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Suppose we have a standardized normal $Z_1 \sim \mathcal{N}(0,1)$, and take the associated lognormal $X = e^{Z_1}$.

Suppose that with have a non-standardized normal $Z_2 \sim \mathcal{N}(\mu,\sigma^2)$ and the associated lognormal $Y = e^{Z_2}$

Is there somewhere an expression in terms of $\mu,\sigma$ for standardisation parameters $a,b \in \mathbb{R}\times\mathbb{R}_{+}$ such that, in distribution,

$$Y = aX+b \text{ ?}$$

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  • $\begingroup$ $Y=e^{Z_n} => Y = e^{aZ_1+b} => Y = (e^{Z_1})^a e^b$ what is wrong with this? $\endgroup$ – quester Oct 2 '19 at 10:33
  • $\begingroup$ This is not linear. I need a linear transformation. $\endgroup$ – Pilou l'Nrv Oct 2 '19 at 10:35
  • $\begingroup$ An immediate implication of @quester's equation is that unless $\sigma=1,$ there exists no such linear transformation. $\endgroup$ – whuber Oct 2 '19 at 12:53
  • $\begingroup$ it's impossible due to left-side restricted support.... only approximations $\endgroup$ – quester Oct 2 '19 at 13:12
  • $\begingroup$ @Quester hints at a neat demonstration of impossibility. When $b\ne 0,$ $aX+b$ either assigns some probability to negative values or fails to assign any probability to small positive values, showing it cannot be lognormal. When $b=0,$ the effect of $a$ is to rescale $X:$ but changing $\mu$ rescales $X$ by $e^\mu.$ Consequently, $a$ is just a reparameterization of $\mu.$ If you could always find such an $a,$ then $\sigma$ would be superfluous: but we know that's not the case. This easily shows why you cannot hope to find such an $a$ and $b$ in general; you can only hope for special cases. $\endgroup$ – whuber Oct 2 '19 at 16:13
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Unless $\sigma=1,$ no such linear transformation exists. This is an immediate consequence of the one-to-one correspondence between $\sigma$ and the skewness of a lognormal variable, because linear transformations do not change the magnitude of the skewness and lognormal variables always have positive skewness.


Consider the skewness of $Y=\exp(Z_2),$ given by

$$\beta_3(Y)= \left(\exp(\sigma^2) + 2\right)\sqrt{\exp(\sigma^2)-1}.$$

The skewness of any variable is the third moment of its standardized version of the variable. Because linear transformations with positive slopes do not change the standardization, they leave the skewness unchanged. Thus, from the two expressions $aX+b=Y=\exp(Z_2)$ and noting $a\gt 0$ we obtain

$$\left(e^1+2\right)\sqrt{e^1-1} = \beta_3(aX+b) = \beta_3(Y) = \left(e^{\sigma^2/2} + 2\right)\sqrt{e^{\sigma^2/2} - 1}.\tag{1}$$

The claim is that $\sigma=1$ is the only (positive) solution to this equation. To see why that is so, for $t\gt 0$ define

$$f(t) = \sqrt{t(t+3)^2}.$$

Writing $\lambda = \exp(\sigma^2)-1$ and $\rho = \exp(1)-1$ (both of which are positive), equation $(1)$ asserts

$$f(\lambda) = f(\rho).$$

However, as a function defined on the positive numbers, $f$ is one-to-one. This is readily seen by noting that the derivative

$$\frac{d}{dt} t(t+3)^2 = 3(t^2 + 4t + 3)$$

can never be zero for $t\gt 0,$ implying $f^2$ is a positive one-to-one function, whence so is $f.$ Consequently $\lambda=\rho,$ showing that $\sigma=1$ is the only solution.

(On the other hand, when $\sigma=1,$ $Z_2 = Z_1 + \mu$ and therefore

$$Y = \exp(Z_2) = \exp(Z_1+\mu) = e^\mu \exp(Z_1) = e^\mu X + 0$$

indeed is a linear transformation of $X$ with $a=e^\mu,b=0.$)

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