# Expectation of the exponential function of Absolute Value of the Difference of Two Double parameter exponentially Distributed Random Variable

Suppose, $$X_1,\ldots, X_n$$ be iid having double parameter Exponential Distribution with common pdf $$f(x)= \dfrac{1}{\sigma} \exp\{ -(x-\mu)/\sigma \} I(x>\mu); \mu \in R, \sigma \in R^+ , n\ge5$$

Let, $$T=\exp(\lvert X_1 - X_3\rvert) - \exp(\lvert X_2 - X_4\rvert)$$

Now I m trying to find $$E[T]$$. But I'm really having no clue how to deal with this!!

• The hard way is to compute the expectations. The easy way is to think about what the iid assumption tells you about the distributions of $|X_1-X_3|$ and $|X_2-X_4|:$ how do they differ? Use your answer to this (very easy) question to reason about how the expectations of their exponentials differ.
– whuber
Oct 19, 2020 at 21:23
• Umm, I think since $X_is$ are iid , $\lvert X_1 - X_3\rvert$ and $\lvert X_2 - X_4\rvert$ would have same distribution. So is the answer is zero?? @whuber Oct 20, 2020 at 4:53
• You also need to make sure that the expectation of each of the two terms at the end are finite for the argument to go through. I believe this requirement will place a bound on $\sigma$ (the answer is different otherwise). Oct 20, 2020 at 8:15
• Your question has $n=4$ exactly so why the inequality $n\ge 5$? Nov 28, 2021 at 20:27

• The distribution of the differences $$X_1 - X_3$$ and $$X_2 - X_4$$ do not depend on $$\mu$$, so we can assume $$\mu=0$$. This reduces to ordinary exponential distributions.
• So, now we need the expectation of the exponential of an exponentially distributed random variable, see Distribution of the exponential of an exponentially distributed random variable?. This expectation will exist, under some condition on your parameter $$\sigma$$. I leave it for you to find the condition! (should be simple)