Covariance and Law of Large numbers
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Say I am taking the average value of the product of two dependent random variables $X$ and $Y$ sampled an infinite amont of times. That is I am computing $lim_{n rightarrow infty} E left[ sum_{i=0}^{n} frac{Y_{i}X_{i}}{n} right]$. Is this the same as computing $lim_{n rightarrow infty} E left[ sum_{i=0}^{n} frac{Y_{i}}{n} right] E left[ sum_{i=0}^{n} frac{X_{i}}{n} right] = E[X] E[Y]$? Assuming $X$ and $Y$ have finite variance.
I know this would not be true if $n$ was small but does law of large numbers make the covariance $0$ in the same way it makes variance $0$?
probability covariance variance law-of-large-numbers
asked Dec 10 '18 at 22:01
TPaulTPaul
428
$endgroup$
add a comment |
$begingroup$
Say I am taking the average value of the product of two dependent random variables $X$ and $Y$ sampled an infinite amont of times. That is I am computing $lim_{n rightarrow infty} E left[ sum_{i=0}^{n} frac{Y_{i}X_{i}}{n} right]$. Is this the same as computing $lim_{n rightarrow infty} E left[ sum_{i=0}^{n} frac{Y_{i}}{n} right] E left[ sum_{i=0}^{n} frac{X_{i}}{n} right] = E[X] E[Y]$? Assuming $X$ and $Y$ have finite variance.
I know this would not be true if $n$ was small but does law of large numbers make the covariance $0$ in the same way it makes variance $0$?
probability covariance variance law-of-large-numbers
asked Dec 10 '18 at 22:01
TPaulTPaul
428
$endgroup$
$begingroup$
Are you sure? Isn't covariance a function of variance and variance goes to zero as n goes to infinity?
$endgroup$
– TPaul
Dec 10 '18 at 22:06
add a comment |
$begingroup$
Say I am taking the average value of the product of two dependent random variables $X$ and $Y$ sampled an infinite amont of times. That is I am computing $lim_{n rightarrow infty} E left[ sum_{i=0}^{n} frac{Y_{i}X_{i}}{n} right]$. Is this the same as computing $lim_{n rightarrow infty} E left[ sum_{i=0}^{n} frac{Y_{i}}{n} right] E left[ sum_{i=0}^{n} frac{X_{i}}{n} right] = E[X] E[Y]$? Assuming $X$ and $Y$ have finite variance.
I know this would not be true if $n$ was small but does law of large numbers make the covariance $0$ in the same way it makes variance $0$?
probability covariance variance law-of-large-numbers
asked Dec 10 '18 at 22:01
TPaulTPaul
428
$endgroup$
Say I am taking the average value of the product of two dependent random variables $X$ and $Y$ sampled an infinite amont of times. That is I am computing $lim_{n rightarrow infty} E left[ sum_{i=0}^{n} frac{Y_{i}X_{i}}{n} right]$. Is this the same as computing $lim_{n rightarrow infty} E left[ sum_{i=0}^{n} frac{Y_{i}}{n} right] E left[ sum_{i=0}^{n} frac{X_{i}}{n} right] = E[X] E[Y]$? Assuming $X$ and $Y$ have finite variance.
I know this would not be true if $n$ was small but does law of large numbers make the covariance $0$ in the same way it makes variance $0$?
probability covariance variance law-of-large-numbers
probability covariance variance law-of-large-numbers
asked Dec 10 '18 at 22:01
TPaulTPaul
428
asked Dec 10 '18 at 22:01
TPaulTPaul
428
edited Dec 10 '18 at 22:08
TPaul
asked Dec 10 '18 at 22:01
TPaulTPaul
428
asked Dec 10 '18 at 22:01
TPaulTPaul
428
asked Dec 10 '18 at 22:01
TPaulTPaul
428
428
$begingroup$
Are you sure? Isn't covariance a function of variance and variance goes to zero as n goes to infinity?
$endgroup$
– TPaul
Dec 10 '18 at 22:06
add a comment |
$begingroup$
Are you sure? Isn't covariance a function of variance and variance goes to zero as n goes to infinity?
