpha = \sqrt{\frac{1-z^2}{z^2}}$ and $\beta = \sqrt{\frac{w^2}{1-w^2}}$, which is very straightforward although cumbersome. And, from here, if you really need the joint PDF, what you need to do is differentiate each of these cases with respect to $w$ and $z$, irrespective of the order.

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  • $\begingroup$ oh my, i didn't realise it would get this much involved and contrived. Thanks for your time. Let me follow your solution first. Question: Why is $T = Y/X$ distributed in $[1/3,1]$? I do not follow the bounds. $\endgroup$ – cgo Sep 19 '18 at 3:53
  • $\begingroup$ max value of y/x is achieved ben y is max, x is min; and vice versa. $\endgroup$ – gunes Sep 19 '18 at 5:05

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