Suppose a fly gets trapped in your tent. Let the tent be represented by the polyhedron P with vertices (0, 0, 0), (0, 1, 0), (1, 1, 0), (1, 0, 0), (0, 1/2, 1), (1, 1/2, 1) and let the position of the fly be represented by the random vector (X, Y, Z). Assume that the position of the fly is uniformly randomly distributed within P.
- Compute the marginal p.d.f. of X
- Compute the marginal p.d.f. of Y
- Compute the marginal p.d.f. of Z
- Are X, Y, Z independent?
- Are X and Y independent?
- Are X and Z independent?
- Are Y and Z independent?
My attempt
In order to calculate the marginal p.d.f.s, we first need to calculate the joint p.d.f. Since it is uniformly distributed, we know that it is just $\frac{1}{volume}$. The volume of our polyhedron is $\frac{1}{2}$ so:
\begin{equation}
f_{X,Y,Z}(x,y,z)=
\begin{cases}
2 & \text{if } (x,y,z)\in D\\
0 & \text{otherwise}
\end{cases}
\end{equation}
For 1., to compute the marginal p.d.f. for X, we need to integrate in terms of $y,z$
so
$$f_X(x)=\int_0^{1/2}\int_0^{2y}2dzdy+\int_{1/2}^1\int_0^{-2y}2dzdy=-1$$
For 2., we integrate in terms of $x,z$ so
$$f_Y(y)=\int^1_0\int^1_02dxdz=2$$
For 3., we integrate in terms of $x,y$, so
$$f_Z(z)=\int^1_0\int^1_02dxdy=2$$
I am unsure about the bounds that I found for these, especially for number 1 becuse I got a negative number.
For independence, we just need to check if the joint p.d.f. is the product of the marginal p.d.f.s.
Best Answer
Please note the second integral in calculation of $f_X(x)$ is not correct.
It should instead be $\displaystyle \int_{1/2}^1 \int_0^{2-2y} 2 \ dz \ dy$.
Instead you can also calculate using single integral if you integrate with respect to $dy$ first.
$\displaystyle f_X(x) = \int_0^1 \int_{z/2}^{(2-z)/2} 2 \ dy \ dz = 1$
Also, there are mistakes in calculations of $f_Y(y)$ and $f_Z(z)$.
$\displaystyle f_Y(y) = \int_0^{(1-|1-2y|)}\int^1_0 2 \ dx \ dz = 2 - 2 |1-2y|$
$\displaystyle f_Z(z) = \int_{z/2}^{(2-z)/2}\int_0^1 2 \ dx \ dy = 2 - 2z$