how to prove $mathbb{E}[UVmid mathcal{B} ] = Umathbb{E}[Vmid mathcal{B} ]$ if $U$ is $mathcal{B}$-measurable?
$begingroup$
How can we prove
$$mathbb{E}[UVmid mathcal{B} ] = Umathbb{E}[Vmid mathcal{B} ]$$
given that both r.v's are positive and $U$ being $mathcal{B}$-measurable?
I use this property pretty often but I don't know how to actually prove it. Either we have to show that $forall B in mathcal{B}$
$$int_{B}UVdmathbb{P} =int_{B}Umathbb{E}[V mid mathcal{B} ] dmathbb{P} $$
or $forall Z text{ that is } mathcal{B}$-measurable and positive
$$mathbb{E}[UVZ] = mathbb{E}[Umathbb{E}[Vmid mathcal{B} ]Z ]$$
but how?
probability-theory conditional-expectation
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add a comment |
$begingroup$
How can we prove
$$mathbb{E}[UVmid mathcal{B} ] = Umathbb{E}[Vmid mathcal{B} ]$$
given that both r.v's are positive and $U$ being $mathcal{B}$-measurable?
I use this property pretty often but I don't know how to actually prove it. Either we have to show that $forall B in mathcal{B}$
$$int_{B}UVdmathbb{P} =int_{B}Umathbb{E}[V mid mathcal{B} ] dmathbb{P} $$
or $forall Z text{ that is } mathcal{B}$-measurable and positive
$$mathbb{E}[UVZ] = mathbb{E}[Umathbb{E}[Vmid mathcal{B} ]Z ]$$
but how?
probability-theory conditional-expectation
$endgroup$
1
$begingroup$
Please replace every $backslash$ bymid
.
$endgroup$
– Did
Jan 8 at 16:52
add a comment |
$begingroup$
How can we prove
$$mathbb{E}[UVmid mathcal{B} ] = Umathbb{E}[Vmid mathcal{B} ]$$
given that both r.v's are positive and $U$ being $mathcal{B}$-measurable?
I use this property pretty often but I don't know how to actually prove it. Either we have to show that $forall B in mathcal{B}$
$$int_{B}UVdmathbb{P} =int_{B}Umathbb{E}[V mid mathcal{B} ] dmathbb{P} $$
or $forall Z text{ that is } mathcal{B}$-measurable and positive
$$mathbb{E}[UVZ] = mathbb{E}[Umathbb{E}[Vmid mathcal{B} ]Z ]$$
but how?
probability-theory conditional-expectation
$endgroup$
How can we prove
$$mathbb{E}[UVmid mathcal{B} ] = Umathbb{E}[Vmid mathcal{B} ]$$
given that both r.v's are positive and $U$ being $mathcal{B}$-measurable?
I use this property pretty often but I don't know how to actually prove it. Either we have to show that $forall B in mathcal{B}$
$$int_{B}UVdmathbb{P} =int_{B}Umathbb{E}[V mid mathcal{B} ] dmathbb{P} $$
or $forall Z text{ that is } mathcal{B}$-measurable and positive
$$mathbb{E}[UVZ] = mathbb{E}[Umathbb{E}[Vmid mathcal{B} ]Z ]$$
but how?
probability-theory conditional-expectation
probability-theory conditional-expectation
edited Jan 8 at 21:17
Shashi
7,1731528
7,1731528
asked Jan 8 at 15:54
rapidracimrapidracim
1,5401319
1,5401319
1
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Please replace every $backslash$ bymid
.
$endgroup$
– Did
Jan 8 at 16:52
add a comment |
1
$begingroup$
Please replace every $backslash$ bymid
.
$endgroup$
– Did
Jan 8 at 16:52
1
1
$begingroup$
Please replace every $backslash$ by
mid
.$endgroup$
– Did
Jan 8 at 16:52
$begingroup$
Please replace every $backslash$ by
mid
.$endgroup$
– Did
Jan 8 at 16:52
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1 Answer
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$begingroup$
Notice that there exists a monotone sequence of simple random variables $U_n$ which are $mathcal B$ - measurable converging point wise to $U$. The random variables $U_n$ can be written as
$$U_n=sum_{k=1}^{N_n}mathbf{1}_{A_{k,n}}$$
where $A_{k, n}inmathcal B $.
Now we have using MCT
begin{align}
int_B UV, dmathbb P&=int_{ B}lim_{ntoinfty} sum_{k=1}^{N_n}mathbf{1}_{A_{k, n}} V, dmathbb P \
&=lim_{ntoinfty} sum_{k=1}^{N_n}int_{A_{k, n} cap B} V, dmathbb P \
&=lim_{ntoinfty} sum_{k=1}^{N_n}int_{A_{k, n} cap B} mathbb E(Vmid mathcal B), dmathbb P\
&=lim_{ntoinfty} sum_{k=1}^{N_n}int_{ B} mathbf{1}_{A_{k,n}} mathbb E(Vmid mathcal B), dmathbb P
end{align}
And this implies indeed using MCT again that...
