# find the probability about sum of random variables

Let $$X_1, X_2, X_3, Y_1, Y_2, Y_3, Z_1, Z_2, Z_3$$ be random variables which have uniform distribution between 0 and 1. It means, the average of $$X_1 = 0.5$$

Let: $$X=X_1 + X_2 + X_3,$$ $$Y=Y_1 + Y_2 + Y_3$$, $$Z=Z_1 + Z_2 + Z_3$$

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 $$\mathbb{P}\left[0.2+X >\max\{Y,Z\}\right]$$?

HINT

1. Let $$X = X_1+X_2+X_3$$. What is the distribution of $$X$$? Well, a direct approach would be to find $$F_X(x) = \mathbb{P}[X_1+X_2+X_3 < x] = \iiint_{[0,1]^3} \mathbb{I}_{[a+b+c which can be translated to a regular volume if you restrict the region of integration so the indicator is always 1.
2. Then, $$Y = Y_1+Y_2+Y_3$$ and $$Z = Z_1+Z_2+Z_3$$ are defined analogously and have the same distribution with pdf $$f(x) = F'(x)$$. It's easy to see $$f(x)$$ only has support for $$x \in [0,3]$$.
3. You want $$\begin{split} \mathbb{P}\left[c+X >\max\{Y,Z\}\right] &= \iiint_{[0,3]^3} \mathbb{I}_{[c+x > \max\{y,z\}]} f(x)f(y)f(z) dxdydz \\ &= \iiint_{[0,3]^3} \mathbb{I}_{[c+x > y]} \mathbb{I}_{[c+x > z]} f(x)f(y)f(z)dxdydz \end{split}$$ which can be similarly manipulated...
• Thank you a lot! – martian03 Jan 28 '19 at 2:46

Thank you @gt6989b

At first, the answer of $$F_X(x)$$ is

$$F(x)=\dfrac{1}{6}x^3$$ when $$0\leq x<1$$

$$F(x)=-\dfrac{1}{3}x^3+\dfrac{3}{2}x^2-\dfrac{3}{2}x+\dfrac{1}{2}$$ when $$1 \leq x <2$$

$$F(x)=1-\dfrac{1}{6}(3-x)^3$$ when $$2\leq x<3$$

and secondly,

$$f(x)=\dfrac{1}{2}x^2$$ when $$0\leq x<1$$

$$f(x)=-x^2 +3x-\dfrac{3}{2}$$ when $$1\leq x <2$$

$$f(x)=\dfrac{1}{2}(3-x)^2$$ when $$2\leq x<3$$

And $$f(y)$$ and $$f(z)$$ follow same way.

Hence, $$\begin{split} \mathbb{P}\left[X >\max\{Y,Z\}\right] &= \iiint_{[0,3]^3} \mathbb{I}_{[x > \max\{y,z\}]} f(x)f(y)f(z) dxdydz \\ &= \dfrac{1}{3}\ \end{split}$$ And I think I can go further. Thank you again.