# Unit circle derived distribution

1. May 2, 2015

### rabbed

Hi

Assume an x-coordinate from the unit circle is picked from a uniform distribution.
This is the outcome of the random variable X with probability density function Xden(x) = 0.5 (-1 < x < 1).
The random variable Y is related to the random variable X by Y = f(X) = √(1-X2) and X = g(Y) = ±√(1-Y2).
What is the probability density function Yden(y)?

Since Yden(y) = Xden(g(y)) / |f'(g(y))|, i first calculate f'(x) = -x/√(1-x2).
Then Yden(y) = Xden( √(1-Y2) ) / f'( √(1-Y2) ) + Xden( -√(1-Y2) ) / f'( -√(1-Y2) )

I get Yden(y) = 0

Where do i go wrong?

Rgds
Rabbed

2. May 2, 2015

### wabbit

Your last formula looks (almost) correct but the result is not zero. What are the values of the Xden(..) in that formula and where did the absolute values go ? Also, do compute f'(g(y)), it will simplify.

As an aside, it's a personnal quirk perhaps but I never use the density formulas, it's easy to mishandle them - using cumulative distribution functions instead, you cannot go wrong.

3. May 2, 2015

### rabbed

Hey

I checked my calculation again (in particular the absolute values) and came up with this:
Yden(y) = √y2 / √(1-y2)
Don't know if that's correct.. i'll try do it the other way around, starting with Yden(y) to get Xden(x) = 0.5

How do you do this using CDF instead?

Rgds
Rabbed

4. May 2, 2015

### wabbit

That looks ok to me, you might replace $\sqrt{y^2}$ with just $y$ though : )

With the CDF $P(Y\leq y)=P(\sqrt{1-X^2}\leq y)=P(|X|\geq\sqrt{1-y^2})=1-\sqrt{1-y^2}$, differentiate and you get your result.
Of more general applicability, differentiate $P(|X|\geq\sqrt{1-y^2})$, this gives you the Y-PDF from the X-PDF without needing the actual X-CDF.

Same thing really - for me it's harder to miss a step this way, I just find it more concrete but it's really a matter of taste, and to be perfectly homest, I'm just too lazy to memorize the PDF formula.

Last edited: May 2, 2015
5. May 4, 2015

### rabbed

Ok, i'll check out that technique, thanks.

So what does the answer tell me, that if i pick an x-coordinate at random between -1 and 1 (with all having equal probability), it's very probable that it's corresponding y-coordinate (on the upper half of the unit circle) is close to 1, and very improbable that the y-coordinate is close to 0?

Rgds
Rabbed

6. May 4, 2015

### wabbit

Something like that, yes, depending how improbable is very improbable :)
This relates to the fact that the half-circle has a horizontal tangent at x=0 and a vertical one at x=1.

7. May 29, 2015

### rabbed

Hi again...

Is this a correct usage of the CDF method?

Y_den(y) =
Y_cdf'(y) =
( P( Y < y ) )' =
( P( f(X) < y ) )' =
( P( sqrt(1-X^2) < y ) )' =
( P( 1-X^2 < y^2 ) )' =
( P( -1+X^2 > -y^2 ) )' =
( P( X^2 > 1-y^2 ) )' =
( P( X > sqrt(1-y^2) ) )' =
( 1 - X_cdf(sqrt(1-y^2) ) )' =
(-X_cdf'(sqrt(1-y^2))) * (-y/sqrt(1-y^2)) =
(-X_pdf(sqrt(1-y^2))) * (-y/sqrt(1-y^2)) =
(-0.5) * (-y/sqrt(1-y^2)) =
0.5*y/sqrt(1-y^2)

Rgds
Rabbed

8. May 29, 2015

### wabbit

Almost. You are off a factor of 2 because of your step $( P( X^2 > 1-y^2 ) )' = ( P( X > \sqrt{1-y^2} ) )'$. Looks like you forgot something here : )

Oh and by the way, do try the tex formatting, it's not difficult (there's a guide in the help/howto section too), and it makes equations much easier to read...

9. May 30, 2015

### rabbed

Do you mean this?
$( P( X^2 > 1-y^2 ) )' = ( P( X > \pm\sqrt{1-y^2} ) )' = ( P( (X > -\sqrt{1-y^2}) OR (X > \sqrt{1-y^2}) ) )' =$
And then i can use the OR/+ formula to continue?

Shouldn't $P( (X > -\sqrt{1-y^2}) OR (X > \sqrt{1-y^2}) ) = P( X > -\sqrt{1-y^2} )$ ?

10. May 30, 2015

### wabbit

Yes the OR was sthe missing part - but it's easy to handle here : from the symmetry of the distribution of X, $P(|X|>x)=2P(X>x)$

Note : it is true of course that in general $P(|X|>x)=P(X>x)+P(X<-x)$ but in this case the two terms are equal, hence the simplification.

11. May 30, 2015

### HallsofIvy

Staff Emeritus
This seems to me to be ambiguous. Yes, the x coordinates of points on the circle all lie between -1 and 1. But does this mean that x is uniformly distributed on -1 to 1, as you are assuming here, or does it mean that x is x coordinate of a point on the unit circle and the points are uniformly distributed?
Those are not at all the same. In the latter case, X is $cos(\theta)$ with $\theta$ uniformly distributed between $0$ and $2\pi$.

12. Jun 13, 2015

### rabbed

Thanks, i've got the same answer with the CDF method now.

HallsofIvy:
Yes, i mean the x-coordinate should be uniformly distributed, not the angle. But I do want to check that also later, to see the difference.

My next concern is how i get from:
Y_pdf(y) = y/√(1-y2)
X = g(Y) = ± √(1-Y2)
to:
X_pdf(x) = 0.5

I think i go wrong here:
( P( ± √(1-Y2) < x ) )' =
( P( -√(1-Y2) < x OR √(1-Y2) < x ) )' =
( P( √(1-Y2) > -x OR √(1-Y2) < x ) )' =
( P( 1-Y2 > x2 OR 1-Y2 < x2 ) )'

Any tips? :)

Rgds
Rabbed

Last edited: Jun 13, 2015
13. Jun 13, 2015

### ChrisVer

Your inequalities are wrong... $x^2>f(y) \Rightarrow x < -\sqrt{f(y)} ~~\text{or}~~ x> + \sqrt{f(y)}$

14. Jun 13, 2015

### ChrisVer

Also when you have something like: $\pm \sqrt{1+Y^2} \le x$ it doesn't really make sense...

edit \*what happened with the square roots and latex?*\

15. Jun 19, 2015

### rabbed

Thanks,

But how do you go from $$X = g(Y) = ± \sqrt{1-Y^2}$$
to insert into the CDF definition $$P(X < x)$$

Should i start by squaring them both, so that i can insert
$$X^2 = g(Y)^2 = (± \sqrt{1-Y^2})^2 = |1-Y^2|$$
into
$$P(X^2 < x^2)$$
and get
$$P(x^2 > |1-Y^2|) = P(x < -\sqrt{|1-Y^2|}\ OR\ x > \sqrt{|1-Y^2|})$$
?

Rgds
Rabbed

16. Jul 24, 2015