You need to figure out how to use MCT yourself though.
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add a comment |
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1 Answer
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$begingroup$
Notice that there exists a monotone sequence of simple random variables $U_n$ which are $mathcal B$ - measurable converging point wise to $U$. The random variables $U_n$ can be written as
$$U_n=sum_{k=1}^{N_n}mathbf{1}_{A_{k,n}}$$
where $A_{k, n}inmathcal B $.
Now we have using MCT
begin{align}
int_B UV, dmathbb P&=int_{ B}lim_{ntoinfty} sum_{k=1}^{N_n}mathbf{1}_{A_{k, n}} V, dmathbb P \
&=lim_{ntoinfty} sum_{k=1}^{N_n}int_{A_{k, n} cap B} V, dmathbb P \
&=lim_{ntoinfty} sum_{k=1}^{N_n}int_{A_{k, n} cap B} mathbb E(Vmid mathcal B), dmathbb P\
&=lim_{ntoinfty} sum_{k=1}^{N_n}int_{ B} mathbf{1}_{A_{k,n}} mathbb E(Vmid mathcal B), dmathbb P
end{align}
And this implies indeed using MCT again that...
You need to figure out how to use MCT yourself though.
$endgroup$
add a comment |
$begingroup$
Notice that there exists a monotone sequence of simple random variables $U_n$ which are $mathcal B$ - measurable converging point wise to $U$. The random variables $U_n$ can be written as
$$U_n=sum_{k=1}^{N_n}mathbf{1}_{A_{k,n}}$$
where $A_{k, n}inmathcal B $.
Now we have using MCT
begin{align}
int_B UV, dmathbb P&=int_{ B}lim_{ntoinfty} sum_{k=1}^{N_n}mathbf{1}_{A_{k, n}} V, dmathbb P \
&=lim_{ntoinfty} sum_{k=1}^{N_n}int_{A_{k, n} cap B} V, dmathbb P \
&=lim_{ntoinfty} sum_{k=1}^{N_n}int_{A_{k, n} cap B} mathbb E(Vmid mathcal B), dmathbb P\
&=lim_{ntoinfty} sum_{k=1}^{N_n}int_{ B} mathbf{1}_{A_{k,n}} mathbb E(Vmid mathcal B), dmathbb P
end{align}
And this implies indeed using MCT again that...
You need to figure out how to use MCT yourself though.
$endgroup$
add a comment |
$begingroup$
Notice that there exists a monotone sequence of simple random variables $U_n$ which are $mathcal B$ - measurable converging point wise to $U$. The random variables $U_n$ can be written as
$$U_n=sum_{k=1}^{N_n}mathbf{1}_{A_{k,n}}$$
where $A_{k, n}inmathcal B $.
Now we have using MCT
begin{align}
int_B UV, dmathbb P&=int_{ B}lim_{ntoinfty} sum_{k=1}^{N_n}mathbf{1}_{A_{k, n}} V, dmathbb P \
&=lim_{ntoinfty} sum_{k=1}^{N_n}int_{A_{k, n} cap B} V, dmathbb P \
&=lim_{ntoinfty} sum_{k=1}^{N_n}int_{A_{k, n} cap B} mathbb E(Vmid mathcal B), dmathbb P\
&=lim_{ntoinfty} sum_{k=1}^{N_n}int_{ B} mathbf{1}_{A_{k,n}} mathbb E(Vmid mathcal B), dmathbb P
end{align}
And this implies indeed using MCT again that...
You need to figure out how to use MCT yourself though.
$endgroup$
Notice that there exists a monotone sequence of simple random variables $U_n$ which are $mathcal B$ - measurable converging point wise to $U$. The random variables $U_n$ can be written as
$$U_n=sum_{k=1}^{N_n}mathbf{1}_{A_{k,n}}$$
where $A_{k, n}inmathcal B $.
Now we have using MCT
begin{align}
int_B UV, dmathbb P&=int_{ B}lim_{ntoinfty} sum_{k=1}^{N_n}mathbf{1}_{A_{k, n}} V, dmathbb P \
&=lim_{ntoinfty} sum_{k=1}^{N_n}int_{A_{k, n} cap B} V, dmathbb P \
&=lim_{ntoinfty} sum_{k=1}^{N_n}int_{A_{k, n} cap B} mathbb E(Vmid mathcal B), dmathbb P\
&=lim_{ntoinfty} sum_{k=1}^{N_n}int_{ B} mathbf{1}_{A_{k,n}} mathbb E(Vmid mathcal B), dmathbb P
end{align}
And this implies indeed using MCT again that...
You need to figure out how to use MCT yourself though.
edited Jan 8 at 21:15
answered Jan 8 at 16:05
ShashiShashi
7,1731528
7,1731528
add a comment |
add a comment |
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– Did
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