If $ X_1, ..., X_n$ is a random sample from $ X \sim N(\mu_1, \sigma^2)$ and $Y_1,..., Y_n$ is a random sample from $Y \sim N(\mu_2, \sigma^2),$ if the samples are independent and $ \sigma^2$ is known, can we say that $\bar{X}-\bar{Y}$ is sufficient for $\mu_1 - \mu_2$ ?

My guess is that it is true. I thought we could define

$$W_i=X_i-Y_i \sim N(\mu_1 - \mu_2, 2 \sigma^2)$$

n independent random variables and use it to show that $\bar{W} = \bar{X} - \bar{Y} $ is sufficient to the mean. Is that correct?


  • $\begingroup$ We can show how the pdf of the difference distribution looks like. Afterwards, use the factorization theorem to prove the sufficiency. $\endgroup$ Nov 15, 2014 at 13:38
  • 2
    $\begingroup$ It's not clear at all to me what it means for a statistic to be "sufficient for" a particular parameter. See my answer here. However, it also seems in the definition proposed here that this definition may depend on the parameterization. $\endgroup$
    – guy
    Nov 15, 2014 at 18:45
  • $\begingroup$ Indeed, the statistic is sufficient when based on the difference sample. But there is no argument for reducing the data to the difference sample. $\endgroup$
    – Xi'an
    Dec 9, 2014 at 13:23
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    $\begingroup$ Consider the case when $X_1,\ldots,X_n$ and $Y_1,\ldots,Y_{2n}$ are your samples... Then what would a sufficient statistic be? There is no rationale in associating $X_i$ with $Y_i$ since the samples are independent. $\endgroup$
    – Xi'an
    Dec 9, 2014 at 13:26
  • $\begingroup$ @Xi'an That is true. I should have been more careful there. $\endgroup$ Apr 11, 2015 at 1:13


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