$endgroup$
– TPaul
Dec 10 '18 at 22:06
$begingroup$
Are you sure? Isn't covariance a function of variance and variance goes to zero as n goes to infinity?
$endgroup$
– TPaul
Dec 10 '18 at 22:06
$begingroup$
Are you sure? Isn't covariance a function of variance and variance goes to zero as n goes to infinity?
$endgroup$
– TPaul
Dec 10 '18 at 22:06
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
From basic properties of expectation,
$$E sum_{i=0}^n frac{Y_i X_i}{n} = frac{1}{n} sum_{i=0}^n E[X_i Y_i] = E[XY]$$
for every $n$. No limits, no law of large numbers.
If you also know $X_i$ and $Y_i$ are uncorrelated (e.g., if they are independent), then $E[X_i Y_i] = E[X_i] E[Y_i]$ and $E[XY] = E[X] E[Y]$.
Under the conditions of the law of large numbers (applied to $XY$), we have
$$sum_{i=0}^n frac{Y_i X_i}{n} to E[XY]$$
almost surely.
Again, if $X$ and $Y$ are uncorrelated, then $E[XY] = E[X] E[Y]$.
answered Dec 10 '18 at 22:06
angryavianangryavian
40.7k23380
$endgroup$
$begingroup$
My question is specifically about correlated values. The difference between $E[XY]$ and $E[X]E[Y]$ is a function of covariance. What does the law of large numbers say about covariance as n grows? Does it go to zero? How can i calculate it?
$endgroup$
– TPaul
Dec 10 '18 at 22:11
$begingroup$
@TPaul If $X$ and $Y$ are zero mean, then the $sum_{i=0}^n X_i Y_i/n$ tends to the covariance $text{Cov}(X,Y) = E[XY]$ almost surely, as my answer implies.
$endgroup$
– angryavian
Dec 10 '18 at 22:24
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Im sorry which part of your answer implies this? Also How does this answer the question i posed about covariance as n grows?
$endgroup$
– TPaul
Dec 10 '18 at 22:45
add a comment |
$begingroup$
I know this would not be true if n was small but does law of large numbers make the covariance 0 in the same way it makes variance 0?
Let me clarify this. LLN doesn't make the variance of the original random variables 0. It makes the variance of a random variable defined as sample average go to 0.
So $X_n$ would still have the same variance as $X_1$ as n goes to infinity. However if we define, $Z = left[ sum_{i=0}^{n} frac{X_{i}}{n} right]$, $E[Z] = E[X_i]$ and $Var(Z) = frac {Var(X)} {n}$. As you see, when n goes to infinity, $Var(Z)$ goes to 0, but Var(X) doesn't change.
Coming back to the original question,
$$lim_{n rightarrow infty} E left[ sum_{i=0}^{n} frac{Y_{i}X_{i}}{n} right] = E[XY]$$
$$E[XY] = E[X]E[Y] + cov(X,Y)$$
Since the $cov(XY)$ isn't going to change no matter the number of n, you cannot calculate the limit the way you proposed in question.
answered Dec 10 '18 at 22:45
OfyaOfya
5198
$endgroup$
$begingroup$
I see. Thank you I think this is where i was stuck.
$endgroup$
– TPaul
Dec 10 '18 at 23:02
$begingroup$
Does this imply $lim_{n rightarrow infty} E[ sum_{i=0}^{n} frac{X_{i}}{n} sum_{i=0}^{n} frac{Y_{i}}{n}] = E[ sum_{i=0}^{n} frac{X_{i}}{n}] E[ sum_{i=0}^{n} frac{Y_{i}}{n}]$ ?
$endgroup$
– TPaul
Dec 10 '18 at 23:05
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
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$begingroup$
From basic properties of expectation,
$$E sum_{i=0}^n frac{Y_i X_i}{n} = frac{1}{n} sum_{i=0}^n E[X_i Y_i] = E[XY]$$
for every $n$. No limits, no law of large numbers.
