Truncated exponential distribution

Questions
1

I need to find the mean of the maximum for n independent identical random variables in a Truncated exponential distribution.

I used the following code to find it for exponential distribution:

ed = {ExponentialDistribution[\[Lambda]], n};
m = Mean[OrderDistribution[ed, 1]] ;
Print[m];

What changes should I make to make the code work on a Truncated exponential distribution on the interval [a,b]?

2

Note: Edited from the original answer.

If you have specific values for n, then the following will work for the mean and variance of the maximum for a sample of size n=3:

trd = TruncatedDistribution[{a, b}, ExponentialDistribution[\[Lambda]]];
otrd = OrderDistribution[{trd, 3}, 3];
FullSimplify[Mean[otrd], Assumptions -> {b > a > 0, \[Lambda] > 0}]
\frac{(6 a \lambda +11) \left(-e^{3 b \lambda }\right)+18 (b \lambda +1) e^{\lambda (a+2 b)}-9 (2 b \lambda +1) e^{\lambda (2 a+b)}+e^{3 a \lambda } (6 b \lambda +2)}{6 \lambda \left(e^{a \lambda }-e^{b \lambda }\right)^3} 
FullSimplify[Variance[otrd], Assumptions -> {b > a > 0, \[Lambda] > 0}]
\frac{-4 e^{3 \lambda (a+b)} (9 \lambda (a-b) (a \lambda -b \lambda +3)+155)+249 e^{2 \lambda (2 a+b)}-42 e^{\lambda (5 a+b)}-6 e^{\lambda (a+5 b)} (6 \lambda (a-b) (3 a \lambda -3 b \lambda +5)+55)+6 e^{2 \lambda (a+2 b)} (6 \lambda (a-b) (3 a \lambda -3 b \lambda +8)+115)+4 e^{6 a \lambda }+49 e^{6 b \lambda }}{36 \lambda ^2 \left(e^{a \lambda }-e^{b \lambda }\right)^6}
3

Thank you for your help, is this only for the mean? I tried to use it to find the variance but it didn’t work.

4

You only asked for the mean. How did it “not work” for the variance? Did you get an error? Can you show what you tried?

5

Yes, you are right, I used n in the variance and it showed the original equation without making any calculations. But when I used a specific value of n, it worked perfectly. Thank you a lot.

Just one more thing if you could help me with, any idea about finding the mean and variance of the maximum for n independent non-identical random variables in a Truncated exponential distribution.

6

You should now have enough to go on to do that yourself. If you get stuck at some point, please show what you’ve tried and explain where you are stuck. I would imagine you’ll get more help that way.