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Let $Y$ be a random sample from $N(\mu,\sigma^2)$ where both $\mu$ and $\sigma^2$ are unknown. Let $\theta$ be the vector of parameters of interest $\theta=(\mu,\sigma^2)$.

I need to find the sufficient statistic for $\theta$ and I know I need to use the factorization theorem for this. Here is what I have done:

$$ \begin{align} f_Y(y;\mu,\sigma^2) &= \left(\frac{1}{\sqrt{2\pi\sigma^2}}\right)^ne^{-\frac{1}{2\sigma^2}\sum\limits_{i=1}^n (y_i-\mu)^2} \\ & = \left(\frac{1}{\sqrt{2\pi\sigma^2}}\right)^n e^{-\frac{1}{2\sigma^2} \left(\sum\limits_{i=1}^n y_i^2 -2\mu \sum\limits_{i=1}^n y_i + n\mu^2 \right) } \\ & = b(h(y),\mu,\sigma^2)c(y) \end{align}$$

where:

  • $c(y)=1$
  • $b(h(y),\mu,\sigma^2) = \left(\frac{1}{\sqrt{2\pi\sigma^2}}\right)^n e^{-\frac{1}{2\sigma^2} \left(\sum\limits_{i=1}^n y_i^2 -2\mu \sum\limits_{i=1}^n y_i + n\mu^2 \right) }$

Now, I am not sure what to put for $h(y)$. I could say:

  1. $h(y)=\left(\sum\limits_{i=1}^n y_i,\, \sum\limits_{i=1}^n y_i^2 \right)$ is a sufficient statistic for $(\mu,\sigma^2)$.

OR

  1. $h(y)=\left( \sum\limits_{i=1}^n y_i^2 ,\, \sum\limits_{i=1}^n y_i \right)$ is a sufficient statistic for $(\mu,\sigma^2)$.

What is the approach to tell which is the correct version to write?

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Both are sufficient statistics. Or perhaps more precisely, either is the minimal sufficient statistic for this family of distributions. If you know the value of either, you can compute the value of the other without knowing $\mu$ or $\sigma^2$. You could also say the pair $\left(\bar Y, \frac 1 n \sum_{i=1}^n (Y_i-\bar Y)^2\right)$ is a sufficient statistic for $(\mu,\sigma)$, where $\bar Y=(Y_1+\cdots+Y_n)/n$.

The phrase "the sufficient statistic" is a bit misleading, implying as it does that there is only one. For example, the whole sample $(Y_1,\ldots,Y_n)$ is a sufficient statistic. But you probably want the minimal sufficient statistic, i.e. a sufficient statistic that is a function of every other sufficient statistic, where the function in no way depends on $(\mu,\sigma^2)$. And there is only one of those, and the pair $\left(\sum_{i=1}^n Y_i,\, \sum_{i=1}^n Y_i^2 \right)$ is it, or the pair $\left( \sum_{i=1}^n Y_i^2,\, \sum_{i=1}^n Y_i \right)$ is it, or the pair $\left(\bar Y, \frac 1 n \sum_{i=1}^n (Y_i-\bar Y)^2\right)$ is it. They are equivalent in that from any of them you can compute the others without knowing $(\mu,\sigma^2)$.

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  • $\begingroup$ So, both 1. and 2. are correct answers? I'm still not convinced (I don't see how to compute the value of the other...). Do mind walking through an example to illustrate this point (and using a simpler distribution if that helps) ? $\endgroup$ – mauna Mar 5 '15 at 17:15
  • $\begingroup$ Suppose $\displaystyle\left(\sum_{i=1}^n y_i, \sum_{i=1}^n y_i^2\right) = (100,22)$. Then $\displaystyle\left(\sum_{i=1}^n y_i^2, \sum_{i=1}^n y_i\right) = (22,100)$. If you know that the first one is $(100,22)$, then you can find that the second one is $(22,100)$. And if you know the second is $(22,100)$, you can find that the first is $(100,22)$. If you know either, you can find the other without knowing the values of $\mu$ or $\sigma^2$, so they are equivalent. ${}\qquad{}$ $\endgroup$ – Michael Hardy Mar 5 '15 at 18:21

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