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.

up vote 3 down vote accepted

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.

  • 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. – cgo Sep 19 at 3:53
  • max value of y/x is achieved ben y is max, x is min; and vice versa. – gunes Sep 19 at 5:05

Your Answer

 
discard

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.