Finding $P(X_1+X_2+X_3+X_4ge3)$ for independent $X_isim U(0,1)$
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How to find $P(X_1 + X_2 + X_3 + X_4 geq 3)$ for uniformly distributed independent random variables $X_1$, $X_2$, $X_3$, $X_4sim U(0,1)$?
It follows from independence that their cumulative density function is 1, but I'm struggling with integration space.
probability-theory probability-distributions uniform-distribution
edited Dec 4 '18 at 15:24
Did
247k23222458
asked Dec 4 '18 at 12:49
AromaTheLoopAromaTheLoop
464
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add a comment |
$begingroup$
How to find $P(X_1 + X_2 + X_3 + X_4 geq 3)$ for uniformly distributed independent random variables $X_1$, $X_2$, $X_3$, $X_4sim U(0,1)$?
It follows from independence that their cumulative density function is 1, but I'm struggling with integration space.
probability-theory probability-distributions uniform-distribution
edited Dec 4 '18 at 15:24
Did
247k23222458
asked Dec 4 '18 at 12:49
AromaTheLoopAromaTheLoop
464
$endgroup$
1
$begingroup$
(Big) Hint: By symmetry, this is equal to $$P(X_1+X_2+X_3+X_4le1)$$ Now, compute the latter...
$endgroup$
– Did
Dec 4 '18 at 15:23
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@Did I can't figure that out. Looks like it gets us to equation $$F(1) = 1 - F(3)$$ but I can't recall any formal symmetry connected with that,
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– AromaTheLoop
Dec 4 '18 at 17:07
1
$begingroup$
Sub-hint: $(1-X_1,1-X_2,1-X_3,1-X_4)$ is distributed like $(X_1,X_2,X_3,X_4)$.
$endgroup$
– Did
Dec 4 '18 at 17:36
$begingroup$
@Did Oh I see, thanks!
$endgroup$
– AromaTheLoop
Dec 4 '18 at 17:54
add a comment |
$begingroup$
How to find $P(X_1 + X_2 + X_3 + X_4 geq 3)$ for uniformly distributed independent random variables $X_1$, $X_2$, $X_3$, $X_4sim U(0,1)$?
It follows from independence that their cumulative density function is 1, but I'm struggling with integration space.
probability-theory probability-distributions uniform-distribution
edited Dec 4 '18 at 15:24
Did
247k23222458
asked Dec 4 '18 at 12:49
AromaTheLoopAromaTheLoop
464
$endgroup$
How to find $P(X_1 + X_2 + X_3 + X_4 geq 3)$ for uniformly distributed independent random variables $X_1$, $X_2$, $X_3$, $X_4sim U(0,1)$?
It follows from independence that their cumulative density function is 1, but I'm struggling with integration space.
probability-theory probability-distributions uniform-distribution
probability-theory probability-distributions uniform-distribution
edited Dec 4 '18 at 15:24
Did
247k23222458
asked Dec 4 '18 at 12:49
AromaTheLoopAromaTheLoop
464
edited Dec 4 '18 at 15:24
Did
247k23222458
asked Dec 4 '18 at 12:49
AromaTheLoopAromaTheLoop
464
edited Dec 4 '18 at 15:24
Did
247k23222458
edited Dec 4 '18 at 15:24
Did
247k23222458
edited Dec 4 '18 at 15:24
Did
247k23222458
247k23222458
asked Dec 4 '18 at 12:49
AromaTheLoopAromaTheLoop
464
asked Dec 4 '18 at 12:49
AromaTheLoopAromaTheLoop
464
asked Dec 4 '18 at 12:49
AromaTheLoopAromaTheLoop
464
464
1
$begingroup$
(Big) Hint: By symmetry, this is equal to $$P(X_1+X_2+X_3+X_4le1)$$ Now, compute the latter...
$endgroup$
– Did
Dec 4 '18 at 15:23
$begingroup$
@Did I can't figure that out. Looks like it gets us to equation $$F(1) = 1 - F(3)$$ but I can't recall any formal symmetry connected with that,
$endgroup$
– AromaTheLoop
Dec 4 '18 at 17:07
1
$begingroup$
Sub-hint: $(1-X_1,1-X_2,1-X_3,1-X_4)$ is distributed like $(X_1,X_2,X_3,X_4)$.
$endgroup$
– Did
Dec 4 '18 at 17:36
$begingroup$
@Did Oh I see, thanks!
$endgroup$
– AromaTheLoop
Dec 4 '18 at 17:54
add a comment |
1
$begingroup$
(Big) Hint: By symmetry, this is equal to $$P(X_1+X_2+X_3+X_4le1)$$ Now, compute the latter...
$endgroup$
– Did
Dec 4 '18 at 15:23
$begingroup$
@Did I can't figure that out. Looks like it gets us to equation $$F(1) = 1 - F(3)$$ but I can't recall any formal symmetry connected with that,
$endgroup$
– AromaTheLoop
Dec 4 '18 at 17:07
1
$begingroup$
Sub-hint: $(1-X_1,1-X_2,1-X_3,1-X_4)$ is distributed like $(X_1,X_2,X_3,X_4)$.
$endgroup$
– Did
Dec 4 '18 at 17:36
$begingroup$
@Did Oh I see, thanks!
