1. ## Expectation question

The function is as in the picture (hopefully it works!)

I need to show the expectation is sigma*exp(mu) but am not sure how. The first thing I notice when I look at that rather complicated function is that it looks a bit like the function of a normal distribution, but help please

2. ## Re: Expectation question

I am not sure if I understood your question in the right way. So, you want to integrate the given function w.r.t. $\displaystyle x?$
Just substitute $\displaystyle y = ln(x)$ (and note that $\displaystyle d y =\frac{1}{x} dx$)

3. ## Re: Expectation question

I need to calculate the expectation, which is the integral of xf(x). So this extra x cancels with the one in the denominator, and then all of that fraction can be taken out as its not dependent on x. To calculate the other part are you suggesting integration by substitution then?

The overall answer should be equal to sigma*e^mu though.

4. ## Re: Expectation question

your $\displaystyle f$ is just the density of a log-normally distributed r.v. $\displaystyle (ln\mathcal{N}(\mu,ln(\sigma^2)))$

and the expectation of a log-normally distributed r.v. with these parameters is given by

$\displaystyle e^{\mu+\frac{1}{2}ln \sigma^2$