If X is log-normal, is: $\frac{a}{\sqrt{b+cX}}$?

Question

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:
$$\mu'=-\frac{\mu+\ln c+b}{2}$$ $$\sigma'^2=\frac{\sigma^2}{4}$$

Answer 1

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.

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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.

Answer 2

For $t\ge 0$

$$P\left\{\frac{a}{\sqrt{b+cX}}\le t\right\}=P\left\{\frac{a^2}{b+cX}\le t^2\right\}=P\left\{X\ge\frac{(a/t)^2-b}{c}\right\}=\int_{\frac{(a/t)^2-b}{c}\vee0}^\infty \frac{1}{x\sigma\sqrt{2\pi}}\exp\left\{-\frac{(\ln(x)-\mu)^2}{\sigma^2}\right\}dx$$

Then for $t\in\left[0,\frac{a}{\sqrt{b}}\right]$

$$f_{X'}(t)=\frac{2a^2}{t^3}\frac{1}{((a/t)^2-b)\sigma\sqrt{2\pi}}\exp\left\{-\left(\ln\left(\frac{(a/t)^2-b}{c}\right)-\mu\right)^2\sigma^{-2}\right\}$$

And if $b=0$,

$$f_{X'}(t)=\frac{1}{t(\sigma/2)\sqrt{2\pi}}\exp\left\{-\frac{(\ln(t)-0.5[ln(a^2/c)-\mu])^2}{(\sigma/2)^2}\right\}$$

which is lognormal with parameters $\mu'=[ln(a^2/c)-\mu]/2$ and $\sigma'=\sigma/2$.