Sum of square product of I.I.D. Normal distribution
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All $a_i$,$b_i$ and $c_i$ $(1<i<10)$ are independent standard normal distributed variable by each other ($a_i$,$b_i$,$c_i$$∼$$N(0,1)$).
and I would like to know about expected value of combination of these variables, which is
$$Ebiggr[sum_{i=0}^{10} (frac{a^2c^2}{a^2+b^2})biggr]$$
I think denominator follows Chi-square distribution
but this equation seems that denominator and numerator are not independent so I have trouble solving this.
what is expected value of above equation?
normal-distribution
asked 8 hours ago
eeqew
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up vote
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All $a_i$,$b_i$ and $c_i$ $(1<i<10)$ are independent standard normal distributed variable by each other ($a_i$,$b_i$,$c_i$$∼$$N(0,1)$).
and I would like to know about expected value of combination of these variables, which is
$$Ebiggr[sum_{i=0}^{10} (frac{a^2c^2}{a^2+b^2})biggr]$$
I think denominator follows Chi-square distribution
but this equation seems that denominator and numerator are not independent so I have trouble solving this.
what is expected value of above equation?
normal-distribution
asked 8 hours ago
eeqew
11
New contributor
eeqew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
All $a_i$,$b_i$ and $c_i$ $(1<i<10)$ are independent standard normal distributed variable by each other ($a_i$,$b_i$,$c_i$$∼$$N(0,1)$).
and I would like to know about expected value of combination of these variables, which is
$$Ebiggr[sum_{i=0}^{10} (frac{a^2c^2}{a^2+b^2})biggr]$$
I think denominator follows Chi-square distribution
but this equation seems that denominator and numerator are not independent so I have trouble solving this.
what is expected value of above equation?
normal-distribution
asked 8 hours ago
eeqew
11
New contributor
eeqew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
All $a_i$,$b_i$ and $c_i$ $(1<i<10)$ are independent standard normal distributed variable by each other ($a_i$,$b_i$,$c_i$$∼$$N(0,1)$).
and I would like to know about expected value of combination of these variables, which is
$$Ebiggr[sum_{i=0}^{10} (frac{a^2c^2}{a^2+b^2})biggr]$$
I think denominator follows Chi-square distribution
but this equation seems that denominator and numerator are not independent so I have trouble solving this.
what is expected value of above equation?
normal-distribution
normal-distribution
asked 8 hours ago
eeqew
11
New contributor
eeqew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
eeqew
11
New contributor
eeqew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
eeqew
11
New contributor
eeqew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 8 hours ago
eeqew
11
asked 8 hours ago
eeqew
11
11
New contributor
eeqew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
eeqew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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1 Answer
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You can do this without any heavy computation. First note that $c_i^{2}$ is independent of $a_i$'s and $b_i$'s. Next note that $Esum frac {a_i^{2}} {a_i^{2}+b_i^{2}} =Esum frac {b_i^{2}} {a_i^{2}+b_i^{2}} $ (in view of the assumptions on $a_i$'s and $b_i$'s) and this two add up to $10$). Hence the answer is $5$.
answered 8 hours ago
Kavi Rama Murthy
39k31748
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
You can do this without any heavy computation. First note that $c_i^{2}$ is independent of $a_i$'s and $b_i$'s. Next note that $Esum frac {a_i^{2}} {a_i^{2}+b_i^{2}} =Esum frac {b_i^{2}} {a_i^{2}+b_i^{2}} $ (in view of the assumptions on $a_i$'s and $b_i$'s) and this two add up to $10$). Hence the answer is $5$.
answered 8 hours ago
Kavi Rama Murthy
39k31748
add a comment |
up vote
1
down vote
You can do this without any heavy computation. First note that $c_i^{2}$ is independent of $a_i$'s and $b_i$'s. Next note that $Esum frac {a_i^{2}} {a_i^{2}+b_i^{2}} =Esum frac {b_i^{2}} {a_i^{2}+b_i^{2}} $ (in view of the assumptions on $a_i$'s and $b_i$'s) and this two add up to $10$). Hence the answer is $5$.
answered 8 hours ago
Kavi Rama Murthy
39k31748
add a comment |
up vote
1
down vote
up vote
1
down vote
You can do this without any heavy computation. First note that $c_i^{2}$ is independent of $a_i$'s and $b_i$'s. Next note that $Esum frac {a_i^{2}} {a_i^{2}+b_i^{2}} =Esum frac {b_i^{2}} {a_i^{2}+b_i^{2}} $ (in view of the assumptions on $a_i$'s and $b_i$'s) and this two add up to $10$). Hence the answer is $5$.
answered 8 hours ago
Kavi Rama Murthy
39k31748
You can do this without any heavy computation. First note that $c_i^{2}$ is independent of $a_i$'s and $b_i$'s. Next note that $Esum frac {a_i^{2}} {a_i^{2}+b_i^{2}} =Esum frac {b_i^{2}} {a_i^{2}+b_i^{2}} $ (in view of the assumptions on $a_i$'s and $b_i$'s) and this two add up to $10$). Hence the answer is $5$.
answered 8 hours ago
Kavi Rama Murthy
39k31748
answered 8 hours ago
Kavi Rama Murthy
39k31748
answered 8 hours ago
Kavi Rama Murthy
39k31748
answered 8 hours ago
Kavi Rama Murthy
39k31748
39k31748
add a comment |
add a comment |
eeqew is a new contributor. Be nice, and check out our Code of Conduct.
eeqew is a new contributor. Be nice, and check out our Code of Conduct.
eeqew is a new contributor. Be nice, and check out our Code of Conduct.
eeqew is a new contributor. Be nice, and check out our Code of Conduct.
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