What is the intuitive meaning of having a linear relationship between the logs of two variables?












13












$begingroup$


I have two variables which don't show much correlation when plotted against each other as is, but a very clear linear relationship when I plot the logs of each variable agains the other.



So I would end up with a model of the type:



$$log(Y) = a log(X) + b$$ , which is great mathematically but doesn't seem to have the explanatory value of a regular linear model.



How can I interpret such a model?










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  • 2




    $begingroup$
    I've nothing substantial to add to the existing answers, but a logarithm in the outcome and the predictor is an elasticity. Searches for that term should find some good resources for interpreting that relationship, which is not very intuitive.
    $endgroup$
    – Upper_Case
    13 hours ago
















13












$begingroup$


I have two variables which don't show much correlation when plotted against each other as is, but a very clear linear relationship when I plot the logs of each variable agains the other.



So I would end up with a model of the type:



$$log(Y) = a log(X) + b$$ , which is great mathematically but doesn't seem to have the explanatory value of a regular linear model.



How can I interpret such a model?










share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    I've nothing substantial to add to the existing answers, but a logarithm in the outcome and the predictor is an elasticity. Searches for that term should find some good resources for interpreting that relationship, which is not very intuitive.
    $endgroup$
    – Upper_Case
    13 hours ago














13












13








13


1



$begingroup$


I have two variables which don't show much correlation when plotted against each other as is, but a very clear linear relationship when I plot the logs of each variable agains the other.



So I would end up with a model of the type:



$$log(Y) = a log(X) + b$$ , which is great mathematically but doesn't seem to have the explanatory value of a regular linear model.



How can I interpret such a model?










share|cite|improve this question











$endgroup$




I have two variables which don't show much correlation when plotted against each other as is, but a very clear linear relationship when I plot the logs of each variable agains the other.



So I would end up with a model of the type:



$$log(Y) = a log(X) + b$$ , which is great mathematically but doesn't seem to have the explanatory value of a regular linear model.



How can I interpret such a model?







regression correlation log






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 10 hours ago









StubbornAtom

2,8371532




2,8371532










asked 15 hours ago









Akaike's ChildrenAkaike's Children

1005




1005








  • 2




    $begingroup$
    I've nothing substantial to add to the existing answers, but a logarithm in the outcome and the predictor is an elasticity. Searches for that term should find some good resources for interpreting that relationship, which is not very intuitive.
    $endgroup$
    – Upper_Case
    13 hours ago














  • 2




    $begingroup$
    I've nothing substantial to add to the existing answers, but a logarithm in the outcome and the predictor is an elasticity. Searches for that term should find some good resources for interpreting that relationship, which is not very intuitive.
    $endgroup$
    – Upper_Case
    13 hours ago








2




2




$begingroup$
I've nothing substantial to add to the existing answers, but a logarithm in the outcome and the predictor is an elasticity. Searches for that term should find some good resources for interpreting that relationship, which is not very intuitive.
$endgroup$
– Upper_Case
13 hours ago




$begingroup$
I've nothing substantial to add to the existing answers, but a logarithm in the outcome and the predictor is an elasticity. Searches for that term should find some good resources for interpreting that relationship, which is not very intuitive.
$endgroup$
– Upper_Case
13 hours ago










4 Answers
4






active

oldest

votes


















15












$begingroup$

You just need to take exponential of both sides of the equation and you will get a potential relation, that may make sense for some data.



$$log(Y) = alog(X) + b$$



$$exp(log(Y)) = exp(a log(X) + b)$$



$$Y = e^bcdot X^a$$



And since $e^b$ is just a parameter that can take any positive value, this model is equivalent to:



$$Y=c cdot X^a$$



It should be noted that model expression should include the error term, and these change of variables has interesting effects on it:



$$log(Y) = a log(X) + b + epsilon$$



$$Y = e^bcdot X^acdot exp(epsilon)$$



That is, your model with a additive errors abiding to the conditions for OLS (normally distributed errors with constant variance) is equivalent to a potential model with multiplicative errors whose logaritm follows a normal distribution with constant variance.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    OP may be interested to know that this distribution has a name, the log-normal: en.wikipedia.org/wiki/Log-normal_distribution
    $endgroup$
    – gardenhead
    11 hours ago






