Mixing
Law of Large Numbers contradicts Central Limit Theorem?
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My text defines the weak law of large numbers:
If $X_1,ldots,X_n$ are IID, then $overline{X} overset{P}{to} mu$.
And the CLT as:
Let $X_1,ldots,X_n$ be IID with mean $mu$ and variance $sigma^2$.
Then:
$$Z_n = frac{sqrt{n}(overline{X}_n - mu)}{sigma} rightsquigarrow Z$$
Where $Z sim N(0,1)$.
The weak law of large numbers says that the sample mean converges in probability to a constant (the population mean). Convergence in probability implies convergence in distribution, so it is also saying that the sample mean converges in distribution to that same constant.
In contrast the central limit theorem appears to be saying that the sample mean converges to a standard normal distribution, not a constant. I recognize that strictly speaking the two expressions are not the same -- but I wouldn't expect subtracting the population mean (a constant) or dividing by the standard deviation (a constant) to change the expression in such a way that it no longer converges to a constant.
The only other difference is the multiplication by $sqrt{n}$. If this is what makes the difference (which seems plausible because it will change the sample mean from being the sum divided by $n$ to the sum divided by $sqrt{n}$), then it seems like we went out of our way to make it so that we would know less about the convergence -- if the point of the CLT is to be able to make probability statements about the sample mean, that seems backwards (we were better off before just using WLLN where knew the number with certainty). What am I missing here?
probability central-limit-theorem law-of-large-numbers
asked Jan 27 at 20:40
Joseph GarvinJoseph Garvin
45928
$endgroup$
add a comment |
$begingroup$
My text defines the weak law of large numbers:
If $X_1,ldots,X_n$ are IID, then $overline{X} overset{P}{to} mu$.
And the CLT as:
Let $X_1,ldots,X_n$ be IID with mean $mu$ and variance $sigma^2$.
Then:
$$Z_n = frac{sqrt{n}(overline{X}_n - mu)}{sigma} rightsquigarrow Z$$
Where $Z sim N(0,1)$.
The weak law of large numbers says that the sample mean converges in probability to a constant (the population mean). Convergence in probability implies convergence in distribution, so it is also saying that the sample mean converges in distribution to that same constant.
In contrast the central limit theorem appears to be saying that the sample mean converges to a standard normal distribution, not a constant. I recognize that strictly speaking the two expressions are not the same -- but I wouldn't expect subtracting the population mean (a constant) or dividing by the standard deviation (a constant) to change the expression in such a way that it no longer converges to a constant.
The only other difference is the multiplication by $sqrt{n}$. If this is what makes the difference (which seems plausible because it will change the sample mean from being the sum divided by $n$ to the sum divided by $sqrt{n}$), then it seems like we went out of our way to make it so that we would know less about the convergence -- if the point of the CLT is to be able to make probability statements about the sample mean, that seems backwards (we were better off before just using WLLN where knew the number with certainty). What am I missing here?
probability central-limit-theorem law-of-large-numbers
asked Jan 27 at 20:40
Joseph GarvinJoseph Garvin
45928
$endgroup$
1
$begingroup$
The CLT is a refinement of the LLN. Namely, the latter says that the sample mean converges to the population mean, and the first gives you a more precise asymptotic result. That is, $bar X_ntomu$ and the difference is actually of the size $frac 1{sqrt n}$. After multiplying the difference by the sharp scale factor $sqrt{n}$ you obtain a limit profile, namely a standard normal profile.
$endgroup$
– GReyes
Jan 27 at 20:51
$begingroup$
The square root of n is the key, it takes the deviations of the sample mean from the population mean and "stretches" them to be of order 1 (neither going to zero nor blowing up) even as n goes to infinity. The advantage is that now you. can make good quantitative estimates of moderate deviations between the sample mean and population mean (moderate meaning of order $n^{-1/2}$).
$endgroup$
– Ian
Jan 27 at 20:52
$begingroup$
Look at the relationship between the CLT, LLN, and LIL here
$endgroup$
– d.k.o.
Jan 27 at 22:52
add a comment |
$begingroup$
My text defines the weak law of large numbers:
If $X_1,ldots,X_n$ are IID, then $overline{X} overset{P}{to} mu$.
