# If the sum, min, max of iid rvs has the same distribution as the rvs, is it sufficient?

It seems intuitive. For example, the $Min(x_1,x_2...x_n)$ where the $x_i's$ are iid $Geo(\theta)$ is also $Geo(\theta)$. This means that the Min, is also a sufficient stat for $\theta$, right?

• The title is unclear. Sufficient for what? Why would you even suspect those specific things imply sufficiency in general? The example in your body text is wrong (the minimum of geometric($\theta$) independent rvs doesn't have the same distribution as the things you took the minimum of) – Glen_b Jul 27 '17 at 0:35
• Sorry about the error. I based it on the distribution map at the back of the textbook. It just says that $min X_i$ of Geo(p) is Geo(p). It did not specify that the parameter is different for this case. – user164144 Jul 27 '17 at 3:57

$$\DeclareMathOperator{geo}{Geo} \DeclareMathOperator{P}{\mathbb{P}}$$ Suppose $X_i\sim \geo(\theta)$ iid for $i\in[n]$. Then: $$\P\{\min_{i\in[n]}X_i> k\}=\prod_{i=1}^n\P\{X_i> k\}=(1-\theta)^{kn}$$ But notice that this implies $\P\{\min_iX_i\le k\}=1-(1-\theta)^{nk}$, where this is exactly the CDF of a $\geo(1-(1-\theta)^n)$-distributed rv, not a $\geo(\theta)$ one. Intuitively, the minimum of $n$ equally distributed, independent things won't usually be distributed the same way, since it's smaller than most of those $n$ things all the time.
Assuming $X_i\sim \geo(\theta)$ are iid for some unknown $\theta$, $\min_i X_i$ is not a sufficient statistic for $\theta$. It suffices to show this for $n=2$. $\P\{X_1=x_1,X_2=x_2\}=(1-\theta)^{x_1+x_2-2}\theta^2$, but this cannot be factored into a product of a function of $x_1,x_2$ alone and $\theta,\min(x_1,x_2)$ alone (take logarithms, the derivative must only be a function of the latter two, but it isn't).