TransformedDistribution[x = Exp[u], u [Distributed] NormalDistribution[[Mu], [Sigma]]]
seems to work fine, producing
LogNormalDistribution[\[Mu], \[Sigma]]
(correct). However I’d rather express mu and sigma in dB and also am interested in having power on the linear side, so wish to scale the argument and use a 10 base for the transformation. Like:
Which returns without evaluation in the same form with u replaced by [FormalX].
Playing with the expression it appears to be the use of 10^ which is the root, replacing 10 with E functions - but is of course not what I want. Using E and adding Log[10] to the exponent doesn’t work, Whereas using E and adding 2.3 to the exponent does…
I think I understand your issue but I think you’d get more and better responses if you use the formatting tools provided. You can also use \TeX commands (surrounding things between dollar signs) when a formula needs to be explicitly clear.
Update: You’ve now used the formatting tools. Thank you!
Well, it’s a bit of a problem for me as I was trying to author a short paper on the derivation and usages of the lognormal power <==> normal in dB transform pair and being able to show their correspondence was the point.True, it’s a bit overkill as the Normal → LogNormal transform seems to reliably evaluated and LogNormal → Normal evaluation with E^ just not with 10^ or Log[10] in the exponent. It’s also a little strange that including a constant or variable it still evaluates, but include the particular constant Log[10] causes it to remain un-evaluated. I suspect that there is some command I’m missing that influences the order or treatment, but trying hold and various version of expand and simplify didn’t work, but since I was guessing about why it didn’t evaluate in the first place, wasn’t likely that adding random commands would address the issue.
By replacing 10^x_ with E^(Hold[Log[10] x) and surrounding with ReleaseHold, it evaluates.
ReleaseHold [
TransformedDistribution[ x = 10^(u /10 ), u [Distributed] NormalDistribution[ mu, sigma] ] ] //. {10^x_ :> E^( Hold[Log[10]] x) }
]
Of course, I’d still be interested in a good explanation as to why it works this way so as perhaps avoid the trial and error approach I used this time.