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I am a user of Keras with Python, if I know some mathematics I am not very good at algebraic transformation.

The context is the following.
I am trying to reproduce the performances from a paper. In this paper, each individual has his own model: a neural network with the pattern 8-5-7-1. All these models reached a MSE of 100.
In my work, I do not use the same variables, so I found a different pattern. I considered 3 individuals (3 models). I have remarked models (of the same pattern) learn better when the skewness of the target is lower. /!\ The target is not transformed at all at this moment.
In concrete terms:

  1. Model 1: skewness 0.41 manage to learn until 100 MSE (reached the goal)
  2. Model 2: skewness 0.66 manage to learn until 200 MSE not less (do not reach the goal)
  3. Model 3: skewness 0.31 manage to learn until 80 MSE (beyond the reached goal)

So this is why I am thinking about making a target engineering by using the logarithm of any base (while it reduces the skewness it is ok). Nevertheless, it transforms the reached metric: 100 MSE as the maximum threshold. And I am pretty sure I cannot just compute log(MSE).
I searched the formula, of course: $ MSE = \frac{1}{n}\sum_{i=1}^{n}(y_{i}-\bar{y}_i)^{2} $

But I am not able with my limited skills to reconsider it correctly. I need help.
So MSE is dependent on the n observations of the dataset, more the logarithm transformation on y. How can I implement a function that will allow transforming the new form of MSE obtained through a logarithmic transformation in order to compare it with the previous goal: 100 MSE on a non-transformed target?

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You have a $Y_{true}$
Then you calculated $Y^{log}_{true}$


So, your predictions are $Y^{log}_{pred}$
Then you can calculate the $MSE^{log}$ but it can't be compared with 100 MSE.
So, you can do a - $base^{MSE^{log}}$
to make it comparabale to 100 MSE.

A better intuitive approach will be to convert back your $Y^{log}_{pred}$
i.e. $base^{Y^{log}_{pred}}$
Then calculate the MSE

The inverse of log is just about using power to the base

import numpy as np
input = np.array([100, 1000])
log_out = np.log10(input)

input_re = 10**(log_out)
input_re
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  • $\begingroup$ Your notation is confusing. To rise the value to a log power where the log has no input let me puzzled. Anyway, considering it is log(y_true) or not, I doubt we can just use the inverse using the base: base ^ log(MSE) is correct, because considering the formula "n" is not concerned by the logarithm. So, the MSE obtained isn't log(MSE) but MSE = Sum[log(y_true)-log(y_pred)]/n. As far I know discussing with mathematician, to claim it's equivalent (not equal) to use base^MSE, it must be demonstrated if the function is bijective. A field that I don't master at all. $\endgroup$
    – AvyWam
    Aug 23 '20 at 13:47

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