$endgroup$
– AromaTheLoop
Dec 4 '18 at 17:54
1
1
$begingroup$
(Big) Hint: By symmetry, this is equal to $$P(X_1+X_2+X_3+X_4le1)$$ Now, compute the latter...
$endgroup$
– Did
Dec 4 '18 at 15:23
$begingroup$
(Big) Hint: By symmetry, this is equal to $$P(X_1+X_2+X_3+X_4le1)$$ Now, compute the latter...
$endgroup$
– Did
Dec 4 '18 at 15:23
$begingroup$
@Did I can't figure that out. Looks like it gets us to equation $$F(1) = 1 - F(3)$$ but I can't recall any formal symmetry connected with that,
$endgroup$
– AromaTheLoop
Dec 4 '18 at 17:07
$begingroup$
@Did I can't figure that out. Looks like it gets us to equation $$F(1) = 1 - F(3)$$ but I can't recall any formal symmetry connected with that,
$endgroup$
– AromaTheLoop
Dec 4 '18 at 17:07
1
1
$begingroup$
Sub-hint: $(1-X_1,1-X_2,1-X_3,1-X_4)$ is distributed like $(X_1,X_2,X_3,X_4)$.
$endgroup$
– Did
Dec 4 '18 at 17:36
$begingroup$
Sub-hint: $(1-X_1,1-X_2,1-X_3,1-X_4)$ is distributed like $(X_1,X_2,X_3,X_4)$.
$endgroup$
– Did
Dec 4 '18 at 17:36
$begingroup$
@Did Oh I see, thanks!
$endgroup$
– AromaTheLoop
Dec 4 '18 at 17:54
$begingroup$
@Did Oh I see, thanks!
$endgroup$
– AromaTheLoop
Dec 4 '18 at 17:54
add a comment |
1 Answer
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Once we get to the step proposed by @Did, we can obtain the solution easily using geometric probability. Our probability here would be the hyper-volume covered by
$$x_1+x_2+x_3+x_4 leq 1 text{ and } 0leq x_i leq 1$$ is exactly the hyper-volume of an $4$-dimensional simplex, which is $dfrac{1}{4!} = dfrac{1}{24}$.
answered Dec 4 '18 at 15:40
Isaac BrowneIsaac Browne
4,59731132
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$begingroup$
Once we get to the step proposed by @Did, we can obtain the solution easily using geometric probability. Our probability here would be the hyper-volume covered by
$$x_1+x_2+x_3+x_4 leq 1 text{ and } 0leq x_i leq 1$$ is exactly the hyper-volume of an $4$-dimensional simplex, which is $dfrac{1}{4!} = dfrac{1}{24}$.
answered Dec 4 '18 at 15:40
Isaac BrowneIsaac Browne
4,59731132
$endgroup$
add a comment |
$begingroup$
Once we get to the step proposed by @Did, we can obtain the solution easily using geometric probability. Our probability here would be the hyper-volume covered by
$$x_1+x_2+x_3+x_4 leq 1 text{ and } 0leq x_i leq 1$$ is exactly the hyper-volume of an $4$-dimensional simplex, which is $dfrac{1}{4!} = dfrac{1}{24}$.
answered Dec 4 '18 at 15:40
Isaac BrowneIsaac Browne
4,59731132
$endgroup$
add a comment |
$begingroup$
Once we get to the step proposed by @Did, we can obtain the solution easily using geometric probability. Our probability here would be the hyper-volume covered by
$$x_1+x_2+x_3+x_4 leq 1 text{ and } 0leq x_i leq 1$$ is exactly the hyper-volume of an $4$-dimensional simplex, which is $dfrac{1}{4!} = dfrac{1}{24}$.
answered Dec 4 '18 at 15:40
Isaac BrowneIsaac Browne
4,59731132
$endgroup$
Once we get to the step proposed by @Did, we can obtain the solution easily using geometric probability. Our probability here would be the hyper-volume covered by
$$x_1+x_2+x_3+x_4 leq 1 text{ and } 0leq x_i leq 1$$ is exactly the hyper-volume of an $4$-dimensional simplex, which is $dfrac{1}{4!} = dfrac{1}{24}$.
answered Dec 4 '18 at 15:40
Isaac BrowneIsaac Browne
4,59731132
answered Dec 4 '18 at 15:40
Isaac BrowneIsaac Browne
4,59731132
answered Dec 4 '18 at 15:40
Isaac BrowneIsaac Browne
4,59731132
answered Dec 4 '18 at 15:40
Isaac BrowneIsaac Browne
4,59731132
4,59731132
add a comment |
add a comment |
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1
$begingroup$
(Big) Hint: By symmetry, this is equal to $$P(X_1+X_2+X_3+X_4le1)$$ Now, compute the latter...
$endgroup$
– Did
Dec 4 '18 at 15:23
$begingroup$
@Did I can't figure that out. Looks like it gets us to equation $$F(1) = 1 - F(3)$$ but I can't recall any formal symmetry connected with that,
$endgroup$
– AromaTheLoop
Dec 4 '18 at 17:07
1
$begingroup$
Sub-hint: $(1-X_1,1-X_2,1-X_3,1-X_4)$ is distributed like $(X_1,X_2,X_3,X_4)$.
$endgroup$
– Did
Dec 4 '18 at 17:36
$begingroup$
@Did Oh I see, thanks!
$endgroup$
– AromaTheLoop
Dec 4 '18 at 17:54