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Suppose I have $X_1 \sim N(\mu_1,\sigma^2_1)$ and $X_2 \sim N(\mu_2,\sigma^2_2)$ (where the distribution is empirically estimated). I believe that $X_1$ and $X_2$ are actually drawn from the same distribution $N_(\mu^*, \sigma^{*2})$. I think the MLE is just to say $\mu^*=\frac{\mu_1+\mu_2}{2}$ and similarly for variance, right?

Context: I want to do a back-of-the-napkin calculation to combine multiple reported statistics. E.g. we have several reports of meat consumption; if I believe they are all drawn from the same underlying distribution I would like to know the best estimator. I've been trying to read some basic tests on meta-analysis as I assume this is a standard procedure, but I can't find anything about this.

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    $\begingroup$ If the distributions have been estimated empirically, were the data used in computing the estimates thrown away? If so, how can you be sure that $X_1$ and $X_2$ seem to have the distributions you think that they have? Do you think that the data used in estimating would be of interest here? For example, if the estimates $\mu_1$ and $\sigma_1^2$ were from $10000$ data points while $\mu_2$ and $\sigma_2^2$ were from $100$ data points, does it make sense to weight them equally in your meta-analysis? $\endgroup$ – Dilip Sarwate Dec 29 '13 at 17:28
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    $\begingroup$ Very closely related: stats.stackexchange.com/questions/24936/…. That is an identical question but the answer focuses on unbiased minimum-variance estimators rather than Maximum Likelihood estimators. $\endgroup$ – whuber Dec 29 '13 at 18:18
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(I expand this answer to cover more completely a setup that occurs frequently, and also, to focus on the estimation of the population variance).

The setup is as follows: from $K$ independent samples that all come from a normal i.i.d. population $N(\mu, \sigma^2)$ and have varying sizes, $n_1+...+n_K = N$, we are not given the actual sample data but only
a) The sample sizes $n_i, i=1,...,k$
b) The sample mean of each sample, $m_i = \frac 1{n_i}\sum_{j=1}^{n_i}x_j$
c) The sample variance of each sample, $v_i = \frac 1{n_i}\sum_{j=1}^{n_i}(x_j-m_i)^2 = \frac 1{n_i}\sum_{j=1}^{n_i}x_j^2-m_i^2$

Note that we consider the maximum likelihood estimator of the population variance, i.e. we divide by $n_i$ and not by $n_i-1$.

We want to derive maximum likelihood estimators for the unknown population parameters, $\mu$, and $\sigma^2$, by using the information we are given.

A) MLE of population mean
Under the maintained hypothesis the sample means are normally distributed, and each sample mean $m_i \sim N(\mu, \frac {\sigma^2}{n_i})$. The samples are independent so the joint density/likelihood function of the sample means is

$$L(\mu, \sigma^2\mid \{m_1,...,m_K\})= \prod_{i=1}^K\frac {\sqrt {n_i}} {\sqrt{2\pi}\sigma}\exp\left\{-\frac 12 \frac {(m_i-\mu)^2}{\sigma^2/n_i}\right\}$$

and the log-likelihood is

$$\ln L(\mu, \sigma^2\mid \{m_1,...,m_K\})= c -K\ln\sigma -\frac 1{2\sigma^2} \sum_{i=1}^Kn_i(m_i-\mu)^2$$

Setting the first derivative of $\ln L$ w.r.t to $\mu$ equal to zero we have

$$\frac {\partial}{\partial \mu} \ln L = 0 \Rightarrow \frac 1{\sigma^2} \sum_{i=1}^Kn_i(m_i-\mu) =0 \Rightarrow \sum_{i=1}^Kn_im_i - \mu\sum_{i=1}^Kn_i =0$$

$$\Rightarrow \hat \mu_{ML} = \sum_{i=1}^K\frac {n_i}{N}m_i = \sum_{i=1}^K\frac {n_i}{N}\left(\frac 1{n_i}\sum_{j=1}^{n_i}x_j\right) = \frac 1N\sum_{j=1}^{N}x_j = \bar X_N \qquad [1]$$

So the MLE of the population mean, weighs each sample mean by the (relative) sample size from which it was derived, becoming a convex combination of the $K$ sample means, and so it ends up numerically equivalent to the full-sample mean we would obtain if we had the original data available and had pooled them in one sample.

Note: although we have $k$ estimates of the population variance available, they do not enter the likelihood, because the maintained assumption is that all samples come from the same population. If instead of $\sigma^2/n_i$ we had included in the likelihood $v_i/n_i$, we would have violated this assumption. A comment that refers to known variances (and provides the correct formula for this case), covers essentially the case where the various samples do not come from the same population (and more over, the variances are known and not estimated).

