# lognormal distribution $z = \frac xy$

I have a problem that says

The random variable $$Y$$ has lognormal distribution with $$u = 2$$ and $$o = 0.4$$. $$z = \frac xy$$. (recall log properties)

Find $$P(Z\leq 6)$$.

The solution begins saying $$\ln Z = \ln X-\ln Y$$ so $$Z$$ is lognormal with $$u = 3-2 = 1$$ and $$o = \sqrt{(0.5)^2 + (0.4)^2} = \sqrt{0.41}$$.

$$P(Z\leq 6) = P\left(Z \leq \frac{\ln6 - 1}{\sqrt{0.41}}\right) = F(1.24) = 0.8925$$

This is confusing to me, I don't know how they got $$u = 3-2 = 1$$ and $$o = \sqrt{(0.5)^2 + (0.4)^2}$$. I have been analyzing the book and my notes in this section, I tried applying the main formula for $$Y\sim \log(u,o^2)$$, and I thought I'd use the given $$2$$ and $$0.4$$ for $$u$$ and $$o$$ in final $$P(Z\leq 6)$$ equation.

I also thought it might be the $$X-Y\sim N(ux-uy,o^2x + o^y)$$ , but then I have no idea why hes using $$ux = 3$$ and $$o^x = 0.5$$.

Can someone please explain the process of how they found the $$u = 1$$ and $$o = \sqrt{0.41}$$ steps?

• Is $X$ of lognormal? What do we know about $X$. – zoli May 8 '17 at 12:00

You're missing some information, but I can infer it from the work that you have.

Suppose $X$ is lognormal with $\mu = 3$ and $\sigma = 0.5$, and $Y$ is lognormal with $\mu = 2$ and $\sigma = 0.4$, and $X$ and $Y$ are independent. Find the distribution of $Z = \dfrac{X}{Y}$.

Recall that if $X^{\prime}$ is normally distributed with $(\mu, \sigma)$, then $X = e^{X^{\prime}}$ has a lognormal distribution with $(\mu, \sigma)$.

It follows that - taking the natural logarithm of both sides - that $\ln(X) = \ln(e^{X^{\prime}}) = X^{\prime}$. Since $X^{\prime}$ is normally distributed with $(\mu, \sigma)$, $\ln(X)$ is also normally distributed with $(\mu, \sigma)$ if $X$ has a lognormal distribution with $(\mu, \sigma)$.

Hence, taking the natural logarithm of both sides of $Z$ gives $$\ln(Z) = \ln\left(\dfrac{X}{Y}\right) = \ln(X)-\ln(Y)\text{.}$$ Since $X$ is lognormal with $\mu = 3$ and $\sigma = 0.5$, $\ln(X)$ is normally distributed with $\mu = 3$ and $\sigma = 0.5$. Since $Y$ is lognormal with $\mu = 2$ and $\sigma = 0.4$, $\ln(Y)$ is normally distributed with $\mu = 2$ and $\sigma = 0.4$.

Hence, $\ln(Z) = \ln(X) - \ln(Y)$ is a difference of normal distributions, which is also normally distributed with \begin{align*} \mathbb{E}[\ln(Z)] &= \mathbb{E}[\ln(X)-\ln(Y)] = \mathbb{E}[\ln(X)] - \mathbb{E}[\ln(Y)] = 3 - 2 = 1 \\ \text{Var}[\ln(Z)] &= \text{Var}[\ln(X)-\ln(Y)] \\ &= \text{Var}[\ln(X)] + \text{Var}[\ln(Y)] \quad \text{(assuming independence)}\\ &= 0.5^2 + 0.4^2 \end{align*} hence $\ln(Z)$ is normal with $(\mu = 1, \sigma = \sqrt{\text{Var}[\ln(Z)]} = \sqrt{0.5^2 + 0.4^2})$. Thus, $e^{\ln(Z)} = Z$ is lognormal with the same $\mu$ and $\sigma$.

• oh wow i forgot that information was at the top of the paper. no wonder i'm so confused. haha i've been up for a while.. thanks – 2316354654 May 8 '17 at 12:06