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Limit of arithmetic mean of series of measurements
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Assume we have a series of measurements $x_i$ of, say, some physical quantity with (unknown) real value $x_r$. In a physics book I found the statement that, if we could exclude systematic errors in our measurement, we have
beginequation
(1)quad x_r=lim_ntoinftyfrac 1nsum_i=1^n x_i.
endequation
There is also some kind of "proof", starting with definitions
beginequation
barx_n:=frac 1nsum_i=1^n x_i,quad
e_i:=x_r-x_i,quad
varepsilon:=x_r-barx,
endequation
that is,
beginequation
varepsilon=frac 1nsum_i=1^n(x_r-x_i)=frac 1nsum_i=1^n e_i.
endequation
Next step is the calculation
beginequation
varepsilon^2=frac1n^2left(sum_i e_iright)^2=frac1n^2sum_i e_i^2+frac1n^2sum_isum_jneq ie_ie_japproxfrac1n^2sum_i e_i^2
endequation
for $ntoinfty$, using that $e_i$ and $e_j$ are statistically independent. I can make no sense of this conclusion, and then, how does this yield the above statement (1)?
limits probability-theory physics means
asked Jul 30 at 9:30
Don Fuchs
234
add a comment |Â
up vote
2
down vote
favorite
Assume we have a series of measurements $x_i$ of, say, some physical quantity with (unknown) real value $x_r$. In a physics book I found the statement that, if we could exclude systematic errors in our measurement, we have
beginequation
(1)quad x_r=lim_ntoinftyfrac 1nsum_i=1^n x_i.
endequation
There is also some kind of "proof", starting with definitions
beginequation
barx_n:=frac 1nsum_i=1^n x_i,quad
e_i:=x_r-x_i,quad
varepsilon:=x_r-barx,
endequation
that is,
beginequation
varepsilon=frac 1nsum_i=1^n(x_r-x_i)=frac 1nsum_i=1^n e_i.
endequation
Next step is the calculation
beginequation
varepsilon^2=frac1n^2left(sum_i e_iright)^2=frac1n^2sum_i e_i^2+frac1n^2sum_isum_jneq ie_ie_japproxfrac1n^2sum_i e_i^2
endequation
for $ntoinfty$, using that $e_i$ and $e_j$ are statistically independent. I can make no sense of this conclusion, and then, how does this yield the above statement (1)?
limits probability-theory physics means
asked Jul 30 at 9:30
Don Fuchs
234
This is basically data reconciliation. Have a look at stats.stackexchange.com/questions/304612/… which is a question of mine. May be, it could help.
– Claude Leibovici
Jul 30 at 10:24
This is essentially an attempt at the Law of Large Numbers
– Robert Israel
Jul 30 at 11:02
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Assume we have a series of measurements $x_i$ of, say, some physical quantity with (unknown) real value $x_r$. In a physics book I found the statement that, if we could exclude systematic errors in our measurement, we have
beginequation
(1)quad x_r=lim_ntoinftyfrac 1nsum_i=1^n x_i.
endequation
There is also some kind of "proof", starting with definitions
beginequation
barx_n:=frac 1nsum_i=1^n x_i,quad
e_i:=x_r-x_i,quad
varepsilon:=x_r-barx,
endequation
that is,
beginequation
varepsilon=frac 1nsum_i=1^n(x_r-x_i)=frac 1nsum_i=1^n e_i.
endequation
Next step is the calculation
beginequation
varepsilon^2=frac1n^2left(sum_i e_iright)^2=frac1n^2sum_i e_i^2+frac1n^2sum_isum_jneq ie_ie_japproxfrac1n^2sum_i e_i^2
endequation
for $ntoinfty$, using that $e_i$ and $e_j$ are statistically independent. I can make no sense of this conclusion, and then, how does this yield the above statement (1)?
limits probability-theory physics means
asked Jul 30 at 9:30
Don Fuchs
234
Assume we have a series of measurements $x_i$ of, say, some physical quantity with (unknown) real value $x_r$. In a physics book I found the statement that, if we could exclude systematic errors in our measurement, we have
beginequation
(1)quad x_r=lim_ntoinftyfrac 1nsum_i=1^n x_i.
endequation
There is also some kind of "proof", starting with definitions
beginequation
barx_n:=frac 1nsum_i=1^n x_i,quad
e_i:=x_r-x_i,quad
varepsilon:=x_r-barx,
endequation
that is,
beginequation
varepsilon=frac 1nsum_i=1^n(x_r-x_i)=frac 1nsum_i=1^n e_i.
