# Which sufficient statistic is for which parameter

Suppose I have n samples $$x_1,x_2,...x_n$$ sampled from a $$\mathcal{N}(\mu,\sigma^2)$$. I need to find the sufficient statistics for $$\mu,\sigma^2$$. I write the likelihood

$$\mathcal{L} = (2 \pi \sigma^2)^{-\frac{n}{2}}\exp\left(-\frac{1}{2\sigma^2}\left(\sum_{i=1}^n x_i^2 + \sum_{i=1}^n \mu^2 - 2\mu\sum_{i=1}^n x_i\right)\right)$$

$$\mathcal{L} = (2 \pi \sigma^2)^{-\frac{n}{2}}\exp\left(-\frac{n\mu^2}{2\sigma^2}\right) \exp\left(-\frac{1}{2\sigma^2}\left(\sum_{i=1}^n x_i^2 - 2\mu\sum_{i=1}^n x_i\right)\right)$$

Then we say that by using factorization theorem $$T_1 = \sum_{i=1}^n x_i^2$$ and $$T_2 = \sum_{i=1}^nx_i$$ are sufficient statistics. I want to ask how we come with that the first sufficient statistic is for $$\mu$$ and other one is for $$\sigma^2$$.

I know like MLE is a function of sufficient statistic with that probably we can say that first one is for $$\mu$$ and other is for $$\sigma$$(not sure about this thing) but what if I don't know the MLE of some distribution, then how to figure out.

• Are you trying to use "MLE" and "likelihood" interchangeably?? BTW, there are some typos in you second expression. You might find it helps to work with the logarithm of the likelihood.
– whuber
Apr 9, 2021 at 20:50
• @whuber No. What I understand is MLE is obtained by differentiating the likelihood and then solving the equation. I have corrected the typos.
– A Q
Apr 9, 2021 at 20:59
• I cannot see what role MLE plays here at all. For addressing questions of sufficient statistics, it's entirely irrelevant.
– whuber
Apr 9, 2021 at 21:01
A sufficient statistic should be understood as function of the sample gathering the important information about the parameter (possibly multidimensional). Note that a sufficient statistic could be multidimensional even when you work with unidimensional parameters. For instance, consider an uniform, $$U(\theta-1, \theta +1).$$ This distribution has just one parameter $$\theta$$ and its (minimal) sufficient statistic is $$(X_{(1)},X_{(n)}).$$
That said, it is very common to obtain a sufficient statistic of the form $$(T_1,\ldots, T_k)$$ when working with a $$k-$$ multidimensional parameter $$(\theta_1,\ldots,\theta_k).$$ An easy way to compute a sufficient statistic for one of the parameters, say $$\theta_1,$$ is fixing as constants the rest of parameters and using factorization theorem. In your case, $$\mathcal{N}(\mu,\sigma^2)$$, say we want to compute a sufficient statistic for $$\mu.$$ We can consider $$\sigma ^2$$ as a known value and the likelihood would be: $$\mathcal{L}=C\cdot \exp \left( - \dfrac{1}{2\sigma ^2} \left( n \mu^2 - 2 \mu \sum_{i=1}^n x_i \right) \right),$$ where $$C$$ is a constant (does not depend on $$\mu$$). Now you can easily apply factorization theorem to see that $$T_2=\sum_{i=1}^n x_i$$ is a sufficient statistic for $$\mu.$$