Correlation and Covariance on Standardized X












0














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







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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
















0














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







share|cite|improve this question
























  • 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














0












0








0







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







share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 29 '18 at 3:21

























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


















  • 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










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
$$operatorname{Cov}(Z_X, Z_Y) = operatorname{Cov}(frac{X-mu_X}{sigma_X}, frac{Y-mu_Y}{sigma_Y}) = frac{operatorname{Cov}(X,Y)}{sigma_X sigma_Y} = operatorname{Corr}(X,Y).$$






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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
    $$operatorname{Cov}(Z_X, Z_Y) = operatorname{Cov}(frac{X-mu_X}{sigma_X}, frac{Y-mu_Y}{sigma_Y}) = frac{operatorname{Cov}(X,Y)}{sigma_X sigma_Y} = operatorname{Corr}(X,Y).$$






    share|cite|improve this 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
      $$operatorname{Cov}(Z_X, Z_Y) = operatorname{Cov}(frac{X-mu_X}{sigma_X}, frac{Y-mu_Y}{sigma_Y}) = frac{operatorname{Cov}(X,Y)}{sigma_X sigma_Y} = operatorname{Corr}(X,Y).$$






      share|cite|improve this 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
        $$operatorname{Cov}(Z_X, Z_Y) = operatorname{Cov}(frac{X-mu_X}{sigma_X}, frac{Y-mu_Y}{sigma_Y}) = frac{operatorname{Cov}(X,Y)}{sigma_X sigma_Y} = operatorname{Corr}(X,Y).$$






        share|cite|improve this answer












        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
        $$operatorname{Cov}(Z_X, Z_Y) = operatorname{Cov}(frac{X-mu_X}{sigma_X}, frac{Y-mu_Y}{sigma_Y}) = frac{operatorname{Cov}(X,Y)}{sigma_X sigma_Y} = operatorname{Corr}(X,Y).$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 29 '18 at 3:44









        angryavian

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