Let $X_1,X_2,X_3$ be iid. U($0,1$) random variables. Then what will be the value of...












-2












$begingroup$


Let $X_1,X_2,X_3$ be iid. U($0,1$) random variables. Then what will be the value of $E(frac{X_1+X_2}{X_1+X_2+X_3}$) ?










share|cite|improve this question











$endgroup$



closed as off-topic by Namaste, 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." – Namaste, Saad, Lord_Farin, Did

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 1




    $begingroup$
    Possible duplicate of $X_1,X_2,X_3 sim^{text{i.i.d}} R(0,1)$. Find $E(frac{X_1+X_2}{X_1+X_2+X_3})$
    $endgroup$
    – StubbornAtom
    Jan 11 at 16:20
















-2












$begingroup$


Let $X_1,X_2,X_3$ be iid. U($0,1$) random variables. Then what will be the value of $E(frac{X_1+X_2}{X_1+X_2+X_3}$) ?










share|cite|improve this question











$endgroup$



closed as off-topic by Namaste, 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." – Namaste, Saad, Lord_Farin, Did

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 1




    $begingroup$
    Possible duplicate of $X_1,X_2,X_3 sim^{text{i.i.d}} R(0,1)$. Find $E(frac{X_1+X_2}{X_1+X_2+X_3})$
    $endgroup$
    – StubbornAtom
    Jan 11 at 16:20














-2












-2








-2





$begingroup$


Let $X_1,X_2,X_3$ be iid. U($0,1$) random variables. Then what will be the value of $E(frac{X_1+X_2}{X_1+X_2+X_3}$) ?










share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 18 at 10:14









md2perpe

8,21111028




8,21111028










asked Jan 11 at 15:37









S.MuniyanS.Muniyan

21




21




closed as off-topic by Namaste, 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." – Namaste, 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 Namaste, 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." – Namaste, Saad, Lord_Farin, Did

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    $begingroup$
    Possible duplicate of $X_1,X_2,X_3 sim^{text{i.i.d}} R(0,1)$. Find $E(frac{X_1+X_2}{X_1+X_2+X_3})$
    $endgroup$
    – StubbornAtom
    Jan 11 at 16:20














  • 1




    $begingroup$
    Possible duplicate of $X_1,X_2,X_3 sim^{text{i.i.d}} R(0,1)$. Find $E(frac{X_1+X_2}{X_1+X_2+X_3})$
    $endgroup$
    – StubbornAtom
    Jan 11 at 16:20








1




1




$begingroup$
Possible duplicate of $X_1,X_2,X_3 sim^{text{i.i.d}} R(0,1)$. Find $E(frac{X_1+X_2}{X_1+X_2+X_3})$
$endgroup$
– StubbornAtom
Jan 11 at 16:20




$begingroup$
Possible duplicate of $X_1,X_2,X_3 sim^{text{i.i.d}} R(0,1)$. Find $E(frac{X_1+X_2}{X_1+X_2+X_3})$
$endgroup$
– StubbornAtom
Jan 11 at 16:20










1 Answer
1






active

oldest

votes


















2












$begingroup$

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 $2/3$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Downvoter, please step ahead and let me know why you did downvote my post.
    $endgroup$
    – Aaron
    Jan 17 at 15:47


















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

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 $2/3$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Downvoter, please step ahead and let me know why you did downvote my post.
    $endgroup$
    – Aaron
    Jan 17 at 15:47
















2












$begingroup$

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 $2/3$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Downvoter, please step ahead and let me know why you did downvote my post.
    $endgroup$
    – Aaron
    Jan 17 at 15:47














2












2








2





$begingroup$

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 $2/3$.






share|cite|improve this answer









$endgroup$



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 $2/3$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 11 at 15:51









AaronAaron

1,912415




1,912415












  • $begingroup$
    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




$begingroup$
Downvoter, please step ahead and let me know why you did downvote my post.
$endgroup$
– Aaron
Jan 17 at 15:47



Popular posts from this blog

Human spaceflight

Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

張江高科駅