Law of iterated expectations applied to a ratio
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Consider the random variables $Y, Z_1, Zequiv(Z_1,dots,Z_n)$ with $Z_1,...,Z_n$ i.i.d.
Is it true that
$$mathbb Eleft(fracYZ_1right)=mathbb Eleft(fracmathbb E(Ymid Z)Z_1 right)$$?
Does the relation hold if we remove i.i.d.?
probability probability-theory expectation conditional-expectation
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up vote
-2
down vote
favorite
Consider the random variables $Y, Z_1, Zequiv(Z_1,dots,Z_n)$ with $Z_1,...,Z_n$ i.i.d.
Is it true that
$$mathbb Eleft(fracYZ_1right)=mathbb Eleft(fracmathbb E(Ymid Z)Z_1 right)$$?
Does the relation hold if we remove i.i.d.?
probability probability-theory expectation conditional-expectation
add a comment |Â
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
Consider the random variables $Y, Z_1, Zequiv(Z_1,dots,Z_n)$ with $Z_1,...,Z_n$ i.i.d.
Is it true that
$$mathbb Eleft(fracYZ_1right)=mathbb Eleft(fracmathbb E(Ymid Z)Z_1 right)$$?
Does the relation hold if we remove i.i.d.?
probability probability-theory expectation conditional-expectation
Consider the random variables $Y, Z_1, Zequiv(Z_1,dots,Z_n)$ with $Z_1,...,Z_n$ i.i.d.
Is it true that
$$mathbb Eleft(fracYZ_1right)=mathbb Eleft(fracmathbb E(Ymid Z)Z_1 right)$$?
Does the relation hold if we remove i.i.d.?
probability probability-theory expectation conditional-expectation
edited Jul 24 at 7:55
asked Jul 24 at 7:23
STF
15319
15319
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add a comment |Â
1 Answer
1
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oldest
votes
up vote
1
down vote
accepted
Yes. Since $frac 1 Z_1$ is measurable with respect to $sigma (Z_1,Z_2,..,Z_n)$ we can write the right side as $$EEleft(frac Y Z_1mid Zright)=Eleft(frac Y Z_1right)$$
What if we remove I.I.d.? Thank you
– STF
Jul 24 at 7:55
1
No need for any independence here. I just used basic properties of conditional expectation valid for any random variables a s long as ll the expectation s exist.
– Kavi Rama Murthy
Jul 24 at 7:59
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Yes. Since $frac 1 Z_1$ is measurable with respect to $sigma (Z_1,Z_2,..,Z_n)$ we can write the right side as $$EEleft(frac Y Z_1mid Zright)=Eleft(frac Y Z_1right)$$
What if we remove I.I.d.? Thank you
– STF
Jul 24 at 7:55
1
No need for any independence here. I just used basic properties of conditional expectation valid for any random variables a s long as ll the expectation s exist.
– Kavi Rama Murthy
Jul 24 at 7:59
add a comment |Â
up vote
1
down vote
accepted
Yes. Since $frac 1 Z_1$ is measurable with respect to $sigma (Z_1,Z_2,..,Z_n)$ we can write the right side as $$EEleft(frac Y Z_1mid Zright)=Eleft(frac Y Z_1right)$$
What if we remove I.I.d.? Thank you
– STF
Jul 24 at 7:55
1
No need for any independence here. I just used basic properties of conditional expectation valid for any random variables a s long as ll the expectation s exist.
– Kavi Rama Murthy
Jul 24 at 7:59
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Yes. Since $frac 1 Z_1$ is measurable with respect to $sigma (Z_1,Z_2,..,Z_n)$ we can write the right side as $$EEleft(frac Y Z_1mid Zright)=Eleft(frac Y Z_1right)$$
Yes. Since $frac 1 Z_1$ is measurable with respect to $sigma (Z_1,Z_2,..,Z_n)$ we can write the right side as $$EEleft(frac Y Z_1mid Zright)=Eleft(frac Y Z_1right)$$
edited Jul 24 at 8:05
Did
242k23208442
242k23208442
answered Jul 24 at 7:34


Kavi Rama Murthy
20.2k2829
20.2k2829
What if we remove I.I.d.? Thank you
– STF
Jul 24 at 7:55
1
No need for any independence here. I just used basic properties of conditional expectation valid for any random variables a s long as ll the expectation s exist.
– Kavi Rama Murthy
Jul 24 at 7:59
add a comment |Â
What if we remove I.I.d.? Thank you
– STF
Jul 24 at 7:55
1
No need for any independence here. I just used basic properties of conditional expectation valid for any random variables a s long as ll the expectation s exist.
– Kavi Rama Murthy
Jul 24 at 7:59
What if we remove I.I.d.? Thank you
– STF
Jul 24 at 7:55
What if we remove I.I.d.? Thank you
– STF
Jul 24 at 7:55
1
1
No need for any independence here. I just used basic properties of conditional expectation valid for any random variables a s long as ll the expectation s exist.
– Kavi Rama Murthy
Jul 24 at 7:59
No need for any independence here. I just used basic properties of conditional expectation valid for any random variables a s long as ll the expectation s exist.
– Kavi Rama Murthy
Jul 24 at 7:59
add a comment |Â
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