Should I subtract lower bound from Gamma distributed data before estimating distribution parameters? I have some real world data that reflects waiting time in a system. As it's about waiting times I assume it's Gamma distributed and visual check (histogram overlaid by a fitted Gamma PDF) shows no contradiction (except for the cases when failures kick in, but that's fine).
The waiting times in the system have a lower bound (i.e. it's technically not possible to achieve connection time under some bound).
(Q1) Assuming the lower bound is known, should it be subtracted from the data to most properly fit Gamma distribution to it?
(Q2) If the lower bound is known "partially" (at least c), should I subtract the lower estimate of it from the data?
I'm asking these questions, because it feels a bit wrong to see non-zero PDF in the "impossible" range (closer to zero). At the same time I can just admit that this is the chosen model, it does not account for the lower bound, but otherwise works good/acceptable, so this is a trade off. (Q3) Would this be an adequate position as well?
Thank you for sharing your knowledge!
 A: Suppose the total waiting time is the sum of ten
exponential components, each with mean $\mu = 5.$ Here are two simple ways
to make an approximate model that accounts for a 
minimum waiting time of $2$ units for each component.
In the R simulation below, $W_1$ adds $2$ to each component (making components non-exponential). By contrast,
$W_2$ increases the mean of each exponential component
by $2.$ 


*

*Means are equal: $E(W_1) = E(W_2) = 70.$ 

*Probabilities of forbidden overall waiting times are similar: $P(W_1 < 20) = 0,$ exactly,  $P(W_2 < 20) < 0.001.$ [But I wonder if it is
realistic to have $P(W_1 < 30) \approx 0.]$

*The variance of $W_2$ is larger. [I wonder which variance is more realistic.]
Histograms of $W_1$ (shifted gamma) and $W_2$ are as follows (R code after):

set.seed(1234)
w1 = replicate(10^5, sum(rexp(10, 1/5) + 2)) 
w2 = replicate(10^5, sum(rexp(10, 1/(5+2))))

mean(w1);  sd(w1);  mean(w1 < 20);  mean(w1 < 30)
[1] 70.01318  # aprx E(W1) = 70
[1] 15.77648  # aprx E(W1) = 15.81
[1] 0         # P(W1 < 20) = 0
[1] 2e-05     # P(W1 < 30) aprx 0. Really?

mean(w2);  sd(w2);  mean(w2 < 20)
[1] 70.10161
[1] 22.18051
[1] 0.00074
[1] 0.01289

par(mfrow=c(2,1)); mx = max(w1,w2)
 hist(w1, prob=T, col="skyblue2", xlim=c(0,mx))
  curve(dgamma(x-20, 10, 1/5), add=T, lwd=2)
  abline(v = 20, col="red")
 hist(w2, prob=T, col="skyblue2", xlim=c(0,mx))
  curve(dgamma(x, 10, 1/7), add=T, lwd=2)
par(mfrow=c(1,1))

