Probability functions in a unit circle

[atrus_ovis]

Homework Statement

Choose a point in the unit axis, say x.Let Y be the distance of that point and the point where thε perpendicular line crosses the unit circle.
Find the density and cumulative functions of Y.

Homework Equations

Basic trigonometry i guess.

The Attempt at a Solution

I don't really have much of a clue.
Y equals r sin(θ) = sin(θ) , since the radius of the unit circle is 1.
I get the feeling that the curve of f(Y) will kind of resemble a normal distribution, since most values of sin(y) are closer to 1 and drop quicker as x approches +1 or -1.

 

[hgfalling]

What's the unit axis? Does this mean "pick a point uniformly at random on [0,1]" ?

I'm not sure exactly what $\theta$ you are using here, but it is true that doing something with $\sin \theta$ for a properly defined $\theta$ could lead you to the right answer.

But maybe it's easier to just work directly in Cartesian coordinates, simply because you are picking a Cartesian coordinate directly with uniform probability, and so you might have some mild unpleasantness in converting the problem to polar coordinates.

Suppose you pick x. Draw the triangle and figure out what the value of y is. Then maybe you can write down the density function, or an integral expression for the cumulative distribution function. Don't worry about trig functions just yet; everything here is algebraic.

 

[atrus_ovis]

It's too late for the homework now, but i'd like to know the solution.

The values of Y are found-able.I couldn't though find a mapping to a probability.
How do you get from a set of (infinite) Y values to a probability of each?

After sniffing around in wiki (yeah i suck that bad in probability theory), i found about the uniform distribution, where each event is equally likely,which i think fits this problem.

 

[hgfalling]

Here's what happens. The probability that Y takes on any particular value is actually zero, because the probability that X takes on any particular value is zero. What we can talk about is the probability that Y takes on a value in an interval.

So, for example, what is the probability that Y is on [0,1] in your problem? Clearly it's 1 because no matter what X is, Y will be on that interval. Now suppose we look at $x=\frac{1}{2}$. If $x=\frac{1}{2}$, then what is Y? Well, we can use the Pythagorean Theorem, and see that

$$y=\sqrt{1^2 - \left(\frac{1}{2}\right)^2} = \frac{\sqrt{3}}{2}$$.

Now it's pretty easy to see that if x is greater than $\frac{1}{2}$, then y is smaller than this, so we know that the probability that $Y < \frac{\sqrt{3}}{2}$ is the same as the probability that $x > \frac{1}{2}$, which is $\frac{1}{2}$. More generally, if we want to know the likelihood that Y is less than some particular value $y_0$, we have that this happens as long as
[tex]x > \sqrt{1-y_0^2} [/itex], which happens with probability 1 - that. So we have the following cumulative distribution function:<br /> <br /> $$P[Y < y] = 1 - \sqrt{1 - y^2}$$<br /> <br /> And the density function is just the derivative of that, which is:<br /> <br /> $$f(y) = \frac{y}{\sqrt{1 - y^2}}$$[/tex]