  • 1




    $begingroup$
    What about the effect of Jensen's inequality? Generally for convex g, $E[g(X)]≥g(E[X])$
    $endgroup$
    – Stats
    10 hours ago





















7












$begingroup$

You can take your model $log(Y)=alog(X)+b$ and calculate the total differential, you will end up with something like :
$$frac{1}YdY=afrac{1}XdX$$
which yields to
$$frac{dY}{dX}frac{X}{Y}=a$$



Hence one simple interpretation of the coefficient $a$ will be the percent change in $Y$ for a percent change in $X$.
This implies furthermore that the variable $Y$ growths at a constant fraction ($a$) of the growth rate of $X$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So if the log-log plot is linear, that would imply a constant growth rate?
    $endgroup$
    – Dimitriy V. Masterov
    14 hours ago










  • $begingroup$
    Not actually, the growth rate of $Y$ will be constant if and only if $a=0$.
    $endgroup$
    – RScrlli
    13 hours ago










  • $begingroup$
    Not over time, the growth rate with respect to the growth in x.
    $endgroup$
    – Dimitriy V. Masterov
    13 hours ago










  • $begingroup$
    reordering doesn't help, i'd remove it
    $endgroup$
    – Aksakal
    13 hours ago










  • $begingroup$
    @DimitriyV.Masterov Ok, then since the $log(Y)$ is linear in $log(X)$ it means that the variable $Y$ grows at a constant fraction of the growth rate of $X$. Is there something wrong with my answer according to you?
    $endgroup$
    – RScrlli
    13 hours ago



















1












$begingroup$

Reconciling the answer by @Rscrill with actual discrete data, consider



$$log(Y_t) = alog(X_t) + b,;;; log(Y_{t-1}) = alog(X_{t-1}) + b$$



$$implies log(Y_t) - log(Y_{t-1}) = aleft[log(X_t)-log(X_{t-1})right]$$



But



$$log(Y_t) - log(Y_{t-1}) = logleft(frac{Y_t}{Y_{t-1}}right) equiv logleft(frac{Y_{t-1}+Delta Y_t}{Y_{t-1}}right) = logleft(1+frac{Delta Y_t}{Y_{t-1}}right)$$



$frac{Delta Y_t}{Y_{t-1}}$ is the percentage change of $Y$ between periods $t-1$ and $t$, or the growth rate of $Y_t$, say $g_{Y_{t}}$. When it is smaller than $0.1$, we have that an acceptable approximation is



$$logleft(1+frac{Delta Y_t}{Y_{t-1}}right) approx frac{Delta Y_t}{Y_{t-1}}=g_{Y_{t}}$$



Therefore we get



$$g_{Y_{t}}approx ag_{X_{t}}$$



which validates in empirical studies the theoretical treatment of @Rscrill.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Intuitively $log$ gives us the order of magnitude of a variable, so we can view the relationship as the orders of magnitudes of the two variables are linearly related. For example, increasing the predictor by one order of magnitude may be associated with an increase of three orders of magnitude of the response.



    When plotting using a log-log plot we see a linear relationship.
    Example I took from Google Images:



    log-log






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      +1 for an intuitive answer to an unintuitive concept. However, the image you have included clearly violates constant error variance across the predictor.
      $endgroup$
      – Frans Rodenburg
      14 mins ago













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    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    15












    $begingroup$

    You just need to take exponential of both sides of the equation and you will get a potential relation, that may make sense for some data.



    $$log(Y) = alog(X) + b$$



    $$exp(log(Y)) = exp(a log(X) + b)$$



    $$Y = e^bcdot X^a$$



    And since $e^b$ is just a parameter that can take any positive value, this model is equivalent to:



    $$Y=c cdot X^a$$



    It should be noted that model expression should include the error term, and these change of variables has interesting effects on it:



    $$log(Y) = a log(X) + b + epsilon$$



    $$Y = e^bcdot X^acdot exp(epsilon)$$



    That is, your model with a additive errors abiding to the conditions for OLS (normally distributed errors with constant variance) is equivalent to a potential model with multiplicative errors whose logaritm follows a normal distribution with constant variance.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      OP may be interested to know that this distribution has a name, the log-normal: en.wikipedia.org/wiki/Log-normal_distribution
      $endgroup$
      – gardenhead
      11 hours ago






