Covariance and Law of Large numbers












0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














0












0








0





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 22:08







TPaul

















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


















  • $begingroup$
    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










2 Answers
2






active

oldest

votes


















2












$begingroup$

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 $E[XY] = E[X] E[Y]$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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



















1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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













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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

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 $E[XY] = E[X] E[Y]$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















2












$begingroup$

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 $E[XY] = E[X] E[Y]$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














2












2








2





$begingroup$

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 $E[XY] = E[X] E[Y]$.






share|cite|improve this answer









$endgroup$



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 $E[XY] = E[X] E[Y]$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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











1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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


















1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















1












1








1





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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




















  • $begingroup$
    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




















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Ellipse (mathématiques)

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