# Determining The Underlying Parameters In Lognormal Distribution

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.

• $E[\eta] \ne 1/E[\phi]$, as the transformation $\eta = 1/\phi$ isn't linear. Dec 12, 2019 at 15:53

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!