# Distribution of min(X+Y,Y+Z,X+Z,Z+V,X+V,Y+V)?

Let $$X,Y,Z,V$$ be i.i.d continuous random variables in an interval $$[a,b]$$. What will be the distribution of $$\min(X+Y,Y+Z,X+Z,Z+V,X+V,Y+V)$$? Assume the distribution of the random variables to be $$F(.)$$, with $$F(a)=0$$ and $$F(b)=1$$.

My attempt:-

Let $$W=\min(X+Y,Y+Z,X+Z, Z+V,X+V,Y+V)$$. We need to find $$P(W\leq w)$$. This means $$P(X+Y \leq w \wedge Y+Z\leq w\wedge X+Z\leq w \wedge Z+V\leq w\wedge X+V\leq w\wedge Y+V\leq w)$$. But these sums are not independent, so I wasn't able to proceed further.

What should be the general approach for such problems?

• Please erase the last probability expression which is not correct. Feb 3 '19 at 15:51

Your notations do not help. Rephrase the problem as $$X_1,\ldots,X_4\stackrel{\text{iid}}{\sim} f(x)$$ and $$Y=\min\{X_1+X_2,X_1+X_3,X_1+X_4,X_2+X_3,X_2+X_4,X_3+X_4\}$$ You can then express $$Y$$ in terms of the order statistics $$X_{(1)}\le X_{(2)}\le X_{(3)}\le X_{(4)}$$ and deduce that $$Y$$ is a specific sum of two of these order statistics, which leads to the conclusion. Indeed, $$Y=X_{(1)}+X_{(2)}$$ and, since $$(X_{(1)},X_{(2)})\sim \frac{4!}{2!}f(x_{(1)})f(x_{(1)})[1-F(x_{(2)})]^2$$ one can derive the distribution of $$Y$$ from a convolution step: $$Y\sim\int_{-\infty}^{\infty} \frac{4!}{2!}f(x_{(1)})f(y-x_{(1)})[1-F(y-x_{(1)})]^2\,\text{d}x_{(1)}$$
• I get your point. I'm supposed to find out the distribution of $X_{(1)} + X_{(2)}$ now. But order statistics are not independent, so do you have any idea what should I do next? Jan 29 '19 at 16:18