If you also know $X_i$ and $Y_i$ are uncorrelated (e.g., if they are independent), then $E[X_i Y_i] = E[X_i] E[Y_i]$ and $E[XY] = E[X] E[Y]$.
Under the conditions of the law of large numbers (applied to $XY$), we have
$$sum_{i=0}^n frac{Y_i X_i}{n} to E[XY]$$
almost surely.
Again, if $X$ and $Y$ are uncorrelated, then $E[XY] = E[X] E[Y]$.
answered Dec 10 '18 at 22:06
angryavianangryavian
40.7k23380
$endgroup$
$begingroup$
My question is specifically about correlated values. The difference between $E[XY]$ and $E[X]E[Y]$ is a function of covariance. What does the law of large numbers say about covariance as n grows? Does it go to zero? How can i calculate it?
$endgroup$
– TPaul
Dec 10 '18 at 22:11
$begingroup$
@TPaul If $X$ and $Y$ are zero mean, then the $sum_{i=0}^n X_i Y_i/n$ tends to the covariance $text{Cov}(X,Y) = E[XY]$ almost surely, as my answer implies.
$endgroup$
– angryavian
Dec 10 '18 at 22:24
$begingroup$
Im sorry which part of your answer implies this? Also How does this answer the question i posed about covariance as n grows?
$endgroup$
– TPaul
Dec 10 '18 at 22:45
add a comment |
$begingroup$
From basic properties of expectation,
$$E sum_{i=0}^n frac{Y_i X_i}{n} = frac{1}{n} sum_{i=0}^n E[X_i Y_i] = E[XY]$$
for every $n$. No limits, no law of large numbers.
If you also know $X_i$ and $Y_i$ are uncorrelated (e.g., if they are independent), then $E[X_i Y_i] = E[X_i] E[Y_i]$ and $E[XY] = E[X] E[Y]$.
Under the conditions of the law of large numbers (applied to $XY$), we have
$$sum_{i=0}^n frac{Y_i X_i}{n} to E[XY]$$
almost surely.
Again, if $X$ and $Y$ are uncorrelated, then $E[XY] = E[X] E[Y]$.
answered Dec 10 '18 at 22:06
angryavianangryavian
40.7k23380
$endgroup$
$begingroup$
My question is specifically about correlated values. The difference between $E[XY]$ and $E[X]E[Y]$ is a function of covariance. What does the law of large numbers say about covariance as n grows? Does it go to zero? How can i calculate it?
$endgroup$
– TPaul
Dec 10 '18 at 22:11
$begingroup$
@TPaul If $X$ and $Y$ are zero mean, then the $sum_{i=0}^n X_i Y_i/n$ tends to the covariance $text{Cov}(X,Y) = E[XY]$ almost surely, as my answer implies.
$endgroup$
– angryavian
Dec 10 '18 at 22:24
$begingroup$
Im sorry which part of your answer implies this? Also How does this answer the question i posed about covariance as n grows?
$endgroup$
– TPaul
Dec 10 '18 at 22:45
add a comment |
$begingroup$
From basic properties of expectation,
$$E sum_{i=0}^n frac{Y_i X_i}{n} = frac{1}{n} sum_{i=0}^n E[X_i Y_i] = E[XY]$$
for every $n$. No limits, no law of large numbers.
If you also know $X_i$ and $Y_i$ are uncorrelated (e.g., if they are independent), then $E[X_i Y_i] = E[X_i] E[Y_i]$ and $E[XY] = E[X] E[Y]$.
Under the conditions of the law of large numbers (applied to $XY$), we have
$$sum_{i=0}^n frac{Y_i X_i}{n} to E[XY]$$
almost surely.
Again, if $X$ and $Y$ are uncorrelated, then $E[XY] = E[X] E[Y]$.
answered Dec 10 '18 at 22:06
angryavianangryavian
40.7k23380
$endgroup$
From basic properties of expectation,
$$E sum_{i=0}^n frac{Y_i X_i}{n} = frac{1}{n} sum_{i=0}^n E[X_i Y_i] = E[XY]$$
for every $n$. No limits, no law of large numbers.