    • 1




      $begingroup$
      What about the effect of Jensen's inequality? Generally for convex g, $E[g(X)]≥g(E[X])$
      $endgroup$
      – Stats
      10 hours ago


















    15












    $begingroup$

    You just need to take exponential of both sides of the equation and you will get a potential relation, that may make sense for some data.



    $$log(Y) = alog(X) + b$$



    $$exp(log(Y)) = exp(a log(X) + b)$$



    $$Y = e^bcdot X^a$$



    And since $e^b$ is just a parameter that can take any positive value, this model is equivalent to:



    $$Y=c cdot X^a$$



    It should be noted that model expression should include the error term, and these change of variables has interesting effects on it:



    $$log(Y) = a log(X) + b + epsilon$$



    $$Y = e^bcdot X^acdot exp(epsilon)$$



    That is, your model with a additive errors abiding to the conditions for OLS (normally distributed errors with constant variance) is equivalent to a potential model with multiplicative errors whose logaritm follows a normal distribution with constant variance.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      OP may be interested to know that this distribution has a name, the log-normal: en.wikipedia.org/wiki/Log-normal_distribution
      $endgroup$
      – gardenhead
      11 hours ago






    • 1




      $begingroup$
      What about the effect of Jensen's inequality? Generally for convex g, $E[g(X)]≥g(E[X])$
      $endgroup$
      – Stats
      10 hours ago
















    15












    15








    15





    $begingroup$

    You just need to take exponential of both sides of the equation and you will get a potential relation, that may make sense for some data.



    $$log(Y) = alog(X) + b$$



    $$exp(log(Y)) = exp(a log(X) + b)$$



    $$Y = e^bcdot X^a$$



    And since $e^b$ is just a parameter that can take any positive value, this model is equivalent to:



    $$Y=c cdot X^a$$



    It should be noted that model expression should include the error term, and these change of variables has interesting effects on it:



    $$log(Y) = a log(X) + b + epsilon$$



    $$Y = e^bcdot X^acdot exp(epsilon)$$



    That is, your model with a additive errors abiding to the conditions for OLS (normally distributed errors with constant variance) is equivalent to a potential model with multiplicative errors whose logaritm follows a normal distribution with constant variance.






    share|cite|improve this answer











    $endgroup$



    You just need to take exponential of both sides of the equation and you will get a potential relation, that may make sense for some data.



    $$log(Y) = alog(X) + b$$



    $$exp(log(Y)) = exp(a log(X) + b)$$



    $$Y = e^bcdot X^a$$



    And since $e^b$ is just a parameter that can take any positive value, this model is equivalent to:



    $$Y=c cdot X^a$$



    It should be noted that model expression should include the error term, and these change of variables has interesting effects on it:



    $$log(Y) = a log(X) + b + epsilon$$



    $$Y = e^bcdot X^acdot exp(epsilon)$$



    That is, your model with a additive errors abiding to the conditions for OLS (normally distributed errors with constant variance) is equivalent to a potential model with multiplicative errors whose logaritm follows a normal distribution with constant variance.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 10 hours ago

























    answered 15 hours ago









    PerePere

    4,5571720




    4,5571720












    • $begingroup$
      OP may be interested to know that this distribution has a name, the log-normal: en.wikipedia.org/wiki/Log-normal_distribution
      $endgroup$
      – gardenhead
      11 hours ago






    • 1




      $begingroup$
      What about the effect of Jensen's inequality? Generally for convex g, $E[g(X)]≥g(E[X])$
      $endgroup$
      – Stats
      10 hours ago




















    • $begingroup$
      OP may be interested to know that this distribution has a name, the log-normal: en.wikipedia.org/wiki/Log-normal_distribution
      $endgroup$
      – gardenhead
      11 hours ago






