I have the following problem:
" Let $ \epsilon $ be a normal random variable with variance $ \sigma^{2} $ and mean $ \sigma^{2}/2$. Then $\phi \equiv e^{\epsilon}$ is a lognormal random variable, $\phi \sim lnN(\sigma^{2}/2, \sigma^{2}) $. What are the parameters of the normal distribution for $\epsilon$ that make the expected value of $\eta = 1/\phi $ equal to 1? "
I am quite new to working with lognormal distributions, but here is my chain of thought below.
The expected value of a lognormal variable is given by: $$ E[\phi] = e^{\mu + \frac{1}{2}\sigma^{2} } $$
letting $ \mu = \frac{\sigma^{2}}{2} $ we would simply get: $$ E[\phi] = e^{\sigma^{2}} $$
then if $ E[\eta] = \frac{1}{E[\phi]} = 1$ it follows that
$$e ^{\sigma^{2}} = 1$$ and thus $$\sigma=0 $$
However, I don't really think that this is correct, because it would imply that phi is lognormally distributed with mean 0 and variance 0.