Why the changing of variance of the response variable (heteroscedasticity) is important for a GLM model e.g. Poisson? Why not just fitting an exponential curve?
The distribution model of the response variable should be appropriate. As you say that the expectation of the response seems to be an exponential function of some other variable, I assume that the response is strictly positive and that its variance increases with its expectation. A normal distribution model would not fit in this case. It would then be better to use a model with Gamma or a log-normal distribution. But you mention Poisson -- is your response a count variable that can be modelles using a Poisson process? If so, you certainly should use a Poisson model anyway. Here you should check that assuming a Poisson process is adequate, in which, for instance, the variance should equal the expectation. Very often, the variance is larger than the expectation, what is then called an over-dispersed Poisson. This should then also be considered in your model, e.g. by using a negative binomial model or a quasi-Poisson model (that includes a parameter that is a coefficient defining the relation between variance and the expectation).
If you are interested only in the "best fit line" but not in inference (strandard errors, confidence interavls, p-values), then the distribution model is actually irrelevant. In this case you can fit an exponential curve using "least-squares".
Another option is to fit the logarithms of the response with a straight-line model assuming normal errors. This is helpful for simple inferences but gives geometric expectations instead of the usual (arithmetic) expectations.
All the above solutions assume that the relationship is of the form E(Y|X) = a*Xb. If there is asome additive constant involved, like E(Y|X) = a*Xb+c, then you need to fit a corresponding nonlinear model. Generalized nonlinear models allow you to specify the distribution model as in generalized linear models.
If you try to fit the curve the linear relationship assumptions will be violated. To predict more accurately a linear relationship is a practical approach that includes a constant rate of change of Y with X. Equal variance of residual is also an assumption to minimize the sum square error.
Xuelei Dai -
Jochen noted that "Very often, the variance is larger than the expectation, what is then called an over-dispersed Poisson." Perhaps that is your issue. You mentioned heteroscedasticity, and when variance goes up with expectation, this is a specific degree of heteroscedasticity. Any regression of form Yi|yi* = yi* + e, where yi* is predicted-yi, should have substantial heteroscedasticity, where variance of Yi goes up with yi*, though it may be modified by other factors such as omitted variables or data quality issues. (See https://www.researchgate.net/publication/320853387_Essential_Heteroscedasticity, with bounds on the coefficient of heteroscedasticity as explained by Ken Brewer,
and
https://www.researchgate.net/publication/354854317_WHEN_WOULD_HETEROSCEDASTICITY_IN_REGRESSION_OCCUR.)
Jochen noted you could have "E(Y|X) = a*X^b+c," in which case yi* = axi^b+c.
Best wishes - Jim
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