    • 1




      $begingroup$
      What about the effect of Jensen's inequality? Generally for convex g, $E[g(X)]≥g(E[X])$
      $endgroup$
      – Stats
      10 hours ago


















    $begingroup$
    OP may be interested to know that this distribution has a name, the log-normal: en.wikipedia.org/wiki/Log-normal_distribution
    $endgroup$
    – gardenhead
    11 hours ago




    $begingroup$
    OP may be interested to know that this distribution has a name, the log-normal: en.wikipedia.org/wiki/Log-normal_distribution
    $endgroup$
    – gardenhead
    11 hours ago




    1




    1




    $begingroup$
    What about the effect of Jensen's inequality? Generally for convex g, $E[g(X)]≥g(E[X])$
    $endgroup$
    – Stats
    10 hours ago






    $begingroup$
    What about the effect of Jensen's inequality? Generally for convex g, $E[g(X)]≥g(E[X])$
    $endgroup$
    – Stats
    10 hours ago















    7












    $begingroup$

    You can take your model $log(Y)=alog(X)+b$ and calculate the total differential, you will end up with something like :
    $$frac{1}YdY=afrac{1}XdX$$
    which yields to
    $$frac{dY}{dX}frac{X}{Y}=a$$



    Hence one simple interpretation of the coefficient $a$ will be the percent change in $Y$ for a percent change in $X$.
    This implies furthermore that the variable $Y$ growths at a constant fraction ($a$) of the growth rate of $X$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      So if the log-log plot is linear, that would imply a constant growth rate?
      $endgroup$
      – Dimitriy V. Masterov
      14 hours ago










    • $begingroup$
      Not actually, the growth rate of $Y$ will be constant if and only if $a=0$.
      $endgroup$
      – RScrlli
      13 hours ago










    • $begingroup$
      Not over time, the growth rate with respect to the growth in x.
      $endgroup$
      – Dimitriy V. Masterov
      13 hours ago










    • $begingroup$
      reordering doesn't help, i'd remove it
      $endgroup$
      – Aksakal
      13 hours ago










    • $begingroup$
      @DimitriyV.Masterov Ok, then since the $log(Y)$ is linear in $log(X)$ it means that the variable $Y$ grows at a constant fraction of the growth rate of $X$. Is there something wrong with my answer according to you?
      $endgroup$
      – RScrlli
      13 hours ago
















    7












    $begingroup$

    You can take your model $log(Y)=alog(X)+b$ and calculate the total differential, you will end up with something like :
    $$frac{1}YdY=afrac{1}XdX$$
    which yields to
    $$frac{dY}{dX}frac{X}{Y}=a$$



    Hence one simple interpretation of the coefficient $a$ will be the percent change in $Y$ for a percent change in $X$.
    This implies furthermore that the variable $Y$ growths at a constant fraction ($a$) of the growth rate of $X$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      So if the log-log plot is linear, that would imply a constant growth rate?
      $endgroup$
      – Dimitriy V. Masterov
      14 hours ago










    • $begingroup$
      Not actually, the growth rate of $Y$ will be constant if and only if $a=0$.
      $endgroup$
      – RScrlli
      13 hours ago










    • $begingroup$
      Not over time, the growth rate with respect to the growth in x.
      $endgroup$
      – Dimitriy V. Masterov
      13 hours ago










    • $begingroup$
      reordering doesn't help, i'd remove it
      $endgroup$
      – Aksakal
      13 hours ago










    • $begingroup$
      @DimitriyV.Masterov Ok, then since the $log(Y)$ is linear in $log(X)$ it means that the variable $Y$ grows at a constant fraction of the growth rate of $X$. Is there something wrong with my answer according to you?
      $endgroup$
      – RScrlli
      13 hours ago














    7












    7








    7





    $begingroup$

    You can take your model $log(Y)=alog(X)+b$ and calculate the total differential, you will end up with something like :
    $$frac{1}YdY=afrac{1}XdX$$
    which yields to
    $$frac{dY}{dX}frac{X}{Y}=a$$



