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 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 at 16:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.