# Probability that the sum of the squares of 2 uniform random variables is less than 1?

1. Dec 12, 2011

### thisguy12

If you were to pick two random numbers on the interval [0,1], what is the probability that the sum of their squares is less than 1? That is, if you let $Y_1$ ~ $U(0,1)$ and $Y_2$ ~ $U(0,1)$, find $P(Y_1^2 + Y^2_2 \leq 1)$. There is also a hint: the substitution $u = 1 - y_1$ may be helpful - look for a beta distribution.

Here's what I've done so far:

I know that the density function for $Y_1$ and $Y_2$ is the same, $f(y_1) = f (y_2) = 1$ on the interval [0,1].

$P(Y_1^2 + Y_2^2 \leq 1) = P(Y_2^2 \leq 1 - Y_1^2) = P(-\sqrt{1 - Y_1^2} \leq Y_2 \leq \sqrt{1 - Y_1^2}) = \int^{\sqrt{1 - Y_1^2}}_{-\sqrt{1 - Y_1^2}} dy_2 = 2\sqrt{1 - Y_1^2}$

And that's where I get stuck. I thought that maybe be a beta distrbution with $\alpha = 3/2$, $\beta = 1$, but the beta function, $\beta(3/2, 1) = 2/3 \neq 1/2$.

2. Dec 13, 2011

### chiro

Re: Probability that the sum of the squares of 2 uniform random variables is less tha

Hey thisguy12 and welcome to the forums.

The way that comes to my mind is to find the PDF of the square of the uniform distribution U(0,1) and then use the convolution theorem to get the CDF for the sum of the two distributions since both are independent.

In terms of getting the PDF of the square of the uniform distribution, you can use a transformation theorem.

Also your PDF expression in the last line is wrong, since for one you can values that are greater than 1 which makes it invalid.

Do you know how to calculate the pdf of a transformed random variable? You don't necessarily need convolution per se for the second part, but its probably a good idea to be able to find the PDF/CDF of your transformed variable to go to the next step.

3. Dec 13, 2011

### mathman

Re: Probability that the sum of the squares of 2 uniform random variables is less tha

One easy way. The joint distribution of X and Y is uniform over the unit square.
The condition X2 + Y2 ≤ 1 is simply putting the variable pair inside a quarter circle of radius 1, so the probability is then the area = π/4.

Last edited: Dec 13, 2011
4. Dec 13, 2011

### thisguy12

Re: Probability that the sum of the squares of 2 uniform random variables is less tha

Okay so using the transformation method, I found the density functions for $U = Y^2_1$ and $W = Y^2_2$.

$h(Y) = U = Y^2_1$
$h^{-1}(U) = Y_1 = \sqrt{U}$
$\frac{d[h^{-1}(u)]}{du} = \frac{1}{2\sqrt{u}}$
$f_U(u) = f_{Y_1}(\sqrt{u}) \frac{1}{2\sqrt{u}} = 1(\frac{1}{2\sqrt{u}})=\frac{1}{2\sqrt{u}}$

Using the same method, I also get $f_W(w) = \frac{1}{2\sqrt{w}}$.

So now I am looking for $P(W + Y \leq 1)$. The joint distribution function of two independent random variables is the product of their two marginal density functions, $f(x_1, x_2) = f_{x_1}(x_1)f_{x_2}(x_2)$.

Thus, $f(u, w) = f_u(u)f_w(w) = (\frac{1}{2\sqrt{u}})(\frac{1}{2\sqrt{w}}) = \frac{1}{4\sqrt{uw}}$. So $P(W + Y \leq 1)$ = $\int^1_0 \int^{1-u}_0 f(u, w) dw du$.

Is this method correct? If so, this integral should give me the correct answer, right?

5. Dec 13, 2011

### chiro

Re: Probability that the sum of the squares of 2 uniform random variables is less tha

That looks pretty good to me.

It would be nice though to get another opinion on this from an experienced member just to be sure.

6. Dec 14, 2011

### mathman

Re: Probability that the sum of the squares of 2 uniform random variables is less tha

If you get π/4 from the double integral, then it is right, although it looks like a difficult way to get an easy answer.