2
$\begingroup$

I have a set of observations of credit loss data, where the mean is 37% and variance 25%. Now, I have to find the distribution and the base assumption is it will follow a beta distribution. the issue is that my alpha and beta derived from mean and variance is being estimated at -0.025012 and -0.042588. I dont understand what to do with the negative values of alpha and beta. The formula which I am using to calculate alpha is mean*(((mean*(1-var))/var)-1) and beta is calculated by (1-mean)(((mean*(1-mean))/var)-1). Please do let me know how can I solve the problem.

$\endgroup$
  • 2
    $\begingroup$ Your results are correct. The interpretation is that there is no Beta distribution with this mean and variance. BTW, variance cannot be "25%"; its units need to be squared percent. If the value of $0.25$ is really the standard deviation, then the matching parameters are $(a,b)$ = $(1.01, 1.72)$. If indeed the values you give are the mean and variance, though, then you ought to change your question to "how can I best estimate parameters of a Beta distribution from data and check the resulting goodness of fit." $\endgroup$ – whuber Mar 20 '13 at 16:20
  • $\begingroup$ @whuber - the variance is indeed 0.25. Thanks for the help. The answer looks simple, but I could not find it online. Can you please tell me where I may get hold of the argument as i plan to refer to it in the model document. Thanks again. $\endgroup$ – Bik Mar 20 '13 at 16:42
  • 1
    $\begingroup$ Could you tell me what you mean by "argument" and what it means to get hold of one? Incidentally, I did not use your formulas; the ones I used are $(a,b)$ = $\left(\frac{m \left(m-2 m^2+m^3-v+m v\right)}{(m-1) v},\frac{m-2 m^2+m^3-v+m v}{v}\right)$ where $m$ is the mean and $v$ the variance. $\endgroup$ – whuber Mar 20 '13 at 17:05
  • $\begingroup$ Well yes, the wrong word was selected. I need to give a reference document which supports the interpretation of negative alpha and beta. Can you please help me with any article available online which you may be of use to me? Reference to any book will do too. Thanks! $\endgroup$ – Bik Mar 20 '13 at 17:50
  • 3
    $\begingroup$ I don't think I have ever seen any document that says "the formula for a Beta distribution makes no sense when either parameter is zero or less," in part because it is clear that the integral diverges in those cases. This is implicit in most treatments that simply declare the parameters must be positive, as in the Wikipedia summary. $\endgroup$ – whuber Mar 20 '13 at 19:36
3
$\begingroup$

Partially answered in comments:

Your results are correct. The interpretation is that there is no Beta distribution with this mean and variance. BTW, variance cannot be "25%"; its units need to be squared percent. If the value of 0.25 is really the standard deviation, then the matching parameters are $(a,b) = (1.01,1.72)$. If indeed the values you give are the mean and variance, though, then you ought to change your question to "how can I best estimate parameters of a Beta distribution from data and check the resulting goodness of fit." – whuber

( @whuber - the variance is indeed 0.25. Thanks for the help. The answer looks simple, but I could not find it online. Can you please tell me where I may get hold of the argument as i plan to refer to it in the model document. – Bik )

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.