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 '19 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 '19 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). $\endgroup$ – martian03 Jan 28 '19 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 '19 at 4:56
  • $\begingroup$ Hey, @LeeDavidChungLin, I cleared the definition of distribution. Can you help me? $\endgroup$ – martian03 Jan 30 '19 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? $\endgroup$ – martian03 Feb 3 '19 at 1:19
  • $\begingroup$ @Matt_F and may I ask you the input formula of Mathematica? $\endgroup$ – martian03 Feb 3 '19 at 11:54
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    $\begingroup$ That’s $\chi$ as the indicator function (en.m.wikipedia.org/wiki/Indicator_function), represented in Mathematica as $\mathtt{Boole}$. $\endgroup$ – user210229 Feb 3 '19 at 17:55
  • $\begingroup$ Thank you @Matt_F $\endgroup$ – martian03 Feb 4 '19 at 1:49

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