endequation
Next step is the calculation
beginequation
varepsilon^2=frac1n^2left(sum_i e_iright)^2=frac1n^2sum_i e_i^2+frac1n^2sum_isum_jneq ie_ie_japproxfrac1n^2sum_i e_i^2
endequation
for $ntoinfty$, using that $e_i$ and $e_j$ are statistically independent. I can make no sense of this conclusion, and then, how does this yield the above statement (1)?
limits probability-theory physics means
asked Jul 30 at 9:30
Don Fuchs
234
asked Jul 30 at 9:30
Don Fuchs
234
asked Jul 30 at 9:30
Don Fuchs
234
asked Jul 30 at 9:30
Don Fuchs
234
234
This is basically data reconciliation. Have a look at stats.stackexchange.com/questions/304612/… which is a question of mine. May be, it could help.
– Claude Leibovici
Jul 30 at 10:24
This is essentially an attempt at the Law of Large Numbers
– Robert Israel
Jul 30 at 11:02
add a comment |Â
This is basically data reconciliation. Have a look at stats.stackexchange.com/questions/304612/… which is a question of mine. May be, it could help.
– Claude Leibovici
Jul 30 at 10:24
This is essentially an attempt at the Law of Large Numbers
– Robert Israel
Jul 30 at 11:02
This is basically data reconciliation. Have a look at stats.stackexchange.com/questions/304612/… which is a question of mine. May be, it could help.
– Claude Leibovici
Jul 30 at 10:24
This is basically data reconciliation. Have a look at stats.stackexchange.com/questions/304612/… which is a question of mine. May be, it could help.
– Claude Leibovici
Jul 30 at 10:24
This is essentially an attempt at the Law of Large Numbers
– Robert Israel
Jul 30 at 11:02
This is essentially an attempt at the Law of Large Numbers
– Robert Israel
Jul 30 at 11:02
add a comment |Â
1 Answer
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1
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This looks an awful lot like the Law of Large Numbers to me.
I'll assume that each of the measurements are independent, identically distributed and (as you say that we can "exclude systematic errors"), each has a mean of $x_r$. So $E(x_i)=x_r$ for each $i$, $E(e_i)=x_r-E(x_i)=0$ for each $i$, and $E(barx_n)=Eleft(frac1nsum_ix_i right)=frac1nsum_iE(x_i)=x_r$ for all $n$.
I'll also edit your book's notation a little: I'll use $varepsilon_n:=x_r-barx_n$ instead of just $varepsilon$ as it seems that the book's use of $varepsilon$ still depends on $n$.
We have that $E(varepsilon_n)=x_r-E(barx_n)=0$ for all $n$. Because of this, $$mathrmVar(varepsilon_n)=E(varepsilon_n^2)-E(varepsilon_n)^2=E(varepsilon_n^2)-0=E(varepsilon_n^2)$$
... so your final line of working suggests that the book wants to be working with the variance of $varepsilon_n$.
Furthermore, if the $x_i$ are independent, then the $e_i$ are also independent, which means that their covariances are zero. So, for all $ineq j$:
$$0=mathrmCov(e_i,e_j)=E(e_ie_j)-E(e_i)E(e_j)=E(e_ie_j)-0=E(e_ie_j)$$
... which suggests why the book ignores the $frac1n^2sum_isum_jneq ie_ie_j$ term.
Putting all of this together (and finding the epectation of your final line of working) we get:
$$
beginalign
mathrmVar(varepsilon_n) &= E(varepsilon_n^2)
\ &= ldots
\ &= frac1n^2sum_i E(e_i^2)
\ &= frac1n^2sum_i mathrmVar(e_i)
\ &= frac1n^2times n times mathrmVar(e_1)
\ &= frac1n mathrmVar(e_1)
endalign$$
where we've used the fact that if the $e_i$ are identically distributed then they all have the same variance.
Intuitively now, you can see that if $n$ gets large then the variance of your error term $varepsilon_n$ becomes small. More rigorously, we could use Chebyshev's inequality to show that $varepsilon_n$ converges in probability to zero and hence that $barx_n$ converges in probability to $x_r$ (your statement (1)).
answered Jul 30 at 11:26
Malkin
1,004421
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
This looks an awful lot like the Law of Large Numbers to me.
I'll assume that each of the measurements are independent, identically distributed and (as you say that we can "exclude systematic errors"), each has a mean of $x_r$. So $E(x_i)=x_r$ for each $i$, $E(e_i)=x_r-E(x_i)=0$ for each $i$, and $E(barx_n)=Eleft(frac1nsum_ix_i right)=frac1nsum_iE(x_i)=x_r$ for all $n$.
