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Let $X_1,...,X_n$ be a random sample of size $n$ from $N(\epsilon,\sigma^2)$ and let $Y_1,...,Y_m$ be a random sample of size $m$ from $N(\eta,2\sigma^2)$.

Find a complete and sufficient statistics for $(\epsilon, \eta, \sigma^2)$. Suppose that we know that $\eta=3\epsilon$. Then use the relevant ones to compute the UMVUE for $\epsilon$ under this condition.

I think that the complete and sufficient statistics are given by $\left< \sum_{i=1}^n x_i, \sum_{j=1}^m y_j, \sum_{i=1}^n x_i^2 \right>$ from the linear exponential family. But I'm not confident. Why should I have chosen $\sum_{i=1}^n x_i^2$ instead of, say, $\sum_{j=1}^m y_j^2$? Are they not both complete and sufficient for $\sigma^2$? Or should instead be using both of the samples somehow?

This question just has me really confused because, in general, I don't know how to take advantage of multiple samples to help me estimate an unknown parameter that appears in both. I suspect that this means that the statistic I provided above for $\sigma^2$ isn't complete and sufficient in this context; but I just don't know.

Similarly, in the second part, one parameter only appears in one sample, and the other only appears in the other. So does new condition really means that there's a different UMVUE? I'm just so confused.

EDIT:

I've written the joint likelihood as: $$L_1(\vec X)L_2(\vec Y)=\frac{\exp\left(-\frac{\sum^n_{i=1}(x_i-\epsilon)^2}{2\sigma^2}-\frac{\sum^m_{j=1}(y_j-\eta)^2}{4\sigma^2} \right)}{(2\pi\sigma^2)^{n/2}(4\pi\sigma^2)^{m/2}}$$

This implies that the joint likelihood can be written in the following linear exponential family form:

$$L_1(\vec X)L_2(\vec Y)=\frac{P(\vec X, \vec Y)e^{\vec T(\vec X, \vec Y) \cdot \vec w(\vec \theta)}}{q(\vec\theta)}$$

Where $\cdot$ represents the dot product, $\vec T(\vec X, \vec Y)=\left< 2\sum_{i=1}^n x_i^2 + \sum_{j=1}^m y_j^2, \sum_{i=1}^n x_i, \sum_{j=1}^m y_j\right>$, $\vec\theta=\left< \sigma^2,\epsilon,\eta\right>$, $\vec w(\vec \theta)=\left<-\frac{1}{4\sigma^2},\frac{\epsilon}{\sigma^2},\frac{\eta}{2\sigma^2}\right>$, $P(\vec X, \vec Y)=1$, and finally $q(\vec \theta)=(2\pi\sigma^2)^{n/2}(4\pi\sigma^2)^{m/2}e^{\frac{n\epsilon^2+m\eta^2}{4\sigma^2}}$.

Hence the joint likelihood can be written in the form of the linear exponential family, which immediately implies that a complete and sufficient statistic for $\left< \sigma^2,\epsilon,\eta \right>$ is given by:

$$\left< 2\sum_{i=1}^n x_i^2 + \sum_{j=1}^m y_j^2, \sum_{i=1}^n x_i, \sum_{j=1}^m y_j\right>$$

Is this correct?

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  • $\begingroup$ This statistics is not sufficient. Use Neymann - Fisher factorization theorem en.wikipedia.org/wiki/Sufficient_statistic#Fisher–Neyman_factorization_theorem $\endgroup$ – NCh Apr 8 '18 at 2:59
  • $\begingroup$ @NCh, could you be a little more clear? Which one, specifically, is not sufficient, or are you referring to all three? Also, I still don't understand how factorization theorem will work with two samples; now I have X and Y, so I don't know the exact form I need to rewrite the PDF in. $\endgroup$ – jippyjoe4 Apr 8 '18 at 6:01
  • $\begingroup$ It is one three-dimensional statistic. It is not sufficient for given three-dimensional parameter. Write joint PDF of $n+m$ samples and make sure that it cannot be expressed in the form $$h(x_1,\ldots,x_n,y_1,\ldots,y_m)\cdot g\left(\sum_{i=1}^n x_i, \sum_{j=1}^m y_j, \sum_{i=1}^n x_i^2, \epsilon, \eta, \sigma^2\right).$$ $\endgroup$ – NCh Apr 8 '18 at 17:01
  • $\begingroup$ @NCh, I think that I understand now, but I'm not certain. I have edited my question with what I've derived so far. Would you please tell me if I'm on the right track? $\endgroup$ – jippyjoe4 Apr 8 '18 at 18:47
  • $\begingroup$ Yes, this is correct. $\endgroup$ – NCh Apr 9 '18 at 1:14

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