Conditional distribution of sum of squared iid uniforms

Question

Is it true, that if $U_{1}$ and $U_{2}$ are iid uniform distributed variables on $\left[-1,1\right]$, then the sum of $U_{1}^{2}$ and $U_{2}^{2}$ is still uniform distributed conditioned on the set that this sum is not greater than $1$? Other words:

$$X\dot{=}U_{1}^{2}+U_{2}^{2}|U_{1}^{2}+U_{2}^{2}\leq1\sim Uni\left[0,1\right]?$$

I've read this in a book and for first glance this hasn't been clear. I guess we have to compare some kind of “conditional charachteristic” functions to decide this question.

Answer 1

Yes, it is true. Let $U_1,U_2\sim U(0,1),Y_1 = U_1^2$, then $$f_Y(y_1)dy_1=f_U(u_1)du_1$$ $$f_Y(y_1)=f_U(u_1)\frac{du_1}{dy_1}=1\cdot \frac{1}{\frac{dy_1}{du_1}}=\frac{1}{2u_1}=\frac{1}{2\sqrt{y_1}},\quad 0< y_1\le 1$$

The density of $X=U_1^2+U_2^2$ is the convolution of $f_Y(y_1)$ with itself, and Wolfram alpha gives this result, from which we can see that the density $f_X(x)$ is constant when $0\le x\le 1$.

$$f_X(x|x\le 1)=\frac{f_X(x)\cdot \mathbf{1}_{[0,1]}(x)}{P(x\le 1)}=\mathbf{1}_{[0,1]}(x)$$

Note: I would like to add that it doesn't matter whether $U_1$ is uniform on $[0,1]$ or $[−1,1]$. If $U_1$ is uniform on $[−1,1]$, then $|U_1|$ is uniform on $[0,1]$, so $U_1^2=|U_1|^2$ still has the same distribution.

Answer 2

A geometric proof: after conditioning, $(U_1, U_2)$ is a point chosen uniformly at random from the unit disk.

So for $0 < x < 1$, the probability $P(U_1^2 + U_2^2 < x)$ is the area of a disk of radius $\sqrt{x}$ divided by the area of the unit disk.

That's $\pi (\sqrt{x})^2/\pi = x$, giving the result.