# Proving the equivalence of a pdf with a log-normal distribution

I have the following function normalised to 1 on $(0, \infty)$: $$g(x) = \frac{e^{-\left(\mu + \frac{\sigma^2}{2}\right)} e^{-\frac{(\mu -\text{Log}[x])^2}{2 \sigma ^2}}}{\sigma \sqrt{2 \pi }}$$ which, after multiplying by $\frac{1}{x}e^{\left(\mu + \frac{\sigma^2}{2}\right)}$ gives a log-normal distribution.

The n-th moment of g is $e^{n \mu + \frac{n^2\sigma ^2}{2} + n \sigma^2}$, which again is the same as for the log-normal to the factor $e^{n \sigma^2}$.

It seems like I might be missing some theory here but how can I show that $g(x)$ and the log-normal distribution are essentially the same? I can make them overlap in the plot...

(EDIT) The reason I think $g(x)$ and the log-normal distribution must be somehow related is the following. Suppose I set the first two moments, $m_1$ and $m_2$, of these distributions to same values. For $g(x)$ I obtain the following scale and shape parameters: $$\mu_g = \log{\frac{m_1^4}{m_2^{3/2}}}\\ \sigma_g = \sqrt{\log{\frac{m_2}{m_1^2}}}$$

For the log-normal distribution I get the following: $$\mu_{LN} = \log{\frac{m_1^2}{m_2^{1/2}}}\\ \sigma_{LN} = \sigma_g$$

After substituting the above parameters to their respective distributions I calculate the ratio: $$\frac{g(x)}{f_{LN}(x)} = x \frac{m_2}{m_1^3} \left( \frac{m_1^2}{m_2} \right)^{\log{ \left( \frac{m_2}{m_1^3} x \right)} / \log{ \left( \frac{m_2}{m_1^2} \right)}} = 1$$ which I find quite surprising especially because the two distribution differ by a factor that is not a constant but depends on $x$. I wonder whether this is a case of a more general rule?

• what do you mean they are essentially the same? they are clearly not? you are multiplying by something which is not a constant. Jan 21, 2014 at 22:05
• @Lost1 I can plot them such that they exactly overlap. Shape parameters are equal and only $\mu$ of g and log-normal differ. Additionally, when I fit the two distributions to the same set of data, the best fit is such that they both exactly overlap. Also the parameter such as Bayesian information criterion for both fits is exactly the same. This made me wonder whether there's something that makes these two distributions connected. Rescaling parameters for instance? Jan 21, 2014 at 22:16
• i dunno how data has anything to do with this? you gave me 2 different density functions and tried to convince me they are the same function. I dunno what you mean by they exactly overlap. they may be very close, but your equations say they are not the same? Jan 21, 2014 at 22:29
• @Lost1 well, it seems they are. See my EDIT in the above question. Jan 22, 2014 at 18:16

For every $a\gt0$ and every $b$, consider the function $h_{a,b}$ defined on $x\gt0$ by $$h_{a,b}(x)=\exp(-a(\log x)^2+(b-1)\log x).$$ Let $h:x\mapsto g(x)/x$ where $g$ is the function in the first displayed formula of the question. Then:
• The function $h$ a multiple of $h_{1/2\sigma^2,\mu/\sigma^2}$.
• For every $c$, $x^ch_{a,b}(x)=h_{a,b+c}(x)$.
Thus, the function $g$ is a multiple of the function $$h_{1/2\sigma^2,1+\mu/\sigma^2}.$$ In other words, $h$ being the lognormal density with parameters $(\mu,\sigma^2)$ implies that $g$ is proportional to the lognormal density with parameters $(\mu+\sigma^2,\sigma^2)$.