B) MLE of population variance
We could derive an MLE of the population variance using the above likelihood as

$$\hat \sigma ^2_{ML} = \frac 1K\sum_{i=1}^Kn_i(m_i-\hat \mu_{ML})^2 \qquad [2]$$

This is a biased estimator (due to the estimation error associated with $\hat \mu_{ML}$), and also, shouldn't we take into account the estimated variances derived from each sample, the $v_i$'s?
We know that $$\frac {n_iv_i}{\sigma^2} \equiv z \sim \chi^2(n_i-1)$$

Then $$v_i = \frac {\sigma^2}{n_i} z \sim \operatorname{Gamma}(k_i,\theta_i),\;\; k_i = \frac {n_i-1}{2},\;\; \theta_i = \frac {2\sigma^2}{n_i}$$

with the Gamma density given by

$$f_{v_i}(v_i) = \frac 1{\Gamma(k_i)\theta_i^{k_i}}v_i^{k_i-1}\exp\left\{-\frac {v_i}{\theta_i}\right\}$$

The $v_i$'s are independent random variables, so they form the following log-likelihood:

$$\ln L_v = c- \sum_{i=1}^Kk_i\ln \theta_i+\sum_{i=1}^K(k_i-1)\ln v_i-\sum_{i=1}^K\frac {v_i}{\theta_i}$$

Note that the shape parameters, $k_i$'s are known, and all $\theta$'s are functions of the same unknown parameter, $\sigma^2$. Setting the first derivative of the log-likelihood w.r.t. $\sigma^2$ equal to zero we get $$\frac {\partial}{\partial \sigma^2} \ln L_v = 0 \Rightarrow -\sum_{i=1}^K\frac {k_i}{\theta_i}\frac {2}{n_i} + \sum_{i=1}^K\frac {v_i}{\theta_i^2}\frac {2}{n_i} =0$$

$$\Rightarrow \frac 1{\sigma^4}\sum_{i=1}^K\frac {n_i^2v_i}{4}\frac {2}{n_i} = \frac 1{\sigma^2}\sum_{i=1}^Kk_i\frac {n_i}{2}\frac {2}{n_i}$$

and simplifying and using also $k_i=(n_i-1)/2$ we obtain

$$\frac 1{2\sigma^2}\sum_{i=1}^Kn_iv_i = \frac 1{2}\sum_{i=1}^K(n_i-1) $$ $$\Rightarrow \hat \sigma^2_{ML}(v) = \sum_{i=1}^{K}\frac {n_i}{N-K} v_i \qquad [3]$$

Note that here, unlike the case of the MLE for the population mean, the sample variances are not combined into a convex combination - and this has the interesting consequence that by combining in this way the biased estimators $v_i$, the estimator $\hat \sigma^2_{ML}(v)$ becomes an unbiased estimator.

By using the scaling and summation properties of the Gamma distribution we have that $$v_i \sim \operatorname{Gamma}\left(\frac {n_i-1}{2},\; \frac {2\sigma^2}{n_i}\right) \Rightarrow \frac {n_i}{N-K} v_i \sim \operatorname{Gamma}\left(\frac {n_i-1}{2},\; \frac {2\sigma^2}{N-K}\right)$$ $$\Rightarrow \sum_{i=1}^{K}\frac {n_i}{N-K} v_i \sim \operatorname{Gamma}\left(\frac {N-K}{2},\; \frac {2\sigma^2}{N-K}\right) $$

and so $$E\left(\hat \sigma^2_{ML}(v)\right) = \frac {N-K}{2}\frac {2\sigma^2}{N-K} =\sigma^2$$

and

$$\operatorname{Var}\left(\hat \sigma^2_{ML}(v)\right) = \frac {N-K}{2}\left(\frac {2\sigma^2}{N-K}\right)^2 = \frac {2\sigma^4}{N-K}$$

Note that this unbiasedness property of $\hat \sigma^2_{ML}(v)$ would not obtain if we have used the unbiased formula (dividing the sum of squared differences in each sample by $n_i-1$ instead of $n_i$) to calculate each sample variance: we need to... "be loyal to ML spirit" from the beginning to be rewarded in the end!

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  • $\begingroup$ Thanks Alecos. So I should throw out my sample variances? That seems weird to me. $\endgroup$ – Xodarap Dec 29 '13 at 20:43
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    $\begingroup$ Your question has no meaningful sample variances: it concerns only the two numbers $X_1$ and $X_2$. If you assume the $\sigma_i$ are known, then the MLE of $\mu$ is easily derived and equals $\left(\sigma_2^2X_1+\sigma_1^2X_2\right)/\left(\sigma_1^2+\sigma_2^2\right).$ $\endgroup$ – whuber Dec 29 '13 at 21:06
  • $\begingroup$ @whuber: what makes a sample variance "meaningful"? I think your equation is line 19 here which I get, but I don't understand the variance equation that follows. $\endgroup$ – Xodarap Dec 29 '13 at 21:38
  • $\begingroup$ @Alecos: in your addendum: if I replace $\sigma^2/n_i$ with $s_i$, then what do I differentiate with respect to? Also if I do that I'm going to lose my $n_i$'s, which seem important. $\endgroup$ – Xodarap Dec 29 '13 at 21:58
  • $\begingroup$ The $n_i$'s and the variance of each sample mean are linked. Also, I hope we are not confusing here the population variance as it is estimated from each sample, with the estimated variance of each sample mean. $\endgroup$ – Alecos Papadopoulos Dec 29 '13 at 22:31

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