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Suppose $X\sim N(0,{\sigma}^2)$ and $Y\sim N(0,{2\sigma}^2)$ . $X_1, ..., X_m$ are the samples from $X$ and $Y_1, ..., Y_n$ are the samples from $Y$. And then combine two samples as a new sample $\{X_1, ..., X_m,Y_1, ..., Y_n\}$. The question is what's the distribution of the new sample? And how to calculate the UMVUE of $\sigma^2$? How to calculate the Fisher Information of $\sigma^2$?

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For some ideas, see this question and its answers on stats.SE. – Dilip Sarwate Jan 7 '13 at 14:41
up vote 1 down vote accepted

Set $$X_{m+1}=\frac{Y_1}{2},...,X_{m+n}=\frac{Y_n}{2}$$ Now you have an i.i.d. sample of size $(m+n)$ from $N(0,\sigma^2)$. The theory in that standard case follows with no modification. An estimator is going to be a function depending on the sample $X_1,...,X_m,Y_1,...,Y_n$ (which is exactly the same as $X_1,...,X_{m+n}$ because you applied a deterministic transformation meaning that if you know the first, then you can get the second with no extra "information"). Now, think about the variane estimator $$ \frac{1}{m + n - 1} \left( \sum_{i = 1}^{m + n} (X_i^{} - \bar{X})^2 \right) $$ where $$ \bar{X} = \frac{1}{m + n} \sum_{i = 1}^{m + n} X_i $$

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Could you please explain what is i.i.d. sample, or give a link? What do you mean by with no modification?Thank you. – maple Jan 7 '13 at 15:31
i.i.d.=independently and identically distributed. I assumed that assumption was missing from your question. (otherwise, you could not derive the likelihood under just normality of the marginals). – Learner Jan 7 '13 at 15:33
Do you know how to answer the question if you just had $(X_1,...,X_n)$ in your sample? – Learner Jan 7 '13 at 15:34
Do you mean that let $$X_{m+1}=\frac{Y_1}{2},...,X_{m+n}=\frac{Y_n}{2}$$ and replace $Y_{i}$ with $X_{m+i}$ ? But it's not the new samples, it has been changed. In fact I don't know the answer when just had $(X_1,...,X_n)$ in the sample. Because, we just learn the distribution of a statistics like $X$ rather than a sample $(X_1,...,X_n)$. Could you please explain it? – maple Jan 7 '13 at 15:48
Note that this answer is valid only if the samples are labelled (ie. if we know which of the $m+n$ samples correspond to which population) – leonbloy Jan 7 '13 at 19:52

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