# Forming Location-Scale Family for Log-Normal

For the random variable X with a log-normal pdf

$f(x)=\frac{1}{\sqrt{2*\pi}}x^{-1}e^{-.5*log(x)^2}$

I am trying to find a location-scale family $h(x)$ such that $h(x)$ has mean 0 and variance 1

Now I know for location-scale family the transformation

$h(x)=\frac{1}{\sigma}f(\frac{x-\mu}{\sigma})$

I found the expected value and variance of $f(x)$ respectively to be $\sqrt{e}$ and $(e-1)e$ but is it just a matter of plugging those in to the above to find what I need? Just a bit confused about what the transformation does. Thanks in advance

• Is this for study purposes, or do you need it for some task? The idea of a location-scale family is not that it has mean 0 and variance 1 (that's one member, not a family). Can you give a definition of $h$ (that isn't contradictory)? You might base the family off a standardized base though, for convenience of interpretation. The usual lognormal is already a scale-shape family, so you'd just need to add a location shift to that. Your formulation assumes a particular shape parameter -- did you want to restrict yourself to only that shape? – Glen_b Feb 4 '15 at 23:32
• Yeah this is a problem in one of my textbooks that I'm trying to work on mostly for study purposes and that was the problem given. I suppose I might just be misunderstanding what it's asking for. – James Snyder Feb 5 '15 at 1:07
• Thanks. [Could you add the self-study tag and see the guidelines in its tag wiki? You may need to add some details to follow the guidelines.] Does the book give a definition of $h$ somewhere else, perhaps when it first mentions location-scale families? – Glen_b Feb 5 '15 at 1:46

I don't think you need to worry about the mean and variance of $f$ as such.
It gives you $f$ and it gives you the transformation (the transformation isn't what you said, the transformation is $Y=\frac{X-\mu}{\sigma}$); $h$ is actually the density of the transformed random variable.
[Presumably you know how to use the usual methods to do the transformation; the Jacobian is just $\frac{1}{\sigma}$. Have you seen the result that for $Y=g(X)$, the density is $f_Y(y)= f_X(g^{-1}(y)) \left|\frac{d g^{-1}(y)}{dy}\right|$? )