I'll also edit your book's notation a little: I'll use $varepsilon_n:=x_r-barx_n$ instead of just $varepsilon$ as it seems that the book's use of $varepsilon$ still depends on $n$.
We have that $E(varepsilon_n)=x_r-E(barx_n)=0$ for all $n$. Because of this, $$mathrmVar(varepsilon_n)=E(varepsilon_n^2)-E(varepsilon_n)^2=E(varepsilon_n^2)-0=E(varepsilon_n^2)$$
... so your final line of working suggests that the book wants to be working with the variance of $varepsilon_n$.
Furthermore, if the $x_i$ are independent, then the $e_i$ are also independent, which means that their covariances are zero. So, for all $ineq j$:
$$0=mathrmCov(e_i,e_j)=E(e_ie_j)-E(e_i)E(e_j)=E(e_ie_j)-0=E(e_ie_j)$$
... which suggests why the book ignores the $frac1n^2sum_isum_jneq ie_ie_j$ term.
Putting all of this together (and finding the epectation of your final line of working) we get:
$$
beginalign
mathrmVar(varepsilon_n) &= E(varepsilon_n^2)
\ &= ldots
\ &= frac1n^2sum_i E(e_i^2)
\ &= frac1n^2sum_i mathrmVar(e_i)
\ &= frac1n^2times n times mathrmVar(e_1)
\ &= frac1n mathrmVar(e_1)
endalign$$
where we've used the fact that if the $e_i$ are identically distributed then they all have the same variance.
Intuitively now, you can see that if $n$ gets large then the variance of your error term $varepsilon_n$ becomes small. More rigorously, we could use Chebyshev's inequality to show that $varepsilon_n$ converges in probability to zero and hence that $barx_n$ converges in probability to $x_r$ (your statement (1)).
answered Jul 30 at 11:26
Malkin
1,004421
add a comment |Â
up vote
1
down vote
accepted
This looks an awful lot like the Law of Large Numbers to me.
I'll assume that each of the measurements are independent, identically distributed and (as you say that we can "exclude systematic errors"), each has a mean of $x_r$. So $E(x_i)=x_r$ for each $i$, $E(e_i)=x_r-E(x_i)=0$ for each $i$, and $E(barx_n)=Eleft(frac1nsum_ix_i right)=frac1nsum_iE(x_i)=x_r$ for all $n$.
I'll also edit your book's notation a little: I'll use $varepsilon_n:=x_r-barx_n$ instead of just $varepsilon$ as it seems that the book's use of $varepsilon$ still depends on $n$.
We have that $E(varepsilon_n)=x_r-E(barx_n)=0$ for all $n$. Because of this, $$mathrmVar(varepsilon_n)=E(varepsilon_n^2)-E(varepsilon_n)^2=E(varepsilon_n^2)-0=E(varepsilon_n^2)$$
... so your final line of working suggests that the book wants to be working with the variance of $varepsilon_n$.
Furthermore, if the $x_i$ are independent, then the $e_i$ are also independent, which means that their covariances are zero. So, for all $ineq j$:
$$0=mathrmCov(e_i,e_j)=E(e_ie_j)-E(e_i)E(e_j)=E(e_ie_j)-0=E(e_ie_j)$$
... which suggests why the book ignores the $frac1n^2sum_isum_jneq ie_ie_j$ term.
Putting all of this together (and finding the epectation of your final line of working) we get:
$$
beginalign
mathrmVar(varepsilon_n) &= E(varepsilon_n^2)
\ &= ldots
\ &= frac1n^2sum_i E(e_i^2)
\ &= frac1n^2sum_i mathrmVar(e_i)
\ &= frac1n^2times n times mathrmVar(e_1)
\ &= frac1n mathrmVar(e_1)
endalign$$
where we've used the fact that if the $e_i$ are identically distributed then they all have the same variance.
Intuitively now, you can see that if $n$ gets large then the variance of your error term $varepsilon_n$ becomes small. More rigorously, we could use Chebyshev's inequality to show that $varepsilon_n$ converges in probability to zero and hence that $barx_n$ converges in probability to $x_r$ (your statement (1)).
answered Jul 30 at 11:26
Malkin
1,004421
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
This looks an awful lot like the Law of Large Numbers to me.
I'll assume that each of the measurements are independent, identically distributed and (as you say that we can "exclude systematic errors"), each has a mean of $x_r$. So $E(x_i)=x_r$ for each $i$, $E(e_i)=x_r-E(x_i)=0$ for each $i$, and $E(barx_n)=Eleft(frac1nsum_ix_i right)=frac1nsum_iE(x_i)=x_r$ for all $n$.
