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Question

# Find the density function of $Y=e^{Z}$, where $Z \sim N\left(\mu, \sigma^{2}\right)$. This is called the lognormal density, since log Y is normally distributed.

Solution

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Given:

$Z\sim N(\mu,\sigma^2)$

$Y=e^Z$

Definition cumulative distribution function of $Y$:

$F_Y(y)=P(Y\leq y)$

$Y$ is defined as the $Y=e^Z$:

$F_Y(y)=P(e^Z\leq y)$

Take the natural logarithm of side of the inequality:

$F_Y(y)=P(\ln e^Z\leq \ln y)$

Use the power property of logarithms $(\ln a^b=b\ln a$):

$F_Y(y)=P(Z\ln e\leq \ln y)$

The natural logarithm of $e$ is equal to 1:

$F_Y(y)=P(Z\leq \ln y)$

Use the definition of the cumulative distribution function of $Z$:

$F_Y(y)=F_Z\left( \ln y\right)$

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