Finding $P(X+Y+Z+U\ge1)$ for $f(x,y,z,u)=\frac{24}{(1+x+y+z+u)^5}1_{x,y,z,u>0}$

Question

Let $f(x,y,z,u)=\frac{24}{(1+x+y+z+u)^5}, x>0,y>0,z>0,u>0$ be a density function. Find $P(X+Y+Z+U \ge 1)$ I have a problem about setting the limits of the integrals under this constraint

Answer 1

Hints:

• Compute $\Pr(X+Y+Z+U \ge 1) = 1-\Pr(X+Y+Z+U \le 1)$.

• For the limits, first consider $X$. Note that $X,Y,Z,U \ge 0$. Consequently, the probability of $X+Y+Z+U \le 1$ is zero when $X > 1$. So the range of $X$ is $0$ to $1$.

• Given the value of $X$, we determine the range of other variables. To this end, observe that $Y+Z+U \le 1-X$. Following the same argument, the range of $Y$ is $0$ to $1-X$.

I leave the calculation of the range of $Z$ and $U$ to you.

Answer 2

It is not difficult to find the distribution of $X+Y+Z+U$ itself.

Transform $(X,Y,Z,U)\mapsto (X_1,X_2,X_3,X_4)$ such that

\begin{align} X_1&=X+Y+Z+U \,, \\X_1X_2&=X+Y+Z \,, \\X_1X_2X_3&=X+Y \,, \\X_1X_2X_3X_4&=X \end{align}

Then $$x,y,z,u>0 \implies x_1>0\,,\,0<x_2,x_3,x_4<1$$

Jacobian determinant of the transformation can be calculated as $$J=-x_1^3x_2^2x_3$$

So joint pdf of $(X_1,X_2,X_3,X_4)$ is given by

\begin{align} f_{X_1,X_2,X_3,X_4}(x_1,x_2,x_3,x_4)&=\frac{24x_1^3x_2^2x_3}{(1+x_1)^5}\mathbf1_{0<x_1<\infty,0<x_2,x_3,x_4<1} \\&=\underbrace{\frac{4x_1^3}{(1+x_1)^5}\mathbf 1_{0<x_1<\infty}}_{f_{X_1}(x_1)}\cdot f_{X_2,X_3,X_4}(x_2,x_3,x_4) \end{align}

Evidently, $X_1$ is independent of $(X_2,X_3,X_4)$.

Now $X_1$ has a Beta distribution of second kind with parameter $(4,1)$, so that in terms of the Beta distribution of first kind: $$\frac{X_1}{1+X_1}\sim \mathsf{Beta}(4,1)$$

Therefore,

$$P(X_1\ge 1)=P\left(\frac{X_1}{1+X_1}\ge \frac12\right)=\int_{1/2}^1 4t^3\,dt=\frac{15}{16}$$