Correlation and Covariance on Standardized X
I am stuck on the following problem:
Let $Z_X$ be the standardized $X$, $Z_X=(X-mu_X)/sigma_X$, and let $Z_Y$ be the standardized $Y$, $Z_Y=(Y-mu_Y)/sigma_Y$. Show that $Corr(X,Y)=Cov(Z_X,Z_Y)=E(Z_XZ_Y)$.
I have tried just plugging in the definition into what we are trying to prove but I don't know how $ Corr(X,Y)=Cov(frac{X-mu_X}{sigma_X},frac{Y-mu_Y}{sigma_Y})=E(frac{X-mu_X}{sigma_X},frac{Y-mu_Y}{sigma_Y}) $ helps. I think I have to use the facts that $Cov(aX+b,cY+d)=acCov(X,Y)$ and $Corr(aX+b,cY+d)=Corr(X,Y)$ when $ a $ and $ c $ the same sign, but I am not sure where exactly. Any suggestions?
statistics random-variables covariance correlation expected-value
asked Nov 29 '18 at 3:10
D. Wei
425
|
show 1 more comment
I am stuck on the following problem:
Let $Z_X$ be the standardized $X$, $Z_X=(X-mu_X)/sigma_X$, and let $Z_Y$ be the standardized $Y$, $Z_Y=(Y-mu_Y)/sigma_Y$. Show that $Corr(X,Y)=Cov(Z_X,Z_Y)=E(Z_XZ_Y)$.
I have tried just plugging in the definition into what we are trying to prove but I don't know how $ Corr(X,Y)=Cov(frac{X-mu_X}{sigma_X},frac{Y-mu_Y}{sigma_Y})=E(frac{X-mu_X}{sigma_X},frac{Y-mu_Y}{sigma_Y}) $ helps. I think I have to use the facts that $Cov(aX+b,cY+d)=acCov(X,Y)$ and $Corr(aX+b,cY+d)=Corr(X,Y)$ when $ a $ and $ c $ the same sign, but I am not sure where exactly. Any suggestions?
statistics random-variables covariance correlation expected-value
asked Nov 29 '18 at 3:10
D. Wei
425
It's not totally clear to me what you are trying to prove. Is there a problem statement missing, before your sentence "I have tried..." ?
– kimchi lover
Nov 29 '18 at 3:15
Sorry, that was a typo. Fixed.
– D. Wei
Nov 29 '18 at 3:16
Is $Z_Y = (Y-mu_Y)/sigma_Y$ not the same as your definition of "let $Z_Y$ be the standardized $Y$"?
– angryavian
Nov 29 '18 at 3:16
Yes it is. The place where I pulled it from only briefly touched on standardization so the author probably felt the need to define it again here.
– D. Wei
Nov 29 '18 at 3:18
I assume the real problem has to do with correlations? Maybe that the correlation of $X$ and $Y$ is the same as that of $Z_X$ and $Z_Y$?
– kimchi lover
Nov 29 '18 at 3:19
|
show 1 more comment
I am stuck on the following problem:
Let $Z_X$ be the standardized $X$, $Z_X=(X-mu_X)/sigma_X$, and let $Z_Y$ be the standardized $Y$, $Z_Y=(Y-mu_Y)/sigma_Y$. Show that $Corr(X,Y)=Cov(Z_X,Z_Y)=E(Z_XZ_Y)$.
I have tried just plugging in the definition into what we are trying to prove but I don't know how $ Corr(X,Y)=Cov(frac{X-mu_X}{sigma_X},frac{Y-mu_Y}{sigma_Y})=E(frac{X-mu_X}{sigma_X},frac{Y-mu_Y}{sigma_Y}) $ helps. I think I have to use the facts that $Cov(aX+b,cY+d)=acCov(X,Y)$ and $Corr(aX+b,cY+d)=Corr(X,Y)$ when $ a $ and $ c $ the same sign, but I am not sure where exactly. Any suggestions?
statistics random-variables covariance correlation expected-value
asked Nov 29 '18 at 3:10
D. Wei
425
I am stuck on the following problem:
Let $Z_X$ be the standardized $X$, $Z_X=(X-mu_X)/sigma_X$, and let $Z_Y$ be the standardized $Y$, $Z_Y=(Y-mu_Y)/sigma_Y$. Show that $Corr(X,Y)=Cov(Z_X,Z_Y)=E(Z_XZ_Y)$.