I'll also edit your book's notation a little: I'll use $varepsilon_n:=x_r-barx_n$ instead of just $varepsilon$ as it seems that the book's use of $varepsilon$ still depends on $n$.
We have that $E(varepsilon_n)=x_r-E(barx_n)=0$ for all $n$. Because of this, $$mathrmVar(varepsilon_n)=E(varepsilon_n^2)-E(varepsilon_n)^2=E(varepsilon_n^2)-0=E(varepsilon_n^2)$$
... so your final line of working suggests that the book wants to be working with the variance of $varepsilon_n$.
Furthermore, if the $x_i$ are independent, then the $e_i$ are also independent, which means that their covariances are zero. So, for all $ineq j$:
$$0=mathrmCov(e_i,e_j)=E(e_ie_j)-E(e_i)E(e_j)=E(e_ie_j)-0=E(e_ie_j)$$
... which suggests why the book ignores the $frac1n^2sum_isum_jneq ie_ie_j$ term.
Putting all of this together (and finding the epectation of your final line of working) we get:
$$
beginalign
mathrmVar(varepsilon_n) &= E(varepsilon_n^2)
\ &= ldots
\ &= frac1n^2sum_i E(e_i^2)
\ &= frac1n^2sum_i mathrmVar(e_i)
\ &= frac1n^2times n times mathrmVar(e_1)
\ &= frac1n mathrmVar(e_1)
endalign$$
where we've used the fact that if the $e_i$ are identically distributed then they all have the same variance.
Intuitively now, you can see that if $n$ gets large then the variance of your error term $varepsilon_n$ becomes small. More rigorously, we could use Chebyshev's inequality to show that $varepsilon_n$ converges in probability to zero and hence that $barx_n$ converges in probability to $x_r$ (your statement (1)).
answered Jul 30 at 11:26
Malkin
1,004421
This looks an awful lot like the Law of Large Numbers to me.
I'll assume that each of the measurements are independent, identically distributed and (as you say that we can "exclude systematic errors"), each has a mean of $x_r$. So $E(x_i)=x_r$ for each $i$, $E(e_i)=x_r-E(x_i)=0$ for each $i$, and $E(barx_n)=Eleft(frac1nsum_ix_i right)=frac1nsum_iE(x_i)=x_r$ for all $n$.
I'll also edit your book's notation a little: I'll use $varepsilon_n:=x_r-barx_n$ instead of just $varepsilon$ as it seems that the book's use of $varepsilon$ still depends on $n$.
We have that $E(varepsilon_n)=x_r-E(barx_n)=0$ for all $n$. Because of this, $$mathrmVar(varepsilon_n)=E(varepsilon_n^2)-E(varepsilon_n)^2=E(varepsilon_n^2)-0=E(varepsilon_n^2)$$
... so your final line of working suggests that the book wants to be working with the variance of $varepsilon_n$.
Furthermore, if the $x_i$ are independent, then the $e_i$ are also independent, which means that their covariances are zero. So, for all $ineq j$:
$$0=mathrmCov(e_i,e_j)=E(e_ie_j)-E(e_i)E(e_j)=E(e_ie_j)-0=E(e_ie_j)$$
... which suggests why the book ignores the $frac1n^2sum_isum_jneq ie_ie_j$ term.
Putting all of this together (and finding the epectation of your final line of working) we get:
$$
beginalign
mathrmVar(varepsilon_n) &= E(varepsilon_n^2)
\ &= ldots
\ &= frac1n^2sum_i E(e_i^2)
\ &= frac1n^2sum_i mathrmVar(e_i)
\ &= frac1n^2times n times mathrmVar(e_1)
\ &= frac1n mathrmVar(e_1)
endalign$$
where we've used the fact that if the $e_i$ are identically distributed then they all have the same variance.
Intuitively now, you can see that if $n$ gets large then the variance of your error term $varepsilon_n$ becomes small. More rigorously, we could use Chebyshev's inequality to show that $varepsilon_n$ converges in probability to zero and hence that $barx_n$ converges in probability to $x_r$ (your statement (1)).
answered Jul 30 at 11:26
Malkin
1,004421
answered Jul 30 at 11:26
Malkin
1,004421
answered Jul 30 at 11:26
Malkin
1,004421
answered Jul 30 at 11:26
Malkin
1,004421
1,004421
add a comment |Â
add a comment |Â
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This is basically data reconciliation. Have a look at stats.stackexchange.com/questions/304612/… which is a question of mine. May be, it could help.
– Claude Leibovici
Jul 30 at 10:24
This is essentially an attempt at the Law of Large Numbers
– Robert Israel
Jul 30 at 11:02