Question

Normalizing a 3D probability distribution

Answer 1

Let X, Y, and Z have the joint probability density function:\[f(x,y,z)= kxy^2z\]Where 0<x, y<1, 0<z<2 , and \[f(x,y,z)=0\]elsewhere. I need to find k such that it's a valid probability distribution function. Here's what I tried:\[k^{-1} = \lim_{a \rightarrow -\infty} \lim_{b \rightarrow \infty} \int\limits_{0}^{2}\int\limits_{a}^{1}\int\limits_{0}^{b}(xy^2z) dx dy dz = \lim_{a \rightarrow -\infty} \lim_{b \rightarrow \infty} \frac{b^2(1-a^3)}{3}\]But this is divergent. What am I doing wrong?

Answer 2

Yikes, thats a painful looking problem lol

Answer 3

I looked at the solutions manual and they used 0 for a and 1 for b, giving an answer of 3 for k. But I don't understand why they used 0 and 1.

Answer 4

Are both \(x\) and \(y\) between 0 and 1, or do you mean \(x>0\) and \(y<1\) individually?

Answer 5

I think you're just misinterpreting the notation. It looks like the question is meant to say
\[f(x,y,z)=\begin{cases}kxy^2z&\text{for }0<x<1,~0<y<1,~0<z<2\\0&\text{elsewhere}\end{cases}\]
and so determining \(k\) is simply a matter of computing the rather simple integral,
\[\int_0^2\int_0^1\int_0^1kxy^2z~dx~dy~dz=1\]

Answer 6

as pointed out above the 0<x, y<1 is 0<x<1 and 0<y<1