I have tried just plugging in the definition into what we are trying to prove but I don't know how $ Corr(X,Y)=Cov(frac{X-mu_X}{sigma_X},frac{Y-mu_Y}{sigma_Y})=E(frac{X-mu_X}{sigma_X},frac{Y-mu_Y}{sigma_Y}) $ helps. I think I have to use the facts that $Cov(aX+b,cY+d)=acCov(X,Y)$ and $Corr(aX+b,cY+d)=Corr(X,Y)$ when $ a $ and $ c $ the same sign, but I am not sure where exactly. Any suggestions?
statistics random-variables covariance correlation expected-value
statistics random-variables covariance correlation expected-value
asked Nov 29 '18 at 3:10
D. Wei
425
asked Nov 29 '18 at 3:10
D. Wei
425
edited Nov 29 '18 at 3:21
asked Nov 29 '18 at 3:10
D. Wei
425
asked Nov 29 '18 at 3:10
D. Wei
425
asked Nov 29 '18 at 3:10
D. Wei
425
425
It's not totally clear to me what you are trying to prove. Is there a problem statement missing, before your sentence "I have tried..." ?
– kimchi lover
Nov 29 '18 at 3:15
Sorry, that was a typo. Fixed.
– D. Wei
Nov 29 '18 at 3:16
Is $Z_Y = (Y-mu_Y)/sigma_Y$ not the same as your definition of "let $Z_Y$ be the standardized $Y$"?
– angryavian
Nov 29 '18 at 3:16
Yes it is. The place where I pulled it from only briefly touched on standardization so the author probably felt the need to define it again here.
– D. Wei
Nov 29 '18 at 3:18
I assume the real problem has to do with correlations? Maybe that the correlation of $X$ and $Y$ is the same as that of $Z_X$ and $Z_Y$?
– kimchi lover
Nov 29 '18 at 3:19
|
show 1 more comment
It's not totally clear to me what you are trying to prove. Is there a problem statement missing, before your sentence "I have tried..." ?
– kimchi lover
Nov 29 '18 at 3:15
Sorry, that was a typo. Fixed.
– D. Wei
Nov 29 '18 at 3:16
Is $Z_Y = (Y-mu_Y)/sigma_Y$ not the same as your definition of "let $Z_Y$ be the standardized $Y$"?
– angryavian
Nov 29 '18 at 3:16
Yes it is. The place where I pulled it from only briefly touched on standardization so the author probably felt the need to define it again here.
– D. Wei
Nov 29 '18 at 3:18
I assume the real problem has to do with correlations? Maybe that the correlation of $X$ and $Y$ is the same as that of $Z_X$ and $Z_Y$?
– kimchi lover
Nov 29 '18 at 3:19
It's not totally clear to me what you are trying to prove. Is there a problem statement missing, before your sentence "I have tried..." ?
– kimchi lover
Nov 29 '18 at 3:15
It's not totally clear to me what you are trying to prove. Is there a problem statement missing, before your sentence "I have tried..." ?
– kimchi lover
Nov 29 '18 at 3:15
Sorry, that was a typo. Fixed.
– D. Wei
Nov 29 '18 at 3:16
Sorry, that was a typo. Fixed.
– D. Wei
Nov 29 '18 at 3:16
Is $Z_Y = (Y-mu_Y)/sigma_Y$ not the same as your definition of "let $Z_Y$ be the standardized $Y$"?
– angryavian
Nov 29 '18 at 3:16
Is $Z_Y = (Y-mu_Y)/sigma_Y$ not the same as your definition of "let $Z_Y$ be the standardized $Y$"?
– angryavian
Nov 29 '18 at 3:16
Yes it is. The place where I pulled it from only briefly touched on standardization so the author probably felt the need to define it again here.
– D. Wei
Nov 29 '18 at 3:18
Yes it is. The place where I pulled it from only briefly touched on standardization so the author probably felt the need to define it again here.
– D. Wei
Nov 29 '18 at 3:18
I assume the real problem has to do with correlations? Maybe that the correlation of $X$ and $Y$ is the same as that of $Z_X$ and $Z_Y$?
– kimchi lover
Nov 29 '18 at 3:19
I assume the real problem has to do with correlations? Maybe that the correlation of $X$ and $Y$ is the same as that of $Z_X$ and $Z_Y$?
