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?

  • $\begingroup$ Please erase the last probability expression which is not correct. $\endgroup$
    – Xi'an
    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: $$Y\sim\int_{-\infty}^{\infty} \frac{4!}{2!}f(x_{(1)})f(y-x_{(1)})[1-F(y-x_{(1)})]^2\,\text{d}x_{(1)}$$

  • 1
    $\begingroup$ 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? $\endgroup$
    – superhulk
    Jan 29 '19 at 16:18

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