1
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ 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 $\Pr(X_1 \gt X_2+X_3 +X_4)$ $\endgroup$ – Henry Jun 25 '17 at 13:20
1
$\begingroup$

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!)

$\endgroup$
0
$\begingroup$

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 $a$.

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

$\endgroup$
  • $\begingroup$ My knowledge about 3D(or more) shapes is limited. Is there a way to solve it with integrals? thanks $\endgroup$ – Yoav Cohen Jun 25 '17 at 13:09
  • $\begingroup$ @YoavCohen Yes, you could calculate it using integration for the 4D shape. $\endgroup$ – Yujie Zha Jun 25 '17 at 13:12
0
$\begingroup$

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 ...

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.