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Reciprocal truncated exponential distribution function

Posted 3 years ago

I want to find the Mean and variance of one random variable having a reciprocal truncated exponential distribution function over the interval [a,b]

I used the following code (n=1)

n=1;
ExpDist = ExponentialDistribution[\[Lambda]];
TruDist = TruncatedDistribution[{1/b, 1/a}, ExpDist];
OrdDist = OrderDistribution[{TruDist, n}, n];
FullSimplify [Mean[OrdDist], Assumptions -> {b > a && a > 0, \[Lambda] > 0}]
FullSimplify [Moment[OrdDist, 2] - Power[Mean[OrdDist], 2], 
 Assumptions -> {b > a && a > 0, \[Lambda] > 0}]

I got wrong result. Any help please

POSTED BY: Dalia Fawaz
3 Replies
Posted 3 years ago

Nothing wrong with using 1/b and 1/a. But, what is wrong with the answer you got?

POSTED BY: Jim Baldwin
Posted 3 years ago

How do you know you got a wrong result?

This question appears nearly identical to your last question but the the truncation (according to your code) being between 1/b and 1/a although your text (as with the previous question) gives the interval being between a and b.

POSTED BY: Jim Baldwin
Posted 3 years ago

I used 1/b and 1/a because I want to find the reciprocal of truncated exponential distribution. I already found theoretical results of the reciprocal and validated them by numerical examples, but this code is giving me different results. I appreciate any help.

POSTED BY: Dalia Fawaz
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