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You would need to have some idea how relapse times are distributed.

For example, if they are exponentially distributed, then the mean of a moderate number of relapse times is roughly 17% of the way from the minimum to the maximum.

Here is a simulation for 200 subjects; results for 100 and were about 19% and results for 500 were about the 15%.

set.seed(2021)
m = 10^5;  v = a = w = numeric(m)
for (i in 1:m) {
 x = rexp(200)
 v[i] = min(x); a[i] = mean(x); w[i] = max(x)
 }
mean(v); mean(a); mean(w)
[1] 0.00499292
[1] 0.9999175
[1] 5.875548
(mean(a)-mean(v))/(mean(w)-mean(v))
[1] 0.1694771

Note: For uniform and normal relapse distributions (both symmetrical), results were about 50%.

You would need to have some idea how relapse times are distributed within the interval.

For example, if they are exponentially distributed, then the mean of a moderate number of relapse times is roughly 17% of the way from the minimum to the maximum.

Here is a simulation for 200 subjects; results for 100 and were about 19% and results for 500 were about the 15%.

set.seed(2021)
m = 10^5;  v = a = w = numeric(m)
for (i in 1:m) {
 x = rexp(200)
 v[i] = min(x); a[i] = mean(x); w[i] = max(x)
 }
mean(v); mean(a); mean(w)
[1] 0.00499292
[1] 0.9999175
[1] 5.875548
(mean(a)-mean(v))/(mean(w)-mean(v))
[1] 0.1694771

Note: For a uniform relapse distribution, results were about 50%.

You would need to have some idea how relapse times are distributed.

For example, if they are exponentially distributed, then the mean of a moderate number of relapse times is about 40% of the way from the minimum to the maximum.

Here is a simulation for 200 subjects; results for 100 and 500 were about the same. (It helps that the exponential distribution has the no-memory property. So, the first relapse doesn't need to coincide with the low end of the interval.)

set.seed(2021)
m = 10^5;  v = a = w = numeric(m)
for (i in 1:m) {
 s = rexp(200)
 v[i] = min(x); a[i] = mean(x); w[i] = max(x)
 }
mean(v); mean(a); mean(w)
[1] 40.64162
[1] 54.0478
[1] 74.16839
(mean(a)-mean(v))/(mean(w)-mean(v))
[1] 0.399865 

You would need to have some idea how relapse times are distributed.

For example, if they are exponentially distributed, then the mean of a moderate number of relapse times is roughly 17% of the way from the minimum to the maximum.

Here is a simulation for 200 subjects; results for 100 and were about 19% and results for 500 were about the 15%.

set.seed(2021)
m = 10^5;  v = a = w = numeric(m)
for (i in 1:m) {
 x = rexp(200)
 v[i] = min(x); a[i] = mean(x); w[i] = max(x)
 }
mean(v); mean(a); mean(w)
[1] 0.00499292
[1] 0.9999175
[1] 5.875548
(mean(a)-mean(v))/(mean(w)-mean(v))
[1] 0.1694771

Note: For uniform and normal relapse distributions (both symmetrical), results were about 50%.

You would need to have some idea how relapse times are distributed.

For example, if they are exponentially distributed, then the mean of a moderate number of relapse times is about 40% of the way from the minimum to the maximum.

Here is a simulation for 200 subjects; results for 100 and 500 were about the same.

set.seed(2021)
m = 10^5;  v = a = w = numeric(m)
for (i in 1:m) {
 s = rexp(200)
 v[i] = min(x); a[i] = mean(x); w[i] = max(x)
 }
mean(v); mean(a); mean(w)
[1] 40.64162
[1] 54.0478
[1] 74.16839
(mean(a)-mean(v))/(mean(w)-mean(v))
[1] 0.399865 

You would need to have some idea how relapse times are distributed.

For example, if they are exponentially distributed, then the mean of a moderate number of relapse times is about 40% of the way from the minimum to the maximum.

Here is a simulation for 200 subjects; results for 100 and 500 were about the same. (It helps that the exponential distribution has the no-memory property. So, the first relapse doesn't need to coincide with the low end of the interval.)

set.seed(2021)
m = 10^5;  v = a = w = numeric(m)
for (i in 1:m) {
 s = rexp(200)
 v[i] = min(x); a[i] = mean(x); w[i] = max(x)
 }
mean(v); mean(a); mean(w)
[1] 40.64162
[1] 54.0478
[1] 74.16839
(mean(a)-mean(v))/(mean(w)-mean(v))
[1] 0.399865