Question

The joint density function for random variables x, y, and z is f(x, y, z)= cxyz if 0 ≤ x ≤ 1, 0 ≤ y ≤ 2, 0 ≤ z ≤ 2, and f(x, y, z) = 0 otherwise. (a) find the value of the constant
c. (b) find p(x ≤ 1, y ≤ 1, z ≤ 1). (c) find p(x + y + z ≤ 1).

Answer 1

In order for $f(x,y,z)$ to be a proper density function, we need to have

$\displaystyle\int_{\mathcal S}f(x,y,z)\,\mathrm dx\,\mathrm dy\,\mathrm dz=1$

where $\mathcal S$ is the support of the joint distribution. Here $\mathcal S$ is the box with dimensions 1x2x2. We have

$\displaystyle\int_{\mathcal S}f(x,y,z)\,\mathrm dx\,\mathrm dy\,\mathrm dz=c\int_{x=0}^{x=1}x\,\mathrm dx\int_{y=0}^{y=2}y\,\mathrm dy\int_{z=0}^{z=2}z\,\mathrm dz$

$=\dfrac{2^2c}2=1\implies c=\dfrac12$

Now since we can write $f_{X,Y,Z}(x,y,z)$ as

$\dfrac{xyz}2=\dfrac x{c_x}\cdot\dfrac y{c_y}\cdot\dfrac z{c_z}$

(where $c_xc_yc_z=2$) we can assign each factor to be the PDF for the corresponding random variable, i.e.

$f_{X,Y,Z}(x,y,z)=f_X(x)\cdot f_Y(y)\cdot f_Z(z)$

which means that $X,Y,Z$ are independent of one another. Then

$P(X\le1,Y\le1,Z\le1)=P(X\le1)\cdot P(Y\le1)\cdot P(Z\le1)$

We can derive the marginal distributions in the same way we solve for $c$ above:

$\displaystyle\int_{x=0}^{x=1}\frac x{c_x}\,\mathrm dx=1\implies c_x=\frac12$

$\implies f_X(x)=\begin{cases}2x&\text{for }0\le x\le1\\0&\text{otherwise}\end{cases}$

$\displaystyle\int_{y=0}^{y=2}\frac y{c_y}\,\mathrm dz=\int_{z=0}^{z=2}\frac z{c_z}\,\mathrm dz=1\implies c_y=c_z=2$

$\implies f_Y(y)=\begin{cases}\dfrac y2&\text{for }0\le y\le2\\\\0&\text{otherwise}\end{cases}$

with $f_Z(z)$ defined similarly, so that

$P(X\le 1,Y\le1,Z\le1)=1\cdot\dfrac14\cdot\dfrac14=\dfrac18$

For the final probability, we condition $Z$ on the event that $X=x$ and $Y=y$:

$P(X+Y+Z\le1)=P(Z\le1-X-Y)$

$=\displaystyle\iint_{(x,y)}P(Z$

$=\displaystyle\int_{x=0}^{x=1}\int_{y=0}^{y=2}F_Z(1-x-y)f_X(x)f_Y(y)\,\mathrm dy\,\mathrm dx$

The CDF of $Z$ is easy to find:

$F_Z(z)=\begin{cases}0&\text{for }z$

So we have

$P(X+Y+Z\le1)=\displaystyle\int_{x=0}^{x=1}\int_{y=0}^{y=2}\frac{(1-x-y)^2}4\cdot2x\cdot\frac y2\,\mathrm dx\,\mathrm dy=\frac{23}{72}$