# sum of random variables that have constraints

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

$$X_1+Y_1+Z_1=1$$

$$X_2+Y_2+Z_2=1$$

$$(X_i,Y_i,Z_i)$$ 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 $$\mathbb{P}\left[0.2+X >\max\{Y,Z\}\right]$$?

• The question needs some information about the distribution and dependence of the random variables involved. – herb steinberg Jan 28 at 4:11
• 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? – d.k.o. Jan 28 at 4:22
• @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). – HoCheol SHIN Jan 28 at 4:40
• 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 $(0,0,2)$. – d.k.o. Jan 28 at 4:56
• Hey, @LeeDavidChungLin, I cleared the definition of distribution. Can you help me? – 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 and for $$c=1/5$$ this is $$30979/67500$$.
• That’s $\chi$ as the indicator function (en.m.wikipedia.org/wiki/Indicator_function), represented in Mathematica as $\mathtt{Boole}$. – Matt F. Feb 3 at 17:55