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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.

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    $\begingroup$ $E[\eta] \ne 1/E[\phi]$, as the transformation $\eta = 1/\phi$ isn't linear. $\endgroup$ – jbowman Dec 12 '19 at 15:53
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From the Wikipedia page on the Lognormal distribution, edited lightly:

If $ \epsilon\sim \operatorname {Lognormal} (\mu ,\sigma ^{2})$ then $\tfrac {1}{\epsilon}\sim \operatorname {Lognormal} (-\mu ,\ \sigma ^{2})$

In your case, since $\mu = \sigma^2/2$, we have $\phi = \tfrac {1}{\epsilon}\sim \operatorname {Lognormal} (-\sigma^2/2 ,\ \sigma ^{2})$, so it follows directly that

$$\mathbb{E}\phi = \operatorname {e}^{-\sigma^2/2 + \sigma^2/2} = \operatorname {e}^0 = 1$$

regardless of the value of $\sigma$!

We can check this interesting result with a little simulation in R:

> sigma2 <- 2
> phi <- exp(rnorm(100000, -sigma2/2, sqrt(sigma2)))
> mean(phi)
[1] 1.003422
> 
> sigma2 <- 3.1415927
> phi <- exp(rnorm(100000, -sigma2/2, sqrt(sigma2)))
> mean(phi)
[1] 1.009064
> 
> sigma2 <- 0.456789
> phi <- exp(rnorm(100000, -sigma2/2, sqrt(sigma2)))
> mean(phi)
[1] 1.001695

Interesting!

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