    Hence one simple interpretation of the coefficient $a$ will be the percent change in $Y$ for a percent change in $X$.
    This implies furthermore that the variable $Y$ growths at a constant fraction ($a$) of the growth rate of $X$.






    share|cite|improve this answer











    $endgroup$



    You can take your model $log(Y)=alog(X)+b$ and calculate the total differential, you will end up with something like :
    $$frac{1}YdY=afrac{1}XdX$$
    which yields to
    $$frac{dY}{dX}frac{X}{Y}=a$$



    Hence one simple interpretation of the coefficient $a$ will be the percent change in $Y$ for a percent change in $X$.
    This implies furthermore that the variable $Y$ growths at a constant fraction ($a$) of the growth rate of $X$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 13 hours ago

























    answered 14 hours ago









    RScrlliRScrlli

    142110




    142110












    • $begingroup$
      So if the log-log plot is linear, that would imply a constant growth rate?
      $endgroup$
      – Dimitriy V. Masterov
      14 hours ago










    • $begingroup$
      Not actually, the growth rate of $Y$ will be constant if and only if $a=0$.
      $endgroup$
      – RScrlli
      13 hours ago










    • $begingroup$
      Not over time, the growth rate with respect to the growth in x.
      $endgroup$
      – Dimitriy V. Masterov
      13 hours ago










    • $begingroup$
      reordering doesn't help, i'd remove it
      $endgroup$
      – Aksakal
      13 hours ago










    • $begingroup$
      @DimitriyV.Masterov Ok, then since the $log(Y)$ is linear in $log(X)$ it means that the variable $Y$ grows at a constant fraction of the growth rate of $X$. Is there something wrong with my answer according to you?
      $endgroup$
      – RScrlli
      13 hours ago


















    • $begingroup$
      So if the log-log plot is linear, that would imply a constant growth rate?
      $endgroup$
      – Dimitriy V. Masterov
      14 hours ago










    • $begingroup$
      Not actually, the growth rate of $Y$ will be constant if and only if $a=0$.
      $endgroup$
      – RScrlli
      13 hours ago










    • $begingroup$
      Not over time, the growth rate with respect to the growth in x.
      $endgroup$
      – Dimitriy V. Masterov
      13 hours ago










    • $begingroup$
      reordering doesn't help, i'd remove it
      $endgroup$
      – Aksakal
      13 hours ago










    • $begingroup$
      @DimitriyV.Masterov Ok, then since the $log(Y)$ is linear in $log(X)$ it means that the variable $Y$ grows at a constant fraction of the growth rate of $X$. Is there something wrong with my answer according to you?
      $endgroup$
      – RScrlli
      13 hours ago
















    $begingroup$
    So if the log-log plot is linear, that would imply a constant growth rate?
    $endgroup$
    – Dimitriy V. Masterov
    14 hours ago




    $begingroup$
    So if the log-log plot is linear, that would imply a constant growth rate?
    $endgroup$
    – Dimitriy V. Masterov
    14 hours ago












    $begingroup$
    Not actually, the growth rate of $Y$ will be constant if and only if $a=0$.
    $endgroup$
    – RScrlli
    13 hours ago




    $begingroup$
    Not actually, the growth rate of $Y$ will be constant if and only if $a=0$.
    $endgroup$
    – RScrlli
    13 hours ago












    $begingroup$
    Not over time, the growth rate with respect to the growth in x.
    $endgroup$
    – Dimitriy V. Masterov
    13 hours ago




    $begingroup$
    Not over time, the growth rate with respect to the growth in x.
    $endgroup$
    – Dimitriy V. Masterov
    13 hours ago












    $begingroup$
    reordering doesn't help, i'd remove it
    $endgroup$
    – Aksakal
    13 hours ago




    $begingroup$
    reordering doesn't help, i'd remove it
    $endgroup$
    – Aksakal
    13 hours ago












    $begingroup$
    @DimitriyV.Masterov Ok, then since the $log(Y)$ is linear in $log(X)$ it means that the variable $Y$ grows at a constant fraction of the growth rate of $X$. Is there something wrong with my answer according to you?
    $endgroup$
    – RScrlli
    13 hours ago




