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
 A: 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!)
A: 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 $a$.
$$\int_0^a\,dx_4\int_0^a\,dx_3\int_0^a\,dx_2\int_{x_2+x_3+x_4}^a\,dx_1$$
A: 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 ...
