probability that $x_1>x_2+x_3+x_4$ for uniform distribution $[0,a]$

How do i found what is the probability for $4$ random variables $x_1,x_2,x_3,x_4$ which are uniformly distributed between $[0,a]$ to exists the following: $x_1>x_2+x_3+x_4$ I tried to find the density function of $y = x_2+x_3+x_4$ but Im not sure its the right direction or how should I do it.

thanks

• Hint: Try to show $\Pr(X_2+X_3 \le y) = \dfrac{y^2}{2a^2}$ when $0 \le y \le a$. Use a similar method to find $\Pr(X_2+X_3 +X_4 \le z)$ when $0 \le z \le a$, and then a related method to find $\Pr(X_1 \gt X_2+X_3 +X_4)$ Jun 25, 2017 at 13:20

Let us denote $X=X_1$, $Y=X_2+X_3+X_4$. Using the law of total probability

\begin{align} P(Y<X)&=\int_0^a P(Y<x)f_X(x)dx\\ &=\frac{1}{a}\int_0^a P(Y<x) dx\\ &=\frac{1}{a}\int_0^a \left(\int_0^x f_Y(y) dy\right)dx\\ \end{align}

So finding the PDF $f_Y(y)$ is the right way to solve the problem. Since $X<a$, you need this function in the interval $[0,a]$ only; to find it you can use convolution twice.

I guess the first convolution (for $Z=X_2+X_3$) should be $f_Z(z)=\frac{z}{a^2} \text{ for } 0\lt z \lt a$, the second one (for $Y=Z+X_4$) should be $f_Y(y)=\frac{y^2}{2a^3}\text{ for } 0\lt y \lt a$ (check it!)

HINT

Assume the four variables are independent. And you could get the probability as:

The volume confined by the hyper-plane of $x_1=x_2+x_3+x_4$ (and upwards w.r.t $x_1$ axis) and the hyper-cube with sides of $a$.

$$\int_0^a\,dx_4\int_0^a\,dx_3\int_0^a\,dx_2\int_{x_2+x_3+x_4}^a\,dx_1$$

• My knowledge about 3D(or more) shapes is limited. Is there a way to solve it with integrals? thanks Jun 25, 2017 at 13:09
• @YoavCohen Yes, you could calculate it using integration for the 4D shape. Jun 25, 2017 at 13:12

The point $(x_1,\,x_2,\,x_3,\,x_4)$ lies within a 4D cube of side $a$.

Fixing a value for $x_1=r$ corresponds to individuate a 3D subcube of side $a$.

$x_2+x_3+x_4<r$ represents the points lying in the 3D cube and below the diagonal plane $x_2+x_3+x_4=r$. Since $0 \le r=x_1 \le a$, the plane is contained inside the cube.

The volume of the resulting right tetrahedron is $1/3r1/2r^2$.

Integrating this for $r=0..a$ and dividing by $a^4$, you can conclude that ...