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A random variable $X$ is logarithmic normally distributed if $\ln X \sim N(\mu, \sigma^2)$. What's the density of the random variable $X$?

About, notation: $X \sim N(\mu, \sigma^2)$ mean that random variable $X$ is normally distributed in interval which begin from $\mu$ to $\sigma^2$.

I solve like this but not sure:

$$f_X(x) = \frac{d}{dx} P(X \leq x) = \frac{d}{dx}P(\ln X \leq \ln x)$$

Now create variable $\Psi$ and $\Gamma$ which are cumulative probability distribution function and also density function of normal distribution $N(0,1)$. Then the above is same as

$$\frac{d}{dx} \Psi\left(\frac{\ln x - \mu}{\sigma}\right) = \Gamma \left(\frac{\ln x - \mu}{\sigma}\right) \cdot \frac{d}{dx} \left(\frac{\ln x - \mu}{\sigma}\right) = \Gamma\left(\frac{\ln x - \mu}{\sigma}\right) \cdot \frac{1}{\sigma x}$$

Then the last thing is density of random variable $X$ or I do it wrong?

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What you've done is correct.

Often the c.d.f. of the standard normal distribution is called $\Phi$ and the density is called $\varphi.$ Thus one has $$ f_X(x) = \varphi\left( \frac {(\ln x) - \mu} \sigma \right) \cdot \frac 1 {\sigma x} \text{ for } x >0. $$

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There must be a simpler way to achieve that. Suppose $F_{N}(v)$ be CDF of a normal distribution with mean $\mu$ and variance $\sigma^2$. Since we want to find CDF(and consequently PDF) of a log-normal variable ,say X, we write: $$X=e^Y$$ where $Y\sim N(\mu,\sigma^2)$ therefore: $$Pr(X<x)=Pr(Y<\ln x)=F_{N}(\ln x)$$ by differentiating we obtain: $$\LARGE f_{X}(x)={d\over{dx}}F_{N}(\ln x)={1\over {x\sigma\sqrt{2\pi}}}{e^{-{{(\ln x-\mu)^2}\over{2\sigma^2}}}}$$

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