If X is log-normal, is: $\frac{a}{\sqrt{b+cX}}$? I am working for the first time with log-normal distributions and I want to verify whether the following statement is true. I am not sure whether all the properties of the log-normal distribution hold also for a 'shifted log-normal distribution'.
If $X$ is a log-normal distribution:
$$X\sim \ln\mathcal{N}(\mu,\sigma^2)$$
Then does it follow that for $a,b,c>0$:
$$\exists\mu',\sigma':\frac{a}{\sqrt{b+cX}}\sim \ln\mathcal{N}(\mu',\sigma'^2)$$  
with:
$$\mu'=-\frac{\mu+\ln c+b}{2}$$  $$\sigma'^2=\frac{\sigma^2}{4}$$

Obviously not:  the support of a log-normal random variable is necessarily on $(0, \infty)$; that is to say, if $Y = \log X \sim \operatorname{Normal}(\mu,\sigma^2)$, then $0 < X < \infty$.  But the transformation $$X' = \frac{a}{\sqrt{b+cX}}$$ is such that the support of $X'$ is bounded above by $$\frac{a}{\sqrt{b}},$$ which occurs as $X \to 0^+$, thus $X'$ has bounded support, and cannot be lognormal.

In the case that $b = 0$, we can rewrite this transformation as $$X' = g(X) = aX^{-1/2}$$ for some suitable constant $a > 0$.  In such a case, what we can do is refer to the distribution of $Y$, which is normal; i.e., $$X' = g(X) = aX^{-1/2} = a(e^Y)^{-1/2} = ae^{-Y/2}.$$  Then taking logarithms gives $$Y' = \log X' = - \frac{Y}{2} + \log a.$$  Consequently, because $Y$ is normal with mean $\mu$ and variance $\sigma^2$, $Y'$ is normal with mean $$\mu' = -\frac{\mu}{2} + \log a$$ and variance $$(\sigma')^2 = \frac{\sigma^2}{4}.$$  This immediately demonstrates that $X'$ is lognormal and furnishes the relevant parameters.
More generally, we can see that for a lognormally distributed variable, any "scale-power" transformation $$X' = aX^m$$ for $m \ne 0$ and $a > 0$ will also result in a lognormal variable, much in the way that any location-scale transformation of a normal variable $$Y' = a + bY$$ yields another normal variable, because $Y = \log X$ transforms the operation of exponentiation into multiplication, and multiplication into addition.
