Model fitting for concave curve: test if asymptote or convergence exist

Question

I am asking a double check if conceptually I set up proper model and proceeding in good direction to drive conclusions.

I have a distribution that looks like this:

I want to test if for increasing x values, the slope of the curve flattens.

I want to test if, for x -> Infinity, either exists an asymptote, or test if the curve could be described or approximated by a linear function, and find the slope of the linear function (see the part of the curve after first 4000, 6000 points).

I would like a double check on what I am doing:

I thought to fit the distribution like this - is it correct?

fModel = Fit[%, {1, x, Exp[x]}, x]

For 10000 points:

7.89305 + 0.000332542 x // for 10000 points, slope is ~ 1/3007

The model fits as a linear function but I puzzled because Exp[x] seems to be completely ignored, despite the initial slope of the curve. How does it actually work this fitting?

When I test the same model with other samplings, 100, 1000 points, it seems to produce results with "similar" fitting, where slope seems to goes proportionally to the number of points :

7.89305 + 0.000332542 x // for 10000 points, slope is ~ 1/3007

5.34334 + 0.00331581 x // for 1000 points , slope is ~ 1/301

2.90882 - 7.12146*10^-44 E^x + 0.0317902 x // for 100 points, E^x is negligible, slope ~ 1/30

Is it sufficient to argue that slope of the curve progressively flattens as number of data tends to infinity?

Is it accurate to say that slope "converges" to a constant value "at a rate of 1/(3 * N) " ?

Answer 1

You ask:

Is it sufficient to argue that the slope of the curve progressively flattens as number of data tends to infinity?

If you set a threshold value epsilon for convergence, usually in Mathematical Analysis books, epsilon is set to a very tiny number and if epsilon>estimated value then you can say that it converges given this specific epsilon. Setting the level of epsilon is entirely up to you.

Answer to your second question:

For the rate, given an epsilon value, you should conduct a regression analysis. Given f(x)

f(x) = C Exp[b x]

take natural Log:

Log[f(x)] = Log[C] + b x + e

where e is residual with E[e]=0 assumed, where E is Expected Value operator. Taking derivative of the Log model with respect to x yields:

b=dLog[f(x)]/dx

which is the rate of change. This implies that Estimated(b) is the answer to your second question.