What is the intuitive meaning of having a linear relationship between the logs of two variables?
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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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add a comment |
$begingroup$
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?
regression correlation log
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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.
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– Upper_Case
8 hours ago
add a comment |
$begingroup$
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?
regression correlation log
$endgroup$
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
regression correlation log
edited 6 hours ago
StubbornAtom
2,8371532
2,8371532
asked 11 hours ago
Akaike's ChildrenAkaike's Children
855
855
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
8 hours ago
add a comment |
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
8 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
8 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
8 hours ago
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
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.
$endgroup$
$begingroup$
OP may be interested to know that this distribution has a name, the log-normal: en.wikipedia.org/wiki/Log-normal_distribution
$endgroup$
– gardenhead
7 hours ago
1
$begingroup$
What about the effect of Jensen's inequality? Generally for convex g, $E[g(X)]≥g(E[X])$
$endgroup$
– Stats
6 hours ago
add a comment |
$begingroup$
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 $X$.
$endgroup$
$begingroup$
So if the log-log plot is linear, that would imply a constant growth rate?
$endgroup$
– Dimitriy V. Masterov
9 hours ago
$begingroup$
Not actually, the growth rate of $Y$ will be constant if and only if $a=0$.
$endgroup$
– RScrlli
8 hours ago
$begingroup$
Not over time, the growth rate with respect to the growth in x.
$endgroup$
– Dimitriy V. Masterov
8 hours ago
$begingroup$
reordering doesn't help, i'd remove it
$endgroup$
– Aksakal
8 hours ago
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@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?
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– RScrlli
8 hours ago
|
show 2 more comments
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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.
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add a comment |
$begingroup$
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:
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add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
$endgroup$
$begingroup$
OP may be interested to know that this distribution has a name, the log-normal: en.wikipedia.org/wiki/Log-normal_distribution
$endgroup$
– gardenhead
7 hours ago
1
$begingroup$
What about the effect of Jensen's inequality? Generally for convex g, $E[g(X)]≥g(E[X])$
$endgroup$
– Stats
6 hours ago
add a comment |
$begingroup$
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.
$endgroup$
$begingroup$
OP may be interested to know that this distribution has a name, the log-normal: en.wikipedia.org/wiki/Log-normal_distribution
$endgroup$
– gardenhead
7 hours ago
1
$begingroup$
What about the effect of Jensen's inequality? Generally for convex g, $E[g(X)]≥g(E[X])$
$endgroup$
– Stats
6 hours ago
add a comment |
$begingroup$
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.
$endgroup$
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.
edited 6 hours ago
answered 11 hours ago
PerePere
4,5471720
4,5471720
$begingroup$
OP may be interested to know that this distribution has a name, the log-normal: en.wikipedia.org/wiki/Log-normal_distribution
$endgroup$
– gardenhead
7 hours ago
1
$begingroup$
What about the effect of Jensen's inequality? Generally for convex g, $E[g(X)]≥g(E[X])$
$endgroup$
– Stats
6 hours ago
add a comment |
$begingroup$
OP may be interested to know that this distribution has a name, the log-normal: en.wikipedia.org/wiki/Log-normal_distribution
$endgroup$
– gardenhead
7 hours ago
1
$begingroup$
What about the effect of Jensen's inequality? Generally for convex g, $E[g(X)]≥g(E[X])$
$endgroup$
– Stats
6 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
7 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
7 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
6 hours ago
$begingroup$
What about the effect of Jensen's inequality? Generally for convex g, $E[g(X)]≥g(E[X])$
$endgroup$
– Stats
6 hours ago
add a comment |
$begingroup$
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 $X$.
$endgroup$
$begingroup$
So if the log-log plot is linear, that would imply a constant growth rate?
$endgroup$
– Dimitriy V. Masterov
9 hours ago
$begingroup$
Not actually, the growth rate of $Y$ will be constant if and only if $a=0$.
$endgroup$
– RScrlli
8 hours ago
$begingroup$
Not over time, the growth rate with respect to the growth in x.
$endgroup$
– Dimitriy V. Masterov
8 hours ago
$begingroup$
reordering doesn't help, i'd remove it
$endgroup$
– Aksakal
8 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
8 hours ago
|
show 2 more comments
$begingroup$
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 $X$.
$endgroup$
$begingroup$
So if the log-log plot is linear, that would imply a constant growth rate?
$endgroup$
– Dimitriy V. Masterov
9 hours ago
$begingroup$
Not actually, the growth rate of $Y$ will be constant if and only if $a=0$.
$endgroup$
– RScrlli
8 hours ago
$begingroup$
Not over time, the growth rate with respect to the growth in x.
$endgroup$
– Dimitriy V. Masterov
8 hours ago
$begingroup$
reordering doesn't help, i'd remove it
$endgroup$
– Aksakal
8 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
8 hours ago
|
show 2 more comments
$begingroup$
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 $X$.
$endgroup$
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 $X$.
edited 8 hours ago
answered 10 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
9 hours ago
$begingroup$
Not actually, the growth rate of $Y$ will be constant if and only if $a=0$.
$endgroup$
– RScrlli
8 hours ago
$begingroup$
Not over time, the growth rate with respect to the growth in x.
$endgroup$
– Dimitriy V. Masterov
8 hours ago
$begingroup$
reordering doesn't help, i'd remove it
$endgroup$
– Aksakal
8 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
8 hours ago
|
show 2 more comments
$begingroup$
So if the log-log plot is linear, that would imply a constant growth rate?
$endgroup$
– Dimitriy V. Masterov
9 hours ago
$begingroup$
Not actually, the growth rate of $Y$ will be constant if and only if $a=0$.
$endgroup$
– RScrlli
8 hours ago
$begingroup$
Not over time, the growth rate with respect to the growth in x.
$endgroup$
– Dimitriy V. Masterov
8 hours ago
$begingroup$
reordering doesn't help, i'd remove it
$endgroup$
– Aksakal
8 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
8 hours ago
$begingroup$
So if the log-log plot is linear, that would imply a constant growth rate?
$endgroup$
– Dimitriy V. Masterov
9 hours ago
$begingroup$
So if the log-log plot is linear, that would imply a constant growth rate?
$endgroup$
– Dimitriy V. Masterov
9 hours ago
$begingroup$
Not actually, the growth rate of $Y$ will be constant if and only if $a=0$.
$endgroup$
– RScrlli
8 hours ago
$begingroup$
Not actually, the growth rate of $Y$ will be constant if and only if $a=0$.
$endgroup$
– RScrlli
8 hours ago
$begingroup$
Not over time, the growth rate with respect to the growth in x.
$endgroup$
– Dimitriy V. Masterov
8 hours ago
$begingroup$
Not over time, the growth rate with respect to the growth in x.
$endgroup$
– Dimitriy V. Masterov
8 hours ago
$begingroup$
reordering doesn't help, i'd remove it
$endgroup$
– Aksakal
8 hours ago
$begingroup$
reordering doesn't help, i'd remove it
$endgroup$
– Aksakal
8 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
8 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
8 hours ago
|
show 2 more comments
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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.
answered 5 hours ago
Alecos PapadopoulosAlecos Papadopoulos
42.7k296195
42.7k296195
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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:
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add a comment |
$begingroup$
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:
$endgroup$
add a comment |
$begingroup$
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:
$endgroup$
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:
answered 2 hours ago
qwrqwr
18011
18011
add a comment |
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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.
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– Upper_Case
8 hours ago