# Math Help - Uniform Distribution

1. ## Uniform Distribution

A circle of radius r has area A=(pi)r^2. If random circle has a radius that is uniformly distributed over the interval (0,1). What are the mean and variance of the area of the circle?

Sol.

First I integrated (pi)r^2 from 0 to 1 and got pi/3. How do we find mean and variance now? thanks

2. Originally Posted by somestudent2
A circle of radius r has area A=(pi)r^2. If random circle has a radius that is uniformly distributed over the interval (0,1). What are the mean and variance of the area of the circle?

Sol.

First I integrated (pi)r^2 from 0 to 1 and got pi/3. How do we find mean and variance now? thanks
Find the mean and variance of the radius first.

3. the mean is 1/2 and variance 1/12 for radius. Do we just substitute that in A=(pi)r^2?

4. Originally Posted by somestudent2
the mean is 1/2 and variance 1/12 for radius. Do we just substitute that in A=(pi)r^2?
I believe so, but don't hold me to that answer just yet...

5. You REALLY should go through it once.

If $Y = \pi r^{2}$

And if r is Uniform on [0,1]

Then f(r) = 1 on [0,1]

Y is defined on $[0,\pi]$

$G(Y) = Pr(Y \leq y) = Pr(\pi r^{2} \leq y) = Pr\left(-\sqrt{\frac{y}{\pi}} \leq r \leq \sqrt{\frac{y}{\pi}}\right) = \int_{-\sqrt{\frac{y}{\pi}}}^{\sqrt{\frac{y}{\pi}}}f(x) dx$

$G(y) = \int_{0}^{\sqrt{\frac{y}{\pi}}}f(x) dx = \int_{0}^{\sqrt{\frac{y}{\pi}}} 1 dx = \sqrt{\frac{y}{\pi}}$

Thus $g(y) = G'(y) = \frac{1}{2\sqrt{\pi}\sqrt{y}}$ on $[0,\pi]$

Checking:

$\int_{0}^{\pi}\frac{1}{2\sqrt{\pi}\sqrt{y}} dy = 1$ -- Good

$E[Y] = \int_{0}^{\pi}\frac{y}{2\sqrt{\pi}\sqrt{y}} dy = \frac{1}{3}\pi$

$E[Y^{2}] = \int_{0}^{\pi}\frac{y^{2}}{2\sqrt{\pi}\sqrt{y}} dy = \frac{1}{5}\pi^{2}$

$Var(y) = \frac{\pi^{2}}{5} - \left(\frac{\pi}{3}\right)^{2} = \frac{4}{45}\pi^{2}$

Of course, if you were paying attention to all that, it may suggest to you the following:

$E[\pi r^{2}] = \int_{0}^{1}\pi r^{2} dr = \frac{1}{3} \pi$

$E\left[\left(\pi r^{2}\right)^{2}\right] = \int_{0}^{1}\left(\pi r^{2}\right)^{2} dr = \frac{1}{5} \pi^{2}$

Those should look familiar.