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I'm trying to calculate the variance of the inverse gamma distribution using the method of movements. According to wikipedia the variance should be:

$$\sigma^2 =\frac{\beta^2}{(\alpha-1)^2(\alpha-2)}$$

Where $\alpha$ is the shape and $\beta$ is the scale of the inverse gamma distribution. When trying this out in R it works reasonably well when $\alpha$ is not to close to 2. For example:

library(MCMCpack) # for the rinvgamma function
a <- 10
b 100

# The variance according to the method of movements
b^2/((a-1)^2*(a-2)) 
## 15.4321

# The variance by generating inverse gamma distributed random numbers and
# calculating the sample variance
var(rinvgamma(n=9999, shape=a, scale=b)) #
## 15.84388

But when $\alpha$ gets close to 2 the method of movements doesn't seem to work anymore. In the following example the sample variance is much smaller than the method of movements variance:

a <- 2.2
b <- 100

# The variance according to the method of movements
b^2/((a-1)^2*(a-2)) 
## 34722.22

# The variance by generating inverse gamma distributed random numbers and
# calculating the sample variance
var(rinvgamma(n=9999, shape=a, scale=b)) # 
##14479.56

Why doesn't the method of movements work? Am I doing something wrong that can be fixed or is there some other way that I can calculate the variance of an inverse gamma distribution?

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    $\begingroup$ With such a huge theoretical variance, you need a quite large sample to estimate it accurately. Even with the normal distribution you will observe discrepancies var(rnorm(10000,100,35000)) (less severe, of course). $\endgroup$ – user10525 Nov 6 '12 at 11:14
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    $\begingroup$ When $\alpha \rightarrow 2$, the variance goes to infinity. You cannot hope to approach "infinity" accurately $\endgroup$ – ocram Nov 6 '12 at 11:15
  • $\begingroup$ @Procrastinator: sorry, we have posted at the same time! $\endgroup$ – ocram Nov 6 '12 at 11:15
  • $\begingroup$ for small $\hat{\alpha}-2$ the variance of the estimate may be infinite (I haven't checked, but I'd expect that to be the case). $\endgroup$ – Glen_b Nov 6 '12 at 11:39
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    $\begingroup$ @RasmusBååth Well, you are not using the method of moments precisely. The expression b^2/((a-1)^2*(a-2)) is actually the theoretical variance. The method of moments consists of solving the system $\dfrac{\beta}{\alpha-1}=\bar{x}$ and $\dfrac{\beta^2}{(\alpha-1)^2(\alpha-2)}=\dfrac{1}{n}\sum_{j=1}^n(x_j-\bar{x})^2$ in terms of $(\alpha,\beta)$. After this, you could plug the solutions $(\hat{\alpha},\hat{\beta})$ into the expression for the variance $\dfrac{\beta^2}{(\alpha-1)^2(\alpha-2)}$ in order to obtain an estimator of this quantity. $\endgroup$ – user10525 Nov 6 '12 at 12:30

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