# Unit circle derived distribution

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

wabbit
Gold Member
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.

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

wabbit
Gold Member
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:
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

wabbit
Gold Member
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.

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

wabbit
Gold Member
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...

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} ) ## ?

wabbit
Gold Member
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.

HallsofIvy
Homework Helper
Hi

Assume an x-coordinate from the unit circle is picked from a uniform distribution.
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 $2\pi$.

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

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

ChrisVer
Gold Member
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?*\

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