And the CLT as:
Let $X_1,ldots,X_n$ be IID with mean $mu$ and variance $sigma^2$.
Then:
$$Z_n = frac{sqrt{n}(overline{X}_n - mu)}{sigma} rightsquigarrow Z$$
Where $Z sim N(0,1)$.
The weak law of large numbers says that the sample mean converges in probability to a constant (the population mean). Convergence in probability implies convergence in distribution, so it is also saying that the sample mean converges in distribution to that same constant.
In contrast the central limit theorem appears to be saying that the sample mean converges to a standard normal distribution, not a constant. I recognize that strictly speaking the two expressions are not the same -- but I wouldn't expect subtracting the population mean (a constant) or dividing by the standard deviation (a constant) to change the expression in such a way that it no longer converges to a constant.
The only other difference is the multiplication by $sqrt{n}$. If this is what makes the difference (which seems plausible because it will change the sample mean from being the sum divided by $n$ to the sum divided by $sqrt{n}$), then it seems like we went out of our way to make it so that we would know less about the convergence -- if the point of the CLT is to be able to make probability statements about the sample mean, that seems backwards (we were better off before just using WLLN where knew the number with certainty). What am I missing here?
probability central-limit-theorem law-of-large-numbers
asked Jan 27 at 20:40
Joseph GarvinJoseph Garvin
45928
$endgroup$
My text defines the weak law of large numbers:
If $X_1,ldots,X_n$ are IID, then $overline{X} overset{P}{to} mu$.
And the CLT as:
Let $X_1,ldots,X_n$ be IID with mean $mu$ and variance $sigma^2$.
Then:
$$Z_n = frac{sqrt{n}(overline{X}_n - mu)}{sigma} rightsquigarrow Z$$
Where $Z sim N(0,1)$.
The weak law of large numbers says that the sample mean converges in probability to a constant (the population mean). Convergence in probability implies convergence in distribution, so it is also saying that the sample mean converges in distribution to that same constant.
In contrast the central limit theorem appears to be saying that the sample mean converges to a standard normal distribution, not a constant. I recognize that strictly speaking the two expressions are not the same -- but I wouldn't expect subtracting the population mean (a constant) or dividing by the standard deviation (a constant) to change the expression in such a way that it no longer converges to a constant.
The only other difference is the multiplication by $sqrt{n}$. If this is what makes the difference (which seems plausible because it will change the sample mean from being the sum divided by $n$ to the sum divided by $sqrt{n}$), then it seems like we went out of our way to make it so that we would know less about the convergence -- if the point of the CLT is to be able to make probability statements about the sample mean, that seems backwards (we were better off before just using WLLN where knew the number with certainty). What am I missing here?
probability central-limit-theorem law-of-large-numbers
probability central-limit-theorem law-of-large-numbers
asked Jan 27 at 20:40
Joseph GarvinJoseph Garvin
45928
asked Jan 27 at 20:40
Joseph GarvinJoseph Garvin
45928
edited Jan 27 at 23:37
Joseph Garvin
asked Jan 27 at 20:40
Joseph GarvinJoseph Garvin
45928
asked Jan 27 at 20:40
Joseph GarvinJoseph Garvin
45928
asked Jan 27 at 20:40
Joseph GarvinJoseph Garvin
45928
45928
1
$begingroup$
The CLT is a refinement of the LLN. Namely, the latter says that the sample mean converges to the population mean, and the first gives you a more precise asymptotic result. That is, $bar X_ntomu$ and the difference is actually of the size $frac 1{sqrt n}$. After multiplying the difference by the sharp scale factor $sqrt{n}$ you obtain a limit profile, namely a standard normal profile.
$endgroup$
– GReyes
Jan 27 at 20:51
$begingroup$
The square root of n is the key, it takes the deviations of the sample mean from the population mean and "stretches" them to be of order 1 (neither going to zero nor blowing up) even as n goes to infinity. The advantage is that now you. can make good quantitative estimates of moderate deviations between the sample mean and population mean (moderate meaning of order $n^{-1/2}$).