– kimchi lover
Nov 29 '18 at 3:19
|
show 1 more comment
1 Answer
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From the formula for covariance, we have
$$operatorname{Cov}(Z_X, Z_Y) = E[Z_X Z_Y] - E[Z_X] E[Z_Y].$$
Note that $E[Z_X]=0$ and $E[Z_Y] = 0$ (why?), so
$$operatorname{Cov}(Z_X, Z_Y) = E[Z_X Z_Y ].$$
Using the facts mentioned at the end of your question, we have
$$operatorname{Cov}(Z_X, Z_Y) = operatorname{Cov}(frac{X-mu_X}{sigma_X}, frac{Y-mu_Y}{sigma_Y}) = frac{operatorname{Cov}(X,Y)}{sigma_X sigma_Y} = operatorname{Corr}(X,Y).$$
answered Nov 29 '18 at 3:44
angryavian
39.2k23280
add a comment |
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1 Answer
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1 Answer
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From the formula for covariance, we have
$$operatorname{Cov}(Z_X, Z_Y) = E[Z_X Z_Y] - E[Z_X] E[Z_Y].$$
Note that $E[Z_X]=0$ and $E[Z_Y] = 0$ (why?), so
$$operatorname{Cov}(Z_X, Z_Y) = E[Z_X Z_Y ].$$
Using the facts mentioned at the end of your question, we have
$$operatorname{Cov}(Z_X, Z_Y) = operatorname{Cov}(frac{X-mu_X}{sigma_X}, frac{Y-mu_Y}{sigma_Y}) = frac{operatorname{Cov}(X,Y)}{sigma_X sigma_Y} = operatorname{Corr}(X,Y).$$
answered Nov 29 '18 at 3:44
angryavian
39.2k23280
add a comment |
From the formula for covariance, we have
$$operatorname{Cov}(Z_X, Z_Y) = E[Z_X Z_Y] - E[Z_X] E[Z_Y].$$
Note that $E[Z_X]=0$ and $E[Z_Y] = 0$ (why?), so
$$operatorname{Cov}(Z_X, Z_Y) = E[Z_X Z_Y ].$$
Using the facts mentioned at the end of your question, we have
$$operatorname{Cov}(Z_X, Z_Y) = operatorname{Cov}(frac{X-mu_X}{sigma_X}, frac{Y-mu_Y}{sigma_Y}) = frac{operatorname{Cov}(X,Y)}{sigma_X sigma_Y} = operatorname{Corr}(X,Y).$$
answered Nov 29 '18 at 3:44
angryavian
39.2k23280
add a comment |
From the formula for covariance, we have
$$operatorname{Cov}(Z_X, Z_Y) = E[Z_X Z_Y] - E[Z_X] E[Z_Y].$$
Note that $E[Z_X]=0$ and $E[Z_Y] = 0$ (why?), so
$$operatorname{Cov}(Z_X, Z_Y) = E[Z_X Z_Y ].$$
Using the facts mentioned at the end of your question, we have
$$operatorname{Cov}(Z_X, Z_Y) = operatorname{Cov}(frac{X-mu_X}{sigma_X}, frac{Y-mu_Y}{sigma_Y}) = frac{operatorname{Cov}(X,Y)}{sigma_X sigma_Y} = operatorname{Corr}(X,Y).$$
answered Nov 29 '18 at 3:44
angryavian
39.2k23280
From the formula for covariance, we have
$$operatorname{Cov}(Z_X, Z_Y) = E[Z_X Z_Y] - E[Z_X] E[Z_Y].$$
Note that $E[Z_X]=0$ and $E[Z_Y] = 0$ (why?), so
$$operatorname{Cov}(Z_X, Z_Y) = E[Z_X Z_Y ].$$
Using the facts mentioned at the end of your question, we have
$$operatorname{Cov}(Z_X, Z_Y) = operatorname{Cov}(frac{X-mu_X}{sigma_X}, frac{Y-mu_Y}{sigma_Y}) = frac{operatorname{Cov}(X,Y)}{sigma_X sigma_Y} = operatorname{Corr}(X,Y).$$
answered Nov 29 '18 at 3:44
angryavian
39.2k23280
answered Nov 29 '18 at 3:44
angryavian
39.2k23280
answered Nov 29 '18 at 3:44
angryavian
39.2k23280
answered Nov 29 '18 at 3:44
angryavian
39.2k23280
39.2k23280
add a comment |
add a comment |
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It's not totally clear to me what you are trying to prove. Is there a problem statement missing, before your sentence "I have tried..." ?
– kimchi lover
Nov 29 '18 at 3:15
Sorry, that was a typo. Fixed.
– D. Wei
Nov 29 '18 at 3:16
Is $Z_Y = (Y-mu_Y)/sigma_Y$ not the same as your definition of "let $Z_Y$ be the standardized $Y$"?
– angryavian
Nov 29 '18 at 3:16
Yes it is. The place where I pulled it from only briefly touched on standardization so the author probably felt the need to define it again here.
– D. Wei
Nov 29 '18 at 3:18
I assume the real problem has to do with correlations? Maybe that the correlation of $X$ and $Y$ is the same as that of $Z_X$ and $Z_Y$?
– kimchi lover
Nov 29 '18 at 3:19