If you also know $X_i$ and $Y_i$ are uncorrelated (e.g., if they are independent), then $E[X_i Y_i] = E[X_i] E[Y_i]$ and $E[XY] = E[X] E[Y]$.
Under the conditions of the law of large numbers (applied to $XY$), we have
$$sum_{i=0}^n frac{Y_i X_i}{n} to E[XY]$$
almost surely.
Again, if $X$ and $Y$ are uncorrelated, then $E[XY] = E[X] E[Y]$.
answered Dec 10 '18 at 22:06
angryavianangryavian
40.7k23380
answered Dec 10 '18 at 22:06
angryavianangryavian
40.7k23380
answered Dec 10 '18 at 22:06
angryavianangryavian
40.7k23380
answered Dec 10 '18 at 22:06
angryavianangryavian
40.7k23380
40.7k23380
$begingroup$
My question is specifically about correlated values. The difference between $E[XY]$ and $E[X]E[Y]$ is a function of covariance. What does the law of large numbers say about covariance as n grows? Does it go to zero? How can i calculate it?
$endgroup$
– TPaul
Dec 10 '18 at 22:11
$begingroup$
@TPaul If $X$ and $Y$ are zero mean, then the $sum_{i=0}^n X_i Y_i/n$ tends to the covariance $text{Cov}(X,Y) = E[XY]$ almost surely, as my answer implies.
$endgroup$
– angryavian
Dec 10 '18 at 22:24
$begingroup$
Im sorry which part of your answer implies this? Also How does this answer the question i posed about covariance as n grows?
$endgroup$
– TPaul
Dec 10 '18 at 22:45
add a comment |
$begingroup$
My question is specifically about correlated values. The difference between $E[XY]$ and $E[X]E[Y]$ is a function of covariance. What does the law of large numbers say about covariance as n grows? Does it go to zero? How can i calculate it?
$endgroup$
– TPaul
Dec 10 '18 at 22:11
$begingroup$
@TPaul If $X$ and $Y$ are zero mean, then the $sum_{i=0}^n X_i Y_i/n$ tends to the covariance $text{Cov}(X,Y) = E[XY]$ almost surely, as my answer implies.
$endgroup$
– angryavian
Dec 10 '18 at 22:24
$begingroup$
Im sorry which part of your answer implies this? Also How does this answer the question i posed about covariance as n grows?
$endgroup$
– TPaul
Dec 10 '18 at 22:45
$begingroup$
My question is specifically about correlated values. The difference between $E[XY]$ and $E[X]E[Y]$ is a function of covariance. What does the law of large numbers say about covariance as n grows? Does it go to zero? How can i calculate it?
$endgroup$
– TPaul
Dec 10 '18 at 22:11
$begingroup$
My question is specifically about correlated values. The difference between $E[XY]$ and $E[X]E[Y]$ is a function of covariance. What does the law of large numbers say about covariance as n grows? Does it go to zero? How can i calculate it?
$endgroup$
– TPaul
Dec 10 '18 at 22:11
$begingroup$
@TPaul If $X$ and $Y$ are zero mean, then the $sum_{i=0}^n X_i Y_i/n$ tends to the covariance $text{Cov}(X,Y) = E[XY]$ almost surely, as my answer implies.
$endgroup$
– angryavian
Dec 10 '18 at 22:24
$begingroup$
@TPaul If $X$ and $Y$ are zero mean, then the $sum_{i=0}^n X_i Y_i/n$ tends to the covariance $text{Cov}(X,Y) = E[XY]$ almost surely, as my answer implies.
$endgroup$
– angryavian
Dec 10 '18 at 22:24
$begingroup$
Im sorry which part of your answer implies this? Also How does this answer the question i posed about covariance as n grows?
$endgroup$
– TPaul
Dec 10 '18 at 22:45
$begingroup$
Im sorry which part of your answer implies this? Also How does this answer the question i posed about covariance as n grows?
$endgroup$
– TPaul
Dec 10 '18 at 22:45
add a comment |
$begingroup$
I know this would not be true if n was small but does law of large numbers make the covariance 0 in the same way it makes variance 0?
