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












1












$begingroup$


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 $P(Wleq w)$. Let $X_2 = x_2,X_3=x_3,X_4=x_4,X_5=x_5$. Then, the problem reduces to



$P(X_1leq w-x_2-x_3,x_2+x_3+x_4leq w,x_3+x_4+x_5leq w,X_6leq w-x_4-x_5|X_2 = x_2,X_3=x_3,X_4=x_4,X_5=x_5)$.



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



$P(X_1leq w-x_2-x_3).P(X_6leq w-x_4-x_5)$ subject to the limits $x_2+x_3+x_4leq w,x_3+x_4+x_5leq w$.



Is my approach correct?






probability statistics probability-distributions self-learning







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    You should go for finding $P(W>w)$ (not $P(Wleq w)$). Note that $P(min(X,Y)>w)=P(X>w,Y>w)$ which is better to handle than $P(Wleq w)=P(Xleq wtext{ or }Yleq w)$.
    $endgroup$
    – drhab
    Jan 30 at 10:24












  • $begingroup$
    @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?
    $endgroup$
    – superhulk
    Jan 30 at 11:15










  • $begingroup$
    Cross-post yet again: stats.stackexchange.com/questions/389864/….
    $endgroup$
    – StubbornAtom
    Jan 30 at 15:55
















1












$begingroup$


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 $P(Wleq w)$. Let $X_2 = x_2,X_3=x_3,X_4=x_4,X_5=x_5$. Then, the problem reduces to



$P(X_1leq w-x_2-x_3,x_2+x_3+x_4leq w,x_3+x_4+x_5leq w,X_6leq w-x_4-x_5|X_2 = x_2,X_3=x_3,X_4=x_4,X_5=x_5)$.



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



$P(X_1leq w-x_2-x_3).P(X_6leq w-x_4-x_5)$ subject to the limits $x_2+x_3+x_4leq w,x_3+x_4+x_5leq w$.



Is my approach correct?






probability statistics probability-distributions self-learning







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    You should go for finding $P(W>w)$ (not $P(Wleq w)$). Note that $P(min(X,Y)>w)=P(X>w,Y>w)$ which is better to handle than $P(Wleq w)=P(Xleq wtext{ or }Yleq w)$.
    $endgroup$
    – drhab
    Jan 30 at 10:24












  • $begingroup$
    @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?
    $endgroup$
    – superhulk
    Jan 30 at 11:15










  • $begingroup$
    Cross-post yet again: stats.stackexchange.com/questions/389864/….
    $endgroup$
    – StubbornAtom
    Jan 30 at 15:55














1












1








1





$begingroup$


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 $P(Wleq w)$. Let $X_2 = x_2,X_3=x_3,X_4=x_4,X_5=x_5$. Then, the problem reduces to



$P(X_1leq w-x_2-x_3,x_2+x_3+x_4leq w,x_3+x_4+x_5leq w,X_6leq w-x_4-x_5|X_2 = x_2,X_3=x_3,X_4=x_4,X_5=x_5)$.



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



$P(X_1leq w-x_2-x_3).P(X_6leq w-x_4-x_5)$ subject to the limits $x_2+x_3+x_4leq w,x_3+x_4+x_5leq w$.



Is my approach correct?






probability statistics probability-distributions self-learning







share|cite|improve this question









$endgroup$




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 $P(Wleq w)$. Let $X_2 = x_2,X_3=x_3,X_4=x_4,X_5=x_5$. Then, the problem reduces to



$P(X_1leq w-x_2-x_3,x_2+x_3+x_4leq w,x_3+x_4+x_5leq w,X_6leq w-x_4-x_5|X_2 = x_2,X_3=x_3,X_4=x_4,X_5=x_5)$.



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



$P(X_1leq w-x_2-x_3).P(X_6leq w-x_4-x_5)$ subject to the limits $x_2+x_3+x_4leq w,x_3+x_4+x_5leq w$.



Is my approach correct?






probability statistics probability-distributions self-learning




probability statistics probability-distributions self-learning






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 30 at 9:55









superhulksuperhulk

1203




1203








  • 2




    $begingroup$
    You should go for finding $P(W>w)$ (not $P(Wleq w)$). Note that $P(min(X,Y)>w)=P(X>w,Y>w)$ which is better to handle than $P(Wleq w)=P(Xleq wtext{ or }Yleq w)$.
    $endgroup$
    – drhab
    Jan 30 at 10:24












  • $begingroup$
    @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?
    $endgroup$
    – superhulk
    Jan 30 at 11:15










  • $begingroup$
    Cross-post yet again: stats.stackexchange.com/questions/389864/….
    $endgroup$
    – StubbornAtom
    Jan 30 at 15:55














  • 2




    $begingroup$
    You should go for finding $P(W>w)$ (not $P(Wleq w)$). Note that $P(min(X,Y)>w)=P(X>w,Y>w)$ which is better to handle than $P(Wleq w)=P(Xleq wtext{ or }Yleq w)$.
    $endgroup$
    – drhab
    Jan 30 at 10:24












  • $begingroup$
    @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?
    $endgroup$
    – superhulk
    Jan 30 at 11:15










  • $begingroup$
    Cross-post yet again: stats.stackexchange.com/questions/389864/….
    $endgroup$
    – StubbornAtom
    Jan 30 at 15:55








2




2




$begingroup$
You should go for finding $P(W>w)$ (not $P(Wleq w)$). Note that $P(min(X,Y)>w)=P(X>w,Y>w)$ which is better to handle than $P(Wleq w)=P(Xleq wtext{ or }Yleq w)$.
$endgroup$
– drhab
Jan 30 at 10:24






$begingroup$
You should go for finding $P(W>w)$ (not $P(Wleq w)$). Note that $P(min(X,Y)>w)=P(X>w,Y>w)$ which is better to handle than $P(Wleq w)=P(Xleq wtext{ or }Yleq w)$.
$endgroup$
– drhab
Jan 30 at 10:24














$begingroup$
@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?
$endgroup$
– superhulk
Jan 30 at 11:15




$begingroup$
@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?
$endgroup$
– superhulk
Jan 30 at 11:15












$begingroup$
Cross-post yet again: stats.stackexchange.com/questions/389864/….
$endgroup$
– StubbornAtom
Jan 30 at 15:55




$begingroup$
Cross-post yet again: stats.stackexchange.com/questions/389864/….
$endgroup$
– StubbornAtom
Jan 30 at 15:55










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