I have two models to predict "Survival" (S in (0;1)) from a "Dose" (D, D≥0):
- a 2-parameter Weibull model:
S(D) = exp( -exp(a*logD-b) )
- and a so-called "multi-target model":
S(D) = 1 - ( 1-exp(-D/t) )^n
where a,b and t,n are the parameters.
A fit of these models to actual data gave strikingly similar predictions, so I wonder if these models are actually two representations of the same thing. But I am unable to show this or to proof that they are not.
I would be grateful for some help. Thanks in advance!
No, these functions are not equivalent.
For example, you can calculate their derivatives with respect to D:
d[Weibull]/dD=exp(-exp(-b)*D^a)*(-exp(-b)*a*D^(a-1))
d[multi-target]/dD=-n*(1-exp(-D/t))^(n-1)*exp(-D/t)/t
As you can see, at D=0, d[Weibull]/dD is either 0 or inf (depending on the sign of (a-1)), but d[multi-target]/dD is non-zero (the sign depends on the sign of t). So these functions have different behavior at D=0, and, thus, are not globally equivalent.
The fact that you get similar predictions from different models may be limited to a certain dataset (that does not "touch" the regions where these models differ significantly) or your datapoints and prediction points may be close enough (in terms of D) and so either model may work as a simple local interpolator/extrapolator.
Great thanks, Konstantin, that helped.
In fact I did calculate the derivatives (used in fitting purposes) but I did not see that I could use them to check if the models are (non-)equivalent. So simple. Thanks again!
After a second though, I was wrong: both functions do have zero derivatives at D=0 (I miscalculated the value of multi-target's derivative at D=0).
Nevertheless, these functions must be not equivalent because it is not possible to "translate" exp(-x^a) into a polynomial of exp(-x) without logarithms (given a1).
To prove the non-equivalence, it is possible to use the following approach: take two arbitrary data-points (D1,S1) and (D2,S2), analytically fit each function to this 2-point-big dataset, and demonstrate that at a third point D3 (not equal to both D1 and D2) these functions are different. I have not tried to do this, but I feel like using a CAS would be very helpful here.
P.S. The "derivatives at D=0" approach seems to be useless because all derivatives (1st, 2nd, 3rd, 4th...) of both functions are either zero or infinite at D=0.
Konstantin, thank you for the revision.
However, as I understand is the derivative of the multi-target model for D=0 zero. I get the solution you posted:
-n*(1-exp(-D/t))^(n-1)*exp(-D/t)/t
and exaluationg this at D=0 gives 0.
However, I can't solve the derivative at D=0 for the Weibull model, it gives:
-exp(-exp(a*logD-b)) * exp(a*logD-b) * a/D
D -> 0: logD -> -Inf:
-exp(-exp(-Inf)) * exp(-Inf) * a/0
-exp(-0) * 0 * a/0
-1 * 0 * Inf
But values of the derivative in close proximity of D=0 are almost 0, and I think that lim(derivative) for D->0 = 0. (I don't know how to show this analytically)
So I get d/dD(D=0) = 0 for both models.
I also tried D-Inf and get the same value of zero for the derivatives of both models, again.
So I followed your hint to fit curves to 2 points. In fact the fitted models are different. Funnily, one shoult not use D3=0 as third point (also not D3=Inf), since the fits will share these values. So both models have 4 points in common (one of them at D=Inf), but never 5 or more.
Here is my quick&dirty R code:
# Points to fit
D
Jochen,
1) The problem in d[multi-target]/dD(D=0) was not the analytical formula, but the fact that (1-exp(-D/t)) is zero at D=0 (that I did not realize last time).
2) Weibull model: transform it into more "sound" form, and it will resolve the "0*inf" uncertainty:
exp(-exp(a*logD-b)) = exp(-D^a/exp(b)) = exp(-(D/t)^a),
where t is equivalent to t from the multi-target model in the sense that it is a scale parameter for D (and here t=exp(b/a)).
3) I am not familiar with R language, but, as I can see, you've done numerical fitting. But in order to strictly prove the non-equivalence of two functions, you need analytical results. On the other hand, from the practical/applied viewpoint, significant numerical difference on one dataset is pretty much enough.
3.5) Of course, D=0 and D=+inf should not be used as comparison points because the values at these points are fixed for both functions by their internal structure. Using two datapoints for the comparison is motivated by the fact that both functions have 2 free parameters.
I think the two are different because if you take natural logs on both sides of each, you get totally different functions. In any case, one depends on a and b, the other on t and n, totally different sets of parameters.
- Badges
- Science topic
- Similar topics
- Mathematics
- Statistics
- Probability
- Probability Theory