Let me clarify this. LLN doesn't make the variance of the original random variables 0. It makes the variance of a random variable defined as sample average go to 0.
So $X_n$ would still have the same variance as $X_1$ as n goes to infinity. However if we define, $Z = left[ sum_{i=0}^{n} frac{X_{i}}{n} right]$, $E[Z] = E[X_i]$ and $Var(Z) = frac {Var(X)} {n}$. As you see, when n goes to infinity, $Var(Z)$ goes to 0, but Var(X) doesn't change.
Coming back to the original question,
$$lim_{n rightarrow infty} E left[ sum_{i=0}^{n} frac{Y_{i}X_{i}}{n} right] = E[XY]$$
$$E[XY] = E[X]E[Y] + cov(X,Y)$$
Since the $cov(XY)$ isn't going to change no matter the number of n, you cannot calculate the limit the way you proposed in question.
answered Dec 10 '18 at 22:45
OfyaOfya
5198
$endgroup$
$begingroup$
I see. Thank you I think this is where i was stuck.
$endgroup$
– TPaul
Dec 10 '18 at 23:02
$begingroup$
Does this imply $lim_{n rightarrow infty} E[ sum_{i=0}^{n} frac{X_{i}}{n} sum_{i=0}^{n} frac{Y_{i}}{n}] = E[ sum_{i=0}^{n} frac{X_{i}}{n}] E[ sum_{i=0}^{n} frac{Y_{i}}{n}]$ ?
$endgroup$
– TPaul
Dec 10 '18 at 23:05
add a comment |
$begingroup$
I know this would not be true if n was small but does law of large numbers make the covariance 0 in the same way it makes variance 0?
Let me clarify this. LLN doesn't make the variance of the original random variables 0. It makes the variance of a random variable defined as sample average go to 0.
So $X_n$ would still have the same variance as $X_1$ as n goes to infinity. However if we define, $Z = left[ sum_{i=0}^{n} frac{X_{i}}{n} right]$, $E[Z] = E[X_i]$ and $Var(Z) = frac {Var(X)} {n}$. As you see, when n goes to infinity, $Var(Z)$ goes to 0, but Var(X) doesn't change.
Coming back to the original question,
$$lim_{n rightarrow infty} E left[ sum_{i=0}^{n} frac{Y_{i}X_{i}}{n} right] = E[XY]$$
$$E[XY] = E[X]E[Y] + cov(X,Y)$$
Since the $cov(XY)$ isn't going to change no matter the number of n, you cannot calculate the limit the way you proposed in question.
answered Dec 10 '18 at 22:45
OfyaOfya
5198
$endgroup$
$begingroup$
I see. Thank you I think this is where i was stuck.
$endgroup$
– TPaul
Dec 10 '18 at 23:02
$begingroup$
Does this imply $lim_{n rightarrow infty} E[ sum_{i=0}^{n} frac{X_{i}}{n} sum_{i=0}^{n} frac{Y_{i}}{n}] = E[ sum_{i=0}^{n} frac{X_{i}}{n}] E[ sum_{i=0}^{n} frac{Y_{i}}{n}]$ ?
$endgroup$
– TPaul
Dec 10 '18 at 23:05
add a comment |
$begingroup$
I know this would not be true if n was small but does law of large numbers make the covariance 0 in the same way it makes variance 0?
Let me clarify this. LLN doesn't make the variance of the original random variables 0. It makes the variance of a random variable defined as sample average go to 0.
So $X_n$ would still have the same variance as $X_1$ as n goes to infinity. However if we define, $Z = left[ sum_{i=0}^{n} frac{X_{i}}{n} right]$, $E[Z] = E[X_i]$ and $Var(Z) = frac {Var(X)} {n}$. As you see, when n goes to infinity, $Var(Z)$ goes to 0, but Var(X) doesn't change.
