Re: st: Random draw from log normal distribution with known mean and sd

[Lulu Zeng <luluzengnz@gmail.com>]

Dear Alfonso and Roger,

Thank you so much for your help. You are right that I incorrectly
calculated the normal mean and sd from the log form. It is now fixed
using the formula you suggested.

Thank you again and have a great rest of the day!


Best Regards,
Lulu


On Tue, Jan 28, 2014 at 12:00 AM, Roger B. Newson
<r.newson@imperial.ac.uk> wrote:
> Another useful source on the lognormal distributon (I find) is Stas
> Kolenikov's page at
>
> http://www.komkon.org/~tacik/science/lognorm.pdf
>
> which presents the useful formulas in one place.
>
> Best wishes
>
> Roger
>
> Roger B Newson BSc MSc DPhil
> Lecturer in Medical Statistics
> Respiratory Epidemiology, Occupational Medicine
> and Public Health Group
> National Heart and Lung Institute
> Imperial College London
> Royal Brompton Campus
> Room 33, Emmanuel Kaye Building
> 1B Manresa Road
> London SW3 6LR
> UNITED KINGDOM
> Tel: +44 (0)20 7352 8121 ext 3381
> Fax: +44 (0)20 7351 8322
> Email: r.newson@imperial.ac.uk
> Web page: http://www.imperial.ac.uk/nhli/r.newson/
> Departmental Web page:
> http://www1.imperial.ac.uk/medicine/about/divisions/nhli/respiration/popgenetics/reph/
>
> Opinions expressed are those of the author, not of the institution.
>
>
> On 27/01/2014 12:51, Alfonso Sánchez-Peñalver wrote:
>>
>> Hi Lulu,
>>
>> please explain how you get the equivalent normal mean and sd of -1.04 and
>> 0.89 from the lognormal mean and sd of -0.22 and 0.74? Because I think that
>> is where the problem is. Check
>> http://en.wikipedia.org/wiki/Log-normal_distribution to see the relationship
>> with the means and the standard deviations. The following seems close enough
>>
>> clear
>> set obs 5000
>> generate n = rnormal(-2.77, 1.58)
>> generate ln = -exp(n)
>> summarize
>>
>> Best,
>>
>> Alfonso
>>
>> On Jan 27, 2014, at 4:15 AM, Lulu Zeng <luluzengnz@gmail.com> wrote:
>>
>>> Dear Statalist,
>>>
>>> I am seeking your help on take random draws from a log normal
>>> distribution (with known mean and sd). I am aware similar question has
>>> been answered on below page but I didn't manage to solve my issue with
>>> this (http://www.stata.com/statalist/archive/2005-04/msg00999.html).
>>>
>>> I am trying to calculate Willingness to Pay (wtp) for a number of
>>> attributes (variables) of a random utility model (mixed logit in my
>>> case).
>>>
>>> wtp for a particular attribute is defined as the ratio of the
>>> coefficient for the attribute (e.g., engine performance) to the
>>> coefficient for the price variable. However, both of the engine
>>> performance and price coefficients are random in my model -
>>> performance is normally distributed & price is lognormal distributed.
>>>
>>> Given the difference in distribution for the two coefficients, I had
>>> to use simulation to work out the wtp. That means - take random draws
>>> from both distribution and divide one by another to work out a
>>> distribution for wtp.
>>>
>>> To achieve this, my first step was to take random draws from my log
>>> normally distributed price coefficient, which has a log mean & log sd
>>> of -0.22 and 0.74 respectively (equivalent to a normal mean & sd of
>>> -1.04 and 0.89 respectively). These figures are the results from my
>>> model (distribution of the coefficient).
>>>
>>> I used below code to take the draw as suggested by the webpage above
>>> (1200 draws):
>>>
>>> gen lognormal = exp(-1.04 + 0.89 * invnorm(uniform()))
>>>
>>> To check, I summed the resulting draws from the above, and the draws a
>>> mean of 0.53 & sd of 0.56. These figures are the same as the -0.22 and
>>> 0.74 I have above in log form, so I thought there must be something
>>> wrong.
>>>
>>> It would be really appreciated if I could have your help on this.
>>>
>>>
>>> Best Regards,
>>> Lulu
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