    $begingroup$
    @DimitriyV.Masterov Ok, then since the $log(Y)$ is linear in $log(X)$ it means that the variable $Y$ grows at a constant fraction of the growth rate of $X$. Is there something wrong with my answer according to you?
    $endgroup$
    – RScrlli
    13 hours ago











    1












    $begingroup$

    Reconciling the answer by @Rscrill with actual discrete data, consider



    $$log(Y_t) = alog(X_t) + b,;;; log(Y_{t-1}) = alog(X_{t-1}) + b$$



    $$implies log(Y_t) - log(Y_{t-1}) = aleft[log(X_t)-log(X_{t-1})right]$$



    But



    $$log(Y_t) - log(Y_{t-1}) = logleft(frac{Y_t}{Y_{t-1}}right) equiv logleft(frac{Y_{t-1}+Delta Y_t}{Y_{t-1}}right) = logleft(1+frac{Delta Y_t}{Y_{t-1}}right)$$



    $frac{Delta Y_t}{Y_{t-1}}$ is the percentage change of $Y$ between periods $t-1$ and $t$, or the growth rate of $Y_t$, say $g_{Y_{t}}$. When it is smaller than $0.1$, we have that an acceptable approximation is



    $$logleft(1+frac{Delta Y_t}{Y_{t-1}}right) approx frac{Delta Y_t}{Y_{t-1}}=g_{Y_{t}}$$



    Therefore we get



    $$g_{Y_{t}}approx ag_{X_{t}}$$



    which validates in empirical studies the theoretical treatment of @Rscrill.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Reconciling the answer by @Rscrill with actual discrete data, consider



      $$log(Y_t) = alog(X_t) + b,;;; log(Y_{t-1}) = alog(X_{t-1}) + b$$



      $$implies log(Y_t) - log(Y_{t-1}) = aleft[log(X_t)-log(X_{t-1})right]$$



      But



      $$log(Y_t) - log(Y_{t-1}) = logleft(frac{Y_t}{Y_{t-1}}right) equiv logleft(frac{Y_{t-1}+Delta Y_t}{Y_{t-1}}right) = logleft(1+frac{Delta Y_t}{Y_{t-1}}right)$$



      $frac{Delta Y_t}{Y_{t-1}}$ is the percentage change of $Y$ between periods $t-1$ and $t$, or the growth rate of $Y_t$, say $g_{Y_{t}}$. When it is smaller than $0.1$, we have that an acceptable approximation is



      $$logleft(1+frac{Delta Y_t}{Y_{t-1}}right) approx frac{Delta Y_t}{Y_{t-1}}=g_{Y_{t}}$$



      Therefore we get



      $$g_{Y_{t}}approx ag_{X_{t}}$$



      which validates in empirical studies the theoretical treatment of @Rscrill.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Reconciling the answer by @Rscrill with actual discrete data, consider



        $$log(Y_t) = alog(X_t) + b,;;; log(Y_{t-1}) = alog(X_{t-1}) + b$$



        $$implies log(Y_t) - log(Y_{t-1}) = aleft[log(X_t)-log(X_{t-1})right]$$



        But



        $$log(Y_t) - log(Y_{t-1}) = logleft(frac{Y_t}{Y_{t-1}}right) equiv logleft(frac{Y_{t-1}+Delta Y_t}{Y_{t-1}}right) = logleft(1+frac{Delta Y_t}{Y_{t-1}}right)$$



        $frac{Delta Y_t}{Y_{t-1}}$ is the percentage change of $Y$ between periods $t-1$ and $t$, or the growth rate of $Y_t$, say $g_{Y_{t}}$. When it is smaller than $0.1$, we have that an acceptable approximation is



        $$logleft(1+frac{Delta Y_t}{Y_{t-1}}right) approx frac{Delta Y_t}{Y_{t-1}}=g_{Y_{t}}$$



        Therefore we get



        $$g_{Y_{t}}approx ag_{X_{t}}$$



        which validates in empirical studies the theoretical treatment of @Rscrill.






