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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?

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

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  • $\begingroup$ 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 $\endgroup$ – 2316354654 May 8 '17 at 12:06

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