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Let $\theta > 0$ be some model parameter for which properties (bias, ...) of an estimate are studied via simulations.

For a given data set, an estimate $\hat{\theta}_i$ of $\theta$ can be obtained by maximising a likelihood function.

As $\theta$ is positive, however, I perform the maximisation over $\eta = \log(\theta) \in \mathbb{R}$ and I set $\hat{\theta} = \exp(\hat{\eta})$.

After running $N$ simulations, how should I compute the mean value,

  • $\bar{\hat{\theta}} = \frac{1}{N} \sum_{i=1}^N \hat{\theta}_i$, or
  • $\bar{\hat{\theta}} = \exp(\frac{1}{N} \sum_{i=1}^N \hat{\eta}_i)$?

Example:

-- Minus log likelihood for a sample from the Exp distribution with mean $\theta = \exp(\eta)$:

minusloglik <- function(eta, sample)
{
  theta <- exp(eta)
  - sum(dexp(x=sample, rate=theta, log=TRUE))
}

-- True value of $\theta$:

theta <- 5.73

-- Simulations:

thetaHat <- etaHat <- rep(NA, 1000)
for(i in 1:1000)
{  
  sample <- rexp(n=100, rate=theta)
  etaHat[i] <- nlm(f=minusloglik, p=0, sample=sample)$estimate
  thetaHat[i] <- exp(etaHat[i])
}

Question:

  • Should I summarise the results as mean(thetaHat) or as exp(mean(etaHat))?
  • Is the answer the same if $\theta$ denotes the variance of a normal distribution?
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    $\begingroup$ Because you seem to have defined $\hat\theta_i=\exp(\hat\eta_i),$ aren't your two calculations identical? $\endgroup$
    – whuber
    Commented Mar 11, 2014 at 19:18
  • 1
    $\begingroup$ @whuber: You are right. This is not what I meant. I did an update. $\endgroup$
    – user7064
    Commented Mar 12, 2014 at 6:02
  • $\begingroup$ OK, now it's clear what the formulas are: one is the arithmetic mean and the other is the geometric mean. But why are you performing these simulations and what are you hoping to learn from them? That information is essential for identifying which formula (if either) would be appropriate. $\endgroup$
    – whuber
    Commented Mar 12, 2014 at 15:02
  • $\begingroup$ @whuber: I've made an edit to make it more transparent. Thank you :-) $\endgroup$
    – user7064
    Commented Mar 13, 2014 at 6:10

1 Answer 1

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If $\frac{1}{N} \sum_{i=1}^N \hat{\eta}_i$ is an unbiased estimate of the common $\eta$*, then exponentiating an unbiased estimate does not yield an unbiased estimate. If you are going to exponentiate the estimate of $\eta$, you need to account for that effect. Some people use Taylor expansions to do a first order correction, or assume a normal distribution for the estimate of $\eta$ and so might seek from that an unbiased estimate of $\exp(\eta)$.

* which it may not be, that depends on the distributional model and link function (and its correctness).

The same problem affects your other estimate - each of the individual estimates is biased, and averaging those biased estimates will also be biased.

In either case you'd need to adjust them.

Personally I wouldn't use either of those estimates anyway.

Consider that the individual estimates of $\eta$ are very unlikely to be equally uncertain. Then it makes no sense to give equal weight to a precise estimate and an imprecise estimate. Normally one would try to weight inversely proportional to variance, where possible.

(Consider an extreme case to illustrate the point. Imagine I have samples of size 10, 10, 10, 10, 10, and 10,000. If I take a plain average of the resulting five estimates of $\eta$, I'm much worse off - on average - than if I just throw away the first four!)

If you're interested in minimum mean square error, or some other property, you should perhaps look to optimize that, but you need to be clear whether the properties you seek are on the estimates of $\theta$ or $\log \theta$.

The properties of the various possible estimates of whatever quantity you're actually interested in can be investigated under various assumptions via simulation.

[The direct answers to your two final questions depend on things you are yet to specify. Please provide additional information.]

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