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Suppose I have independent pairs $(x_i,y_i)$ $i=1,2...n$

Where $y_i=\theta x_i+\epsilon_i$ and the $x_i's$ and $\epsilon_i's$ are iid $\sim N(0,1)$

and the likelihood function for $\theta$ is given by:

$L(\theta|(x_i,y_i)'s)=(\frac{1}{2\pi})^ne^{-\frac12\sum x_i^2-\frac12\sum (y_i-\theta x_i)^2}$

How do you find a minimal sufficient stat for $\theta$? How do you show that this statistic is incomplete?

I'm having problems trying to apply the factorization theorem to the bivariate pdf. I cannot see how I can isolate a sufficient stat let alone check if it is minimally sufficient.

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It might help if you expand the square in the exponential:

$$ L(\theta ; \textbf{x},\textbf{y}) = \left(\frac{1}{2\pi}\right)^ne^{-\frac{1}{2}(\sum x_i^2+\sum y_i^2) -\frac{1}{2}(2\theta\sum x_iy_i + \theta^2\sum x_i^2)} $$

  1. From here, you can apply the factorization theorem to find a (two-dimensional) sufficient statistic, $(T_1,T_2)$ for $\theta$.

  2. To show minimal sufficiency, suppose $\textbf{X'}$ and $\textbf{Y'}$ were another sample from the same distribution. If the ratio of pdfs, $\dfrac{L(\theta ; \textbf{x},\textbf{y})}{L(\theta ; \textbf{x'},\textbf{y'})}$ is independent of $\theta$ iff $T_1(\textbf{X},\textbf{Y}) = T_1(\textbf{X'},\textbf{Y'})$ and $T_2(\textbf{X},\textbf{Y}) = T_2(\textbf{X'},\textbf{Y'})$, then $(T_1,T_2)$ is minimal sufficient.

  3. To show that $(T_1, T_2)$ is not complete, you need to construct a function $g(T_1,T_2)$ such that $\mathbb{E}[g(T_1,T_2)] = 0$, but $g(T_1,T_2)$ is not identically $0$. As a hint, what is $\mathbb{E}[X^2 - 1]$?

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    $\begingroup$ I actually did do that but did not recognize that I could have a 2-dimensional statistic until you pointed it out. Thanks! So my sufficient statistic is $(T_1,T_2)$ where $T_1=\sum x_iy_i$ and $T_2=\sum x_i^2$, right? $\endgroup$ – user164144 Jul 14 '17 at 6:42

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