        share|cite|improve this answer









        $endgroup$



        Reconciling the answer by @Rscrill with actual discrete data, consider



        $$log(Y_t) = alog(X_t) + b,;;; log(Y_{t-1}) = alog(X_{t-1}) + b$$



        $$implies log(Y_t) - log(Y_{t-1}) = aleft[log(X_t)-log(X_{t-1})right]$$



        But



        $$log(Y_t) - log(Y_{t-1}) = logleft(frac{Y_t}{Y_{t-1}}right) equiv logleft(frac{Y_{t-1}+Delta Y_t}{Y_{t-1}}right) = logleft(1+frac{Delta Y_t}{Y_{t-1}}right)$$



        $frac{Delta Y_t}{Y_{t-1}}$ is the percentage change of $Y$ between periods $t-1$ and $t$, or the growth rate of $Y_t$, say $g_{Y_{t}}$. When it is smaller than $0.1$, we have that an acceptable approximation is



        $$logleft(1+frac{Delta Y_t}{Y_{t-1}}right) approx frac{Delta Y_t}{Y_{t-1}}=g_{Y_{t}}$$



        Therefore we get



        $$g_{Y_{t}}approx ag_{X_{t}}$$



        which validates in empirical studies the theoretical treatment of @Rscrill.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 9 hours ago









        Alecos PapadopoulosAlecos Papadopoulos

        42.7k296195




        42.7k296195























            1












            $begingroup$

            Intuitively $log$ gives us the order of magnitude of a variable, so we can view the relationship as the orders of magnitudes of the two variables are linearly related. For example, increasing the predictor by one order of magnitude may be associated with an increase of three orders of magnitude of the response.



            When plotting using a log-log plot we see a linear relationship.
            Example I took from Google Images:



            log-log






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              +1 for an intuitive answer to an unintuitive concept. However, the image you have included clearly violates constant error variance across the predictor.
              $endgroup$
              – Frans Rodenburg
              14 mins ago


















            1












            $begingroup$

            Intuitively $log$ gives us the order of magnitude of a variable, so we can view the relationship as the orders of magnitudes of the two variables are linearly related. For example, increasing the predictor by one order of magnitude may be associated with an increase of three orders of magnitude of the response.



            When plotting using a log-log plot we see a linear relationship.
            Example I took from Google Images:



            log-log






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              +1 for an intuitive answer to an unintuitive concept. However, the image you have included clearly violates constant error variance across the predictor.
              $endgroup$
              – Frans Rodenburg
              14 mins ago
















            1












            1








            1





            $begingroup$

            Intuitively $log$ gives us the order of magnitude of a variable, so we can view the relationship as the orders of magnitudes of the two variables are linearly related. For example, increasing the predictor by one order of magnitude may be associated with an increase of three orders of magnitude of the response.



            When plotting using a log-log plot we see a linear relationship.
            Example I took from Google Images:



            log-log






            share|cite|improve this answer









            $endgroup$



            Intuitively $log$ gives us the order of magnitude of a variable, so we can view the relationship as the orders of magnitudes of the two variables are linearly related. For example, increasing the predictor by one order of magnitude may be associated with an increase of three orders of magnitude of the response.



            When plotting using a log-log plot we see a linear relationship.
            Example I took from Google Images:



            log-log







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 6 hours ago









            qwrqwr

            19011




            19011












            • $begingroup$
              +1 for an intuitive answer to an unintuitive concept. However, the image you have included clearly violates constant error variance across the predictor.
              $endgroup$
              – Frans Rodenburg
              14 mins ago




















            • $begingroup$
              +1 for an intuitive answer to an unintuitive concept. However, the image you have included clearly violates constant error variance across the predictor.
              $endgroup$
              – Frans Rodenburg
              14 mins ago


















            $begingroup$
            +1 for an intuitive answer to an unintuitive concept. However, the image you have included clearly violates constant error variance across the predictor.
            $endgroup$
            – Frans Rodenburg
            14 mins ago






            $begingroup$
            +1 for an intuitive answer to an unintuitive concept. However, the image you have included clearly violates constant error variance across the predictor.
            $endgroup$
            – Frans Rodenburg
            14 mins ago




















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