Predict probability from 1 point I would like to predict the probability of success (completion) of a task from a single point with some 95%CI.
Every year I get monthly %task completion. If in May I have 70% completion I'm sure that the task will finish till Dec. Same, if in Nov I have 30% completion I won't expect completion till Dec.
Since I get monthly measurements I could have a time-series but for now I'm just wondering how to do that with a single point.
In my head the probability to get completion should be following something like a sigmoid curve. Now, if in June I have finished the 30% what are the chances to complete the task by Dec? (with some CI)
Update following Dave2e's comment
The space is from 0% in Jan to 100% in Dec however e.g. in June I can't know what will happen in Dec. So my actual points are two, 0% in Jan and X% in June. Indeed, many curves can fit but as I'm expecting a sigmoid function to be the true one shouldn't this help?
I would use the following model. How could I fit this two known point (Jan & June) and get an estimation for the last point-Dec?
x <-  seq(1,12, length.out = 100)
beta = .5 
alpha = -3
log_odds <- exp(alpha+beta * x)/(1+exp(alpha+beta * x))
plot(x, y = log_odds)


2nd update
Indeed, my dataset is small that is why my hope to get an estimate is by choosing the non-linear model.
For now I'm using the above function with a fixed alpha so I need to estimate the beta,
If in Sep I got a completion of p=50%
p = .5
t = 9
f <- function(betax) { (log(p/(1-p)) - alpha)/t} #it's the logit function for x = month nr.
beta <- nlm(f,c(.5))$minimum 

x <-  seq(1,12, length.out = 100)
alpha = -3
log_odds <- exp(alpha+beta * x)/(1+exp(alpha+beta * x))
plot(x, y = log_odds, ylim = c(0, 1))
abline(a=0, b= 1/12, col = 'red')

Gives the below,

The output
data.frame(x, log_odds) %>% filter(x %in% c(9, 12))

Confirms that the in Sep I have completion rate of 50% and it gives an estimate for Dec.

Thanks

You don't have enough data. When you only have very little data, you should stick to very basic models because they are less likely to give strange results and would be easier to interpret. In your case, you have only two points, so the simplest thing you can do is use linear interpolation. Yes, this model would be wrong, because at some point it would start predicting probabilities higher than one, but it would be a useful one to solve your problem.
It wouldn't give you the prediction intervals. Using linear interpolation would be equivalent to linear regression, for linear regression, depending on implementation and how you would use it, it would return "not a number" for the intervals, or maybe zero width intervals, or maybe something else, because you don't have enough data. A more realistic answer would be that the intervals are something very wide and unlikely useful.
Linear interpolation would give you a good rough guess and you probably can't expect to get much more than this without cheating yourself. One alternative could be if you have some prior knowledge and could formulate it as a Bayesian model, so the result would depend not only on those two points, but also the prior, but that's a different story.
