Let $X_1, X_2, Y_1, Y_2, Z_1, Z_2$ be non-negative random variables which have following constraints.



$(X_i,Y_i,Z_i)$ is uniformly distributed. It means, the point of $(X_i,Y_i,Z_i)$ is uniformly distributed on the surface of the triangle with the vertex of (1,0,0)(0,1,0)(0,0,1).

Surely $(X_1,Y_1,Z_1)$ and $(X_2,Y_2,Z_2)$ are independent.

let: $X=X_1 + X_2 $, $Y=Y_1 + Y_2 $, $Z=Z_1 + Z_2 $

In this case, the probability of $\{X$ is bigger than $Y$ and $Z$ both$\}$ would be $\dfrac{1}{3}$.

My question is:

What is the probability of "$c+X$ is bigger than $Y$ and $Z$" when $c$ is a constant"?

For example: what is $\mathbb{P}\left[0.2+X >\max\{Y,Z\}\right]$?

  • 1
    $\begingroup$ The question needs some information about the distribution and dependence of the random variables involved. $\endgroup$ – herb steinberg Jan 28 at 4:11
  • $\begingroup$ As you've noticed, the joint distribution of $(X_j,Y_j,Z_j)$ is concentrated on the triangle with vertices $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$. Can you specify this distribution (e.g. uniform)? Are $(X_1,Y_1,Z_1)$ and $(X_2,Y_2,Z_2)$ i.i.d? $\endgroup$ – d.k.o. Jan 28 at 4:22
  • $\begingroup$ @d.k.o. Yes. the distribution is uniform on the area of the triangle of (1,0,0) , (0,1,0), and (0,0,1). $\endgroup$ – HoCheol SHIN Jan 28 at 4:40
  • $\begingroup$ OK. Then you may find the density of the sum $(X_1,Y_1,Z_1)+(X_2,Y_2,Z_2)$. Assuming independence, I believe it is a pyramid whose base is the triangle with vertices $(2,0,0)$, $(0,2,0)$, and $(0,0,2)$. $\endgroup$ – d.k.o. Jan 28 at 4:56
  • $\begingroup$ Hey, @LeeDavidChungLin, I cleared the definition of distribution. Can you help me? $\endgroup$ – HoCheol SHIN Jan 30 at 5:19

Letting $f(y_1,z_1,y_2,z_2)={\large\chi}_{\ c+(1-y_1-z_1)+(1-y_2-z_2)\ > \ \max(y_1+y_2,z_1+z_2)}\ $, the probability is $$ 4 \int_{y_1=0}^1\int_{z_1=0}^{1-y_1} \int_{y_2=0}^1\int_{z_2=0}^{1-y_2} f(y_1,z_1,y_2,z_2)\ dy_1\ dz_1 \ dy_2 \ dz_2 $$ The $4$ arises because the probability measures are $2\ dy_1 \ dz_1$ and $2\ dy_2 \ dz_2$ to make $\int_{y_1=0}^1\int_{z_1=0}^{1-y_1} 2\ dy_1 \ dz_1= \int_{y_2=0}^1\int_{z_2=0}^{1-y_2} 2\ dy_2 \ dz_2=1$.

According to Mathematica, the result is $$\frac{36+64c+24c^2-24c^3-c^4}{108}\ \text{ if }\ 0<c<1$$ and for $c=1/5$ this is $30979/67500$.

  • $\begingroup$ @Matt_F. What is χ ? Is it the Chi in the Chi-squared distribution? $\endgroup$ – HoCheol SHIN Feb 3 at 1:19
  • $\begingroup$ @Matt_F and may I ask you the input formula of Mathematica? $\endgroup$ – HoCheol SHIN Feb 3 at 11:54
  • 1
    $\begingroup$ That’s $\chi$ as the indicator function (en.m.wikipedia.org/wiki/Indicator_function), represented in Mathematica as $\mathtt{Boole}$. $\endgroup$ – user210229 Feb 3 at 17:55
  • $\begingroup$ Thank you @Matt_F $\endgroup$ – HoCheol SHIN Feb 4 at 1:49

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.