Let $X_1,X_2,X_3$ be iid. U($0,1$) random variables. Then what will be the value of...
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Let $X_1,X_2,X_3$ be iid. U($0,1$) random variables. Then what will be the value of $E(frac{X_1+X_2}{X_1+X_2+X_3}$) ?
uniform-distribution expected-value
edited Jan 18 at 10:14
md2perpe
7,80111028
asked Jan 11 at 15:37
S.MuniyanS.Muniyan
11
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closed as off-topic by amWhy, Saad, StubbornAtom, Lord_Farin, Did Jan 11 at 20:18
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Saad, Lord_Farin, Did
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
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Let $X_1,X_2,X_3$ be iid. U($0,1$) random variables. Then what will be the value of $E(frac{X_1+X_2}{X_1+X_2+X_3}$) ?
uniform-distribution expected-value
edited Jan 18 at 10:14
md2perpe
7,80111028
asked Jan 11 at 15:37
S.MuniyanS.Muniyan
11
$endgroup$
closed as off-topic by amWhy, Saad, StubbornAtom, Lord_Farin, Did Jan 11 at 20:18
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Saad, Lord_Farin, Did
If this question can be reworded to fit the rules in the help center, please edit the question.
1
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Possible duplicate of $X_1,X_2,X_3 sim^{text{i.i.d}} R(0,1)$. Find $E(frac{X_1+X_2}{X_1+X_2+X_3})$
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– StubbornAtom
Jan 11 at 16:20
add a comment |
$begingroup$
Let $X_1,X_2,X_3$ be iid. U($0,1$) random variables. Then what will be the value of $E(frac{X_1+X_2}{X_1+X_2+X_3}$) ?
uniform-distribution expected-value
edited Jan 18 at 10:14
md2perpe
7,80111028
asked Jan 11 at 15:37
S.MuniyanS.Muniyan
11
$endgroup$
Let $X_1,X_2,X_3$ be iid. U($0,1$) random variables. Then what will be the value of $E(frac{X_1+X_2}{X_1+X_2+X_3}$) ?
uniform-distribution expected-value
uniform-distribution expected-value
edited Jan 18 at 10:14
md2perpe
7,80111028
asked Jan 11 at 15:37
S.MuniyanS.Muniyan
11
edited Jan 18 at 10:14
md2perpe
7,80111028
asked Jan 11 at 15:37
S.MuniyanS.Muniyan
11
edited Jan 18 at 10:14
md2perpe
7,80111028
edited Jan 18 at 10:14
md2perpe
7,80111028
edited Jan 18 at 10:14
md2perpe
7,80111028
7,80111028
asked Jan 11 at 15:37
S.MuniyanS.Muniyan
11
asked Jan 11 at 15:37
S.MuniyanS.Muniyan
11
asked Jan 11 at 15:37
S.MuniyanS.Muniyan
11
11
closed as off-topic by amWhy, Saad, StubbornAtom, Lord_Farin, Did Jan 11 at 20:18
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Saad, Lord_Farin, Did
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by amWhy, Saad, StubbornAtom, Lord_Farin, Did Jan 11 at 20:18
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Saad, Lord_Farin, Did
If this question can be reworded to fit the rules in the help center, please edit the question.
1
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Possible duplicate of $X_1,X_2,X_3 sim^{text{i.i.d}} R(0,1)$. Find $E(frac{X_1+X_2}{X_1+X_2+X_3})$
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– StubbornAtom
Jan 11 at 16:20
add a comment |
1
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Possible duplicate of $X_1,X_2,X_3 sim^{text{i.i.d}} R(0,1)$. Find $E(frac{X_1+X_2}{X_1+X_2+X_3})$
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– StubbornAtom
Jan 11 at 16:20
1
1
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Possible duplicate of $X_1,X_2,X_3 sim^{text{i.i.d}} R(0,1)$. Find $E(frac{X_1+X_2}{X_1+X_2+X_3})$
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– StubbornAtom
Jan 11 at 16:20
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Possible duplicate of $X_1,X_2,X_3 sim^{text{i.i.d}} R(0,1)$. Find $E(frac{X_1+X_2}{X_1+X_2+X_3})$
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– StubbornAtom
Jan 11 at 16:20
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1 Answer
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So, convince yourself that random variables $Y_i$, defined through $Y_itriangleq displaystylefrac{X_i}{X_1+X_2+X_3}$ for $i =1,2,3$ are identically distributed, due to symmetry. Hence, $mathbb{E}[Y_1]=mathbb{E}[Y_2]=mathbb{E}[Y_3]$. Now, $Y_1+Y_2+Y_3=1$, implying that $mathbb{E}[Y_1+Y_2+Y_3]=1implies mathbb{E}[Y_i]=1/3$ for every $i $.
