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

Question

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!!

Answer 1

Hints: You can fill in the details!

• 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.

• The distribution of the difference of two iid exponential distributions is a Laplace distribution.

• The absolute value of a Laplace distributed random variable have an exponential distribution.

• 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)

• Assuming the condition from the former bullet point is fulfilled, you ask for the expectation of the difference of two iid variables. That is zero!