Conditional probability of triple joint density

Question

The joint density function of X,Y and Z is $$f(x,y,z) = \begin{cases}6(z-y)& \text{for } 0<x<1, \; 0<y<z<1,\\ 0 &\text{otherwise} \end{cases}$$

So finding the probability $P(X + Y + Z \le 1)$ gives us $$P(X + Y + Z \le 1) = \int_0^1 \int_0^{1-x} \int_0^{1-x-y} 6(z-y) \ \mathrm dz \ \mathrm dy \ \mathrm dx$$

For a conditional density $P(X \le Z | X + Y + Z \le 1)$ this is equal to $$\frac{P(X \le Z , X + Y + Z \le 1)}{P(X + Y + Z \le 1)}$$

I am unsure what bounds of integration to use for the top part of this.

Answer 1

Since $y < z$, your denominator could be $$P(X + Y + Z \le 1) = \int_{x=0}^1 \int_{z=0}^{1-x} \int_{y=0}^{\min(z,1-x-z)} 6(z-y) \ \mathrm dy \ \mathrm dz \ \mathrm dx $$ or $$P(X + Y + Z \le 1) = \int_{z=0}^{1} \int_{y=0}^{\min(z,1-z)} \int_{x=0}^{1-y-z} 6(z-y) \ \mathrm dx \ \mathrm dy \ \mathrm dz $$

Your numerator could then be $$P(X\le Z, X + Y + Z \le 1) = \int_{z=0}^{1} \int_{y=0}^{\min(z,1-z)} \int_{x=0}^{\min(z,1-y-z)} 6(z-y) \ \mathrm dx \ \mathrm dy \ \mathrm dz $$

Answer 2

i) Bounds for $P(X + Y + Z \le 1)$

We also have $0 \le y \le z, 0 \le z \le 1, 0 \le x \le 1$.

Intersection of $y = z$ and $x+y+z = 1$ is given by $\displaystyle y = \frac{1-x}{2}$

So the bounds should be,

$y \leq z \leq (1-x-y), 0 \leq y \leq \frac{1-x}{2}, 0 \leq x \leq 1$. The integral will be,

$\displaystyle \int_0^{1} \int_0^{(1-x)/2} \int_y^{1-x-y} 6(z-y) \ dz \ dy \ dx$

ii) Now for second integral we have additional condition of $x \le z$. We will have to split the integral into two - for $y \ge x$ and for $x \ge y$.

Intersection of $x = z, y = z, x + y + z = 1$ is $x = y = z = \displaystyle \frac{1}{3}$

For $y \ge x$,

Upper bound of $y$ is intersection of $x+y+z = 1$ and $y = z$. So the integral is,

$I_1 = \displaystyle \int_0^{1/3} \int_{x}^{(1-x)/2} \int_y^{1-x-y} 6(z-y) \ dz \ dy \ dx$

For $x \ge y$,

Upper bound of $x$ is intersection of $x+y+z = 1$ and $x = z$. So the integral is,

$I_2 = \displaystyle \int_0^{1/3} \int_{y}^{(1-y)/2} \int_x^{1-x-y} 6(z-y) \ dz \ dx \ dy$