How to find density of one random variable if joint pdf contains $3$ random variables?

Question

My book asks me the following question:

Let $X, Y, Z$ have joint density $f(x, y, z) = 6$, for $0 < x < y < z < 1$ and $f(x, y, z) = 0$ otherwise.

a. Are $X$ and $Y$ and $Z$ independent?

I wrote no because it doesn't seem like the region of $0 < x < y < z < 1$ is rectangular so they aren't independent.

b. Find the density $f(x)$ of X.

c. Find the density $f (y)$ of Y .

d. Find the density $f(z)$ of Z

I need help finding the density of one random variable for example $f(x)$ if the joint pdf of three random variables are given.

Can someone give me the formula or methodology to do so? If I can figure out how to do one, I can definitely do the rest.

Thank you for any hints!

Answer 1

The power density function of $X$ $f_X$ is given by :

$$ f_X(x) = \int_{(y,z)\in \mathbb{R}^2} f(x,y,z)dydz$$

By Fubini-Tonelli, you can integrate either w.r.t. $y$ or $z$ first, so you get

$$ f_X(x) = \int_{y\in \mathbb{R}} \int_{z\in \mathbb{R}} f(x,y,z)dzdy = \int_{y\in \mathbb{R}} \int_{z\in \mathbb{R}} 6 \times \textbf{1}_{\{0 < x < y < z < 1\}}dzdy $$

Using the definition of $f$, only taking into account the regions of the plane where it is non-zero :

$$ f_X(x) = \int_{y = x}^1 \int_{z = y}^1 6dzdy = \int_{y = x}^1 6 \times (1-y)dy$$

And finally you get

$$ f_X(x) = 3 \times (1 - x)^2 $$ (you can check that this is indeed a density function on $[0,1]$)

Now you have to proceed in a similar way to compute $f_Y$ and $f_Z$. Hope that helps.