Coming back to the original question,
$$lim_{n rightarrow infty} E left[ sum_{i=0}^{n} frac{Y_{i}X_{i}}{n} right] = E[XY]$$
$$E[XY] = E[X]E[Y] + cov(X,Y)$$
Since the $cov(XY)$ isn't going to change no matter the number of n, you cannot calculate the limit the way you proposed in question.
answered Dec 10 '18 at 22:45
OfyaOfya
5198
$endgroup$
I know this would not be true if n was small but does law of large numbers make the covariance 0 in the same way it makes variance 0?
Let me clarify this. LLN doesn't make the variance of the original random variables 0. It makes the variance of a random variable defined as sample average go to 0.
So $X_n$ would still have the same variance as $X_1$ as n goes to infinity. However if we define, $Z = left[ sum_{i=0}^{n} frac{X_{i}}{n} right]$, $E[Z] = E[X_i]$ and $Var(Z) = frac {Var(X)} {n}$. As you see, when n goes to infinity, $Var(Z)$ goes to 0, but Var(X) doesn't change.
Coming back to the original question,
$$lim_{n rightarrow infty} E left[ sum_{i=0}^{n} frac{Y_{i}X_{i}}{n} right] = E[XY]$$
$$E[XY] = E[X]E[Y] + cov(X,Y)$$
Since the $cov(XY)$ isn't going to change no matter the number of n, you cannot calculate the limit the way you proposed in question.
answered Dec 10 '18 at 22:45
OfyaOfya
5198
answered Dec 10 '18 at 22:45
OfyaOfya
5198
answered Dec 10 '18 at 22:45
OfyaOfya
5198
answered Dec 10 '18 at 22:45
OfyaOfya
5198
5198
$begingroup$
I see. Thank you I think this is where i was stuck.
$endgroup$
– TPaul
Dec 10 '18 at 23:02
$begingroup$
Does this imply $lim_{n rightarrow infty} E[ sum_{i=0}^{n} frac{X_{i}}{n} sum_{i=0}^{n} frac{Y_{i}}{n}] = E[ sum_{i=0}^{n} frac{X_{i}}{n}] E[ sum_{i=0}^{n} frac{Y_{i}}{n}]$ ?
$endgroup$
– TPaul
Dec 10 '18 at 23:05
add a comment |
$begingroup$
I see. Thank you I think this is where i was stuck.
$endgroup$
– TPaul
Dec 10 '18 at 23:02
$begingroup$
Does this imply $lim_{n rightarrow infty} E[ sum_{i=0}^{n} frac{X_{i}}{n} sum_{i=0}^{n} frac{Y_{i}}{n}] = E[ sum_{i=0}^{n} frac{X_{i}}{n}] E[ sum_{i=0}^{n} frac{Y_{i}}{n}]$ ?
$endgroup$
– TPaul
Dec 10 '18 at 23:05
$begingroup$
I see. Thank you I think this is where i was stuck.
$endgroup$
– TPaul
Dec 10 '18 at 23:02
$begingroup$
I see. Thank you I think this is where i was stuck.
$endgroup$
– TPaul
Dec 10 '18 at 23:02
$begingroup$
Does this imply $lim_{n rightarrow infty} E[ sum_{i=0}^{n} frac{X_{i}}{n} sum_{i=0}^{n} frac{Y_{i}}{n}] = E[ sum_{i=0}^{n} frac{X_{i}}{n}] E[ sum_{i=0}^{n} frac{Y_{i}}{n}]$ ?
$endgroup$
– TPaul
Dec 10 '18 at 23:05
$begingroup$
Does this imply $lim_{n rightarrow infty} E[ sum_{i=0}^{n} frac{X_{i}}{n} sum_{i=0}^{n} frac{Y_{i}}{n}] = E[ sum_{i=0}^{n} frac{X_{i}}{n}] E[ sum_{i=0}^{n} frac{Y_{i}}{n}]$ ?
$endgroup$
– TPaul
Dec 10 '18 at 23:05
add a comment |
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$begingroup$
Are you sure? Isn't covariance a function of variance and variance goes to zero as n goes to infinity?
$endgroup$
– TPaul
Dec 10 '18 at 22:06