$endgroup$
– Ian
Jan 27 at 20:52
$begingroup$
Look at the relationship between the CLT, LLN, and LIL here
$endgroup$
– d.k.o.
Jan 27 at 22:52
add a comment |
1
$begingroup$
The CLT is a refinement of the LLN. Namely, the latter says that the sample mean converges to the population mean, and the first gives you a more precise asymptotic result. That is, $bar X_ntomu$ and the difference is actually of the size $frac 1{sqrt n}$. After multiplying the difference by the sharp scale factor $sqrt{n}$ you obtain a limit profile, namely a standard normal profile.
$endgroup$
– GReyes
Jan 27 at 20:51
$begingroup$
The square root of n is the key, it takes the deviations of the sample mean from the population mean and "stretches" them to be of order 1 (neither going to zero nor blowing up) even as n goes to infinity. The advantage is that now you. can make good quantitative estimates of moderate deviations between the sample mean and population mean (moderate meaning of order $n^{-1/2}$).
$endgroup$
– Ian
Jan 27 at 20:52
$begingroup$
Look at the relationship between the CLT, LLN, and LIL here
$endgroup$
– d.k.o.
Jan 27 at 22:52
1
1
$begingroup$
The CLT is a refinement of the LLN. Namely, the latter says that the sample mean converges to the population mean, and the first gives you a more precise asymptotic result. That is, $bar X_ntomu$ and the difference is actually of the size $frac 1{sqrt n}$. After multiplying the difference by the sharp scale factor $sqrt{n}$ you obtain a limit profile, namely a standard normal profile.
$endgroup$
– GReyes
Jan 27 at 20:51
$begingroup$
The CLT is a refinement of the LLN. Namely, the latter says that the sample mean converges to the population mean, and the first gives you a more precise asymptotic result. That is, $bar X_ntomu$ and the difference is actually of the size $frac 1{sqrt n}$. After multiplying the difference by the sharp scale factor $sqrt{n}$ you obtain a limit profile, namely a standard normal profile.
$endgroup$
– GReyes
Jan 27 at 20:51
$begingroup$
The square root of n is the key, it takes the deviations of the sample mean from the population mean and "stretches" them to be of order 1 (neither going to zero nor blowing up) even as n goes to infinity. The advantage is that now you. can make good quantitative estimates of moderate deviations between the sample mean and population mean (moderate meaning of order $n^{-1/2}$).
$endgroup$
– Ian
Jan 27 at 20:52
$begingroup$
The square root of n is the key, it takes the deviations of the sample mean from the population mean and "stretches" them to be of order 1 (neither going to zero nor blowing up) even as n goes to infinity. The advantage is that now you. can make good quantitative estimates of moderate deviations between the sample mean and population mean (moderate meaning of order $n^{-1/2}$).
$endgroup$
– Ian
Jan 27 at 20:52
$begingroup$
Look at the relationship between the CLT, LLN, and LIL here
$endgroup$
– d.k.o.
Jan 27 at 22:52
$begingroup$
Look at the relationship between the CLT, LLN, and LIL here
$endgroup$
– d.k.o.
Jan 27 at 22:52
add a comment |
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$begingroup$
The CLT is a refinement of the LLN. Namely, the latter says that the sample mean converges to the population mean, and the first gives you a more precise asymptotic result. That is, $bar X_ntomu$ and the difference is actually of the size $frac 1{sqrt n}$. After multiplying the difference by the sharp scale factor $sqrt{n}$ you obtain a limit profile, namely a standard normal profile.
$endgroup$
– GReyes
Jan 27 at 20:51
$begingroup$
The square root of n is the key, it takes the deviations of the sample mean from the population mean and "stretches" them to be of order 1 (neither going to zero nor blowing up) even as n goes to infinity. The advantage is that now you. can make good quantitative estimates of moderate deviations between the sample mean and population mean (moderate meaning of order $n^{-1/2}$).
$endgroup$
– Ian
Jan 27 at 20:52
$begingroup$
Look at the relationship between the CLT, LLN, and LIL here
$endgroup$
– d.k.o.
Jan 27 at 22:52