From here, your answer is $2/3$.
answered Jan 11 at 15:51
AaronAaron
1,828415
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Downvoter, please step ahead and let me know why you did downvote my post.
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– Aaron
Jan 17 at 15:47
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
So, convince yourself that random variables $Y_i$, defined through $Y_itriangleq displaystylefrac{X_i}{X_1+X_2+X_3}$ for $i =1,2,3$ are identically distributed, due to symmetry. Hence, $mathbb{E}[Y_1]=mathbb{E}[Y_2]=mathbb{E}[Y_3]$. Now, $Y_1+Y_2+Y_3=1$, implying that $mathbb{E}[Y_1+Y_2+Y_3]=1implies mathbb{E}[Y_i]=1/3$ for every $i $.
From here, your answer is $2/3$.
answered Jan 11 at 15:51
AaronAaron
1,828415
$endgroup$
$begingroup$
Downvoter, please step ahead and let me know why you did downvote my post.
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– Aaron
Jan 17 at 15:47
add a comment |
$begingroup$
So, convince yourself that random variables $Y_i$, defined through $Y_itriangleq displaystylefrac{X_i}{X_1+X_2+X_3}$ for $i =1,2,3$ are identically distributed, due to symmetry. Hence, $mathbb{E}[Y_1]=mathbb{E}[Y_2]=mathbb{E}[Y_3]$. Now, $Y_1+Y_2+Y_3=1$, implying that $mathbb{E}[Y_1+Y_2+Y_3]=1implies mathbb{E}[Y_i]=1/3$ for every $i $.
From here, your answer is $2/3$.
answered Jan 11 at 15:51
AaronAaron
1,828415
$endgroup$
$begingroup$
Downvoter, please step ahead and let me know why you did downvote my post.
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– Aaron
Jan 17 at 15:47
add a comment |
$begingroup$
So, convince yourself that random variables $Y_i$, defined through $Y_itriangleq displaystylefrac{X_i}{X_1+X_2+X_3}$ for $i =1,2,3$ are identically distributed, due to symmetry. Hence, $mathbb{E}[Y_1]=mathbb{E}[Y_2]=mathbb{E}[Y_3]$. Now, $Y_1+Y_2+Y_3=1$, implying that $mathbb{E}[Y_1+Y_2+Y_3]=1implies mathbb{E}[Y_i]=1/3$ for every $i $.
From here, your answer is $2/3$.
answered Jan 11 at 15:51
AaronAaron
1,828415
$endgroup$
So, convince yourself that random variables $Y_i$, defined through $Y_itriangleq displaystylefrac{X_i}{X_1+X_2+X_3}$ for $i =1,2,3$ are identically distributed, due to symmetry. Hence, $mathbb{E}[Y_1]=mathbb{E}[Y_2]=mathbb{E}[Y_3]$. Now, $Y_1+Y_2+Y_3=1$, implying that $mathbb{E}[Y_1+Y_2+Y_3]=1implies mathbb{E}[Y_i]=1/3$ for every $i $.
From here, your answer is $2/3$.
answered Jan 11 at 15:51
AaronAaron
1,828415
answered Jan 11 at 15:51
AaronAaron
1,828415
answered Jan 11 at 15:51
AaronAaron
1,828415
answered Jan 11 at 15:51
AaronAaron
1,828415
1,828415
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Downvoter, please step ahead and let me know why you did downvote my post.
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– Aaron
Jan 17 at 15:47
add a comment |
$begingroup$
Downvoter, please step ahead and let me know why you did downvote my post.
$endgroup$
– Aaron
Jan 17 at 15:47
$begingroup$
Downvoter, please step ahead and let me know why you did downvote my post.
$endgroup$
– Aaron
Jan 17 at 15:47
$begingroup$
Downvoter, please step ahead and let me know why you did downvote my post.
$endgroup$
– Aaron
Jan 17 at 15:47
add a comment |
1
$begingroup$
Possible duplicate of $X_1,X_2,X_3 sim^{text{i.i.d}} R(0,1)$. Find $E(frac{X_1+X_2}{X_1+X_2+X_3})$
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– StubbornAtom
Jan 11 at 16:20