$E[Z]$ when $X$ follows an exponential distribution and $Z = \exp\{X\}$.

Question

I am trying to find the expected value of an exponential distrubtion $E(Z)$. I am told $X$ is exponential distributed with the parameter $\lambda = 4$, and I am told $Z = \exp(X)$. I am having trouble solving for the density function of $Z$, even though I know the form of the density function for $X$.

Answer 1

How to find the density of $Z$?

Notice that $Z = \exp\{X\}$ is stricly increasing, and hence we can use the one-to-one change of variable; $X = \log Z$ and $$f_Z(z) = \frac{f_X(\log z)}{\left|\frac{dz}{dx}\right|_{\log z}}.$$

Or you can use the cdf method and compute $$P(Z\leq z) = P(\exp\{X\}\leq z).$$

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How to find $E[Z]$?

To compute the expectation of $Z$, you can proceed as usual. Let $g(X) = \exp\{X\}$. Then $$E[Z] = E[\exp\{X\}] =\int_0^\infty g(x)\cdot f_X(x)\,dx = \int_0^\infty e^x\cdot \lambda e^{-\lambda x}\,dx.$$

It is also possible to find the density of $Z$ above and do the the usual $E[Z]$.

Answer 2

There is no need to find the distribution function of $Z$. By the Law of the Unconscious Statistician, $$E(Z)=\int_0^\infty (e^x)(\lambda e^{-\lambda x})\,dx.$$ In our case we are integrating $4e^{-3x}$.