Solved – Negative parameters of beta distribution

distributions

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.

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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 )

Related Solutions

Beta Distribution – Calculating Parameters using Mean and Variance

I set$$\mu=\frac{\alpha}{\alpha+\beta}$$and$$\sigma^2=\frac{\alpha\beta}{(\alpha+\beta)^2(\alpha+\beta+1)}$$and solved for $\alpha$ and $\beta$. My results show that$$\alpha=\left(\frac{1-\mu}{\sigma^2}-\frac{1}{\mu}\right)\mu^2$$and$$\beta=\alpha\left(\frac{1}{\mu}-1\right)$$

I've written up some R code to estimate the parameters of the Beta distribution from a given mean, mu, and variance, var:

estBetaParams <- function(mu, var) {
  alpha <- ((1 - mu) / var - 1 / mu) * mu ^ 2
  beta <- alpha * (1 / mu - 1)
  return(params = list(alpha = alpha, beta = beta))
}

There's been some confusion around the bounds of $\mu$ and $\sigma^2$ for any given Beta distribution, so let's make that clear here.

$\mu=\frac{\alpha}{\alpha+\beta}\in\left(0, 1\right)$
$\sigma^2=\frac{\alpha\beta}{\left(\alpha+\beta\right)^2\left(\alpha+\beta+1\right)}=\frac{\mu\left(1-\mu\right)}{\alpha+\beta+1}<\frac{\mu\left(1-\mu\right)}{1}=\mu\left(1-\mu\right)\in\left(0,0.5^2\right)$

Solved – Estimating parameters in a truncated negative binomial distribution

Are you thinking that somehow the inequality (≥) factors in to the calculation? It does not. That inequality is just telling you you are talking about a truncated distribution.

In each case, mean and variance, you have three items:

X=Y=Z

Lop off the "X=", and use the remaining items. You'll need to solve for the two unknowns, alpha and beta, which you do using Newton's method.

I actually need to do that very thing, so I'll let you know how it goes.

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Related Solutions

Beta Distribution – Calculating Parameters using Mean and Variance

Solved – Estimating parameters in a truncated negative binomial distribution

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