What's the meaning of multiplicative errors and additive errors?
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Such skewed, thick-tailed data suggest a model with multiplicative errors instead of additive errors. A standard solution is to transform the dependent variable by taking the natural logarithm.
Can anyone explain multiplicative errors and additive errors here?
Many thanks in advance!
statistics
asked Nov 11 at 16:11
Yao Zhao
215
add a comment |
up vote
0
down vote
favorite
Such skewed, thick-tailed data suggest a model with multiplicative errors instead of additive errors. A standard solution is to transform the dependent variable by taking the natural logarithm.
Can anyone explain multiplicative errors and additive errors here?
Many thanks in advance!
statistics
asked Nov 11 at 16:11
Yao Zhao
215
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Such skewed, thick-tailed data suggest a model with multiplicative errors instead of additive errors. A standard solution is to transform the dependent variable by taking the natural logarithm.
Can anyone explain multiplicative errors and additive errors here?
Many thanks in advance!
statistics
asked Nov 11 at 16:11
Yao Zhao
215
Such skewed, thick-tailed data suggest a model with multiplicative errors instead of additive errors. A standard solution is to transform the dependent variable by taking the natural logarithm.
Can anyone explain multiplicative errors and additive errors here?
Many thanks in advance!
statistics
statistics
asked Nov 11 at 16:11
Yao Zhao
215
asked Nov 11 at 16:11
Yao Zhao
215
asked Nov 11 at 16:11
Yao Zhao
215
asked Nov 11 at 16:11
Yao Zhao
215
asked Nov 11 at 16:11
Yao Zhao
215
215
add a comment |
add a comment |
1 Answer
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There is not enough context here, but here is a general explanation:
Let's say you are trying to measure an underlying signal $X$ and the observed signal $Y$ is related to the actual signal by $Y = f(X) + epsilon$, where $f(cdot)$ is some function (either known or estimated) and $epsilon$ is the measurement error or noise. In this case, the error is additive, because it adds to the model $Y = f(X)$.
In an alternative scenario, consider that $Y$ and $X$ are related by $Y = g(X)epsilon$. In this case, the error term $epsilon$ is multiplicative, because it multiplies with the model $Y = g(X)$. By applying the log transformation $log(Y) = log(g(X)) + log(epsilon)$, we are back to an additive error framework.
answered Nov 21 at 6:29
Aditya Dua
6458
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
There is not enough context here, but here is a general explanation:
Let's say you are trying to measure an underlying signal $X$ and the observed signal $Y$ is related to the actual signal by $Y = f(X) + epsilon$, where $f(cdot)$ is some function (either known or estimated) and $epsilon$ is the measurement error or noise. In this case, the error is additive, because it adds to the model $Y = f(X)$.
In an alternative scenario, consider that $Y$ and $X$ are related by $Y = g(X)epsilon$. In this case, the error term $epsilon$ is multiplicative, because it multiplies with the model $Y = g(X)$. By applying the log transformation $log(Y) = log(g(X)) + log(epsilon)$, we are back to an additive error framework.
answered Nov 21 at 6:29
Aditya Dua
6458
add a comment |
up vote
2
down vote
There is not enough context here, but here is a general explanation:
Let's say you are trying to measure an underlying signal $X$ and the observed signal $Y$ is related to the actual signal by $Y = f(X) + epsilon$, where $f(cdot)$ is some function (either known or estimated) and $epsilon$ is the measurement error or noise. In this case, the error is additive, because it adds to the model $Y = f(X)$.
In an alternative scenario, consider that $Y$ and $X$ are related by $Y = g(X)epsilon$. In this case, the error term $epsilon$ is multiplicative, because it multiplies with the model $Y = g(X)$. By applying the log transformation $log(Y) = log(g(X)) + log(epsilon)$, we are back to an additive error framework.
answered Nov 21 at 6:29
Aditya Dua
6458
add a comment |
up vote
2
down vote
up vote
2
down vote
There is not enough context here, but here is a general explanation:
Let's say you are trying to measure an underlying signal $X$ and the observed signal $Y$ is related to the actual signal by $Y = f(X) + epsilon$, where $f(cdot)$ is some function (either known or estimated) and $epsilon$ is the measurement error or noise. In this case, the error is additive, because it adds to the model $Y = f(X)$.
In an alternative scenario, consider that $Y$ and $X$ are related by $Y = g(X)epsilon$. In this case, the error term $epsilon$ is multiplicative, because it multiplies with the model $Y = g(X)$. By applying the log transformation $log(Y) = log(g(X)) + log(epsilon)$, we are back to an additive error framework.
answered Nov 21 at 6:29
Aditya Dua
6458
There is not enough context here, but here is a general explanation:
Let's say you are trying to measure an underlying signal $X$ and the observed signal $Y$ is related to the actual signal by $Y = f(X) + epsilon$, where $f(cdot)$ is some function (either known or estimated) and $epsilon$ is the measurement error or noise. In this case, the error is additive, because it adds to the model $Y = f(X)$.
In an alternative scenario, consider that $Y$ and $X$ are related by $Y = g(X)epsilon$. In this case, the error term $epsilon$ is multiplicative, because it multiplies with the model $Y = g(X)$. By applying the log transformation $log(Y) = log(g(X)) + log(epsilon)$, we are back to an additive error framework.
answered Nov 21 at 6:29
Aditya Dua
6458
answered Nov 21 at 6:29
Aditya Dua
6458
answered Nov 21 at 6:29
Aditya Dua
6458
answered Nov 21 at 6:29
Aditya Dua
6458
6458
add a comment |
add a comment |
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