# Distribution of $\min(X_1+X_2+X_3,X_2+X_3+X_4,X_3+X_4+X_5,X_4+X_5+X_6)$

Let $$X_{i}\,(i=1,2,3,4,5,6)$$ be i.i.d. continuous random variables with distribution $$F( \cdot )$$ and density $$f( \cdot )$$. What will be the distribution of $$\min(X_1+X_2+X_3,~ X_2+X_3+X_4,~ X_3+X_4+X_5,~ X_4+X_5+X_6)~?$$

Here is my attempt-

Let $$W=\min(X_1+X_2+X_3,X_2+X_3+X_4,X_3+X_4+X_5,X_4+X_5+X_6)$$. We need to find $$P(W\leq w)$$. Let $$X_2 = x_2,X_3=x_3,X_4=x_4,X_5=x_5$$. Then, the problem reduces to

P\left( \begin{aligned} &X_1 \leq w-x_2-x_3, \\ &x_2+x_3+x_4\leq w, \\ &x_3+x_4+x_5\leq w, \\ &X_6\leq w-x_4-x_5 \end{aligned}~\middle|~ \begin{aligned} X_2 &= x_2, \\ X_3 &= x_3, \\ X_4 &= x_4, \\ X_5 &=x_5 \end{aligned}\right)~

Since all $$X_i$$ are independent, then we can write the above expression as

$$P(X_1\leq w-x_2-x_3) \cdot P(X_6\leq w-x_4-x_5)$$ subject to the limits $$x_2+x_3+x_4\leq w,x_3+x_4+x_5\leq w~.$$

Is my approach correct?

• You should go for finding $P(W>w)$ (not $P(W\leq w)$). Note that $P(\min(X,Y)>w)=P(X>w,Y>w)$ which is better to handle than $P(W\leq w)=P(X\leq w\text{ or }Y\leq w)$. – drhab Jan 30 at 10:24
• @drhab I get your point. But I'm a little confused about the limits of the integrals. Can you please help me out with that? – superhulk Jan 30 at 11:15
• Cross-post yet again: stats.stackexchange.com/questions/389864/…. – StubbornAtom Jan 30 at 15:55