1. ## Gamma Distribution

The radii of craters are exponentially distributed with mean 10 feet. Find the mean and variance of the areas.

I know I'm going to have to use an integral, but how do I know what beta will be? And where will the mean come into play in the integral?

Thanks!

2. Originally Posted by fulltwist8
The radii of craters are exponentially distributed with mean 10 feet. Find the mean and variance of the areas.

I know I'm going to have to use an integral, but how do I know what beta will be? And where will the mean come into play in the integral?

Thanks!
For starters, the pdf of the radius is given here and it should be clear how the mean comes into play .....

Now you need to think about the mean and the standard deviation corresponding to the pdf for $\pi r^2 = (\sqrt{\pi} r)^2$ .....

3. still confused... and now there's a lambda involved?

4. Originally Posted by fulltwist8
still confused... and now there's a lambda involved?
$\lambda = \frac{1}{\mu}$.

$\mu = 10 \Rightarrow \lambda = \frac{1}{10}$.

Therefore the pdf for the radius is $f(r) = \frac{1}{10}\, e^{-r/10}$.

5. Originally Posted by mr fantastic
$\lambda = \frac{1}{\mu}$.

$\mu = 10 \Rightarrow \lambda = \frac{1}{10}$.

Therefore the pdf for the radius is $f(r) = \frac{1}{10}\, e^{-r/10}$.
This is probably just an academic exercise now ....... Anyway, for what it's worth, here's one way of getting the probability distribution function of $R^2$:

Note: Once the pdf is known, the mean and variance of $R^2$ can (in principle) be calculated.

Let $Y = R^2$ and let the pdf of Y be g(y).

Let $G(y) = \Pr(Y \leq y)$. By definition, $g(y) = \frac{dG}{dy}$.

$Y < y \Rightarrow R^2 < y \Rightarrow -\sqrt{y} < R < \sqrt{y}$.

Therefore:

$G(y) = \Pr(Y \leq y) = \Pr(-\sqrt{y} < R < \sqrt{y}) = \int_{-\sqrt{y}}^{\sqrt{y}} f(r) \, dr$.

But $f(r) = \frac{1}{10}\, e^{-r/10}$ for $r \geq 0$ and zero elsewhere.

Therefore:

$G(y) = \Pr(Y \leq y) = \int_{0}^{\sqrt{y}} \frac{1}{10}\, e^{-r/10} \, dr = -\left[ e^{-r/10}\right]_{0}^{\sqrt{y}} = 1 - e^{-\sqrt{y}/10}$.

Therefore:

$g(y) = \frac{dG}{dy} = \frac{1}{20} \, \frac{1}{\sqrt{y}} \, e^{-\sqrt{y}/10}$.

$Y = R^2$ and $r \geq 0 \Rightarrow y \geq 0$.

Therefore:

${\color{red}g(y) = \frac{1}{20} \, \frac{1}{\sqrt{y}} \, e^{-\sqrt{y}/10}}$ for ${\color{red}y > 0}$ and zero otherwise

since $\frac{1}{\sqrt{y}} \, e^{-\sqrt{y}/10}$ is undefined for y = 0.

$E(Y) = E(R^2) = \int_{-\infty}^{\infty} y \, g(y) \, dy = \frac{1}{20} \, \int_{0}^{\infty} \sqrt{y} \, e^{-\sqrt{y}/10} \, dy = 200$.

$E(Y^2) = \int_{-\infty}^{\infty} y^2 \, g(y) \, dy = \frac{1}{20} \, \int_{0}^{\infty} y^{3/2} \, e^{-\sqrt{y}/10} \, dy = 240000$.

$Var[Y] = Var[R^2] = E[Y^2] - (E[Y])^2 = 240000 - 40000 = 200000$.

$sd_Y = 447$, which can't be right so there's probably a small mistake somewhere ....

Then $E[\pi R^2] = \pi E[R^2]$ and $Var[\pi R^2] = \pi^2 Var[R^2]$ using the usual formula.