# Sufficient statistic for normal distribution with unknown mean and known variance

Let $$X$$ be from a normal distribution $$N(\theta,1)$$.

a) Find a sufficient statistic for $$\theta$$.
b) Is $$S_n^2$$ a sufficient statistic for $$\theta$$?

For part a)

Since the joint p.d.f is $$1 \over (2\pi)^{n/2}$$, $$e^{{-1 \over 2}\sum(x_i-\theta)^2}$$, I can say that $$\sum X_i$$ is a sufficient statistic for $$\theta$$ because $$e^{{-1 \over 2}\sum(x_i-\theta)^2}$$ depends on X only through the values of $$\sum X_i$$ right? Because if I know the value of $$\sum X_i$$, then I know $$\sum X_i^2$$ as well.

For part b)

Expanding the joint p.d.f as $$\frac{1}{(2\pi)^{n/2}}e^{{-1 \over 2}\sum(x_i-\theta)^2} = \frac{1}{(2\pi)^{n/2}}e^{{-1 \over 2}\sum(x_i- \bar x + \bar x-\theta)^2} = \frac{1}{(2\pi)^{n/2}}e^{{-1 \over 2}\Big[\sum(x_i- \bar x)^2+n(\bar x-\theta)^2\Big]} = \frac{1}{(2\pi)^{n/2}}e^{{-1 \over 2}\Big[{\sum(x_i- \bar x)^2 \over n-1}n-1+n(\bar x-\theta)^2\Big]}$$.

Now can I say $$S_n^2$$ is a sufficient statistic for $$\theta$$ . Is it a problem that I have $$\bar x$$ in the function $$g(S_n^2,\theta)$$?. Because $$\bar x$$ is a particular value I thought $$g(S_n^2,\theta)$$ depends on $$\theta$$ only through the values of $$S_n^2$$.

• "$e^{{-1 \over 2}\sum(x_i-\theta)^2}$ depends on X only through the values of $\sum X_i$ right?" No, not right, $e^{{-1 \over 2}\sum(x_i-\theta)^2}$ does not depend on X only through the values of $\sum X_i$. "Because if I know the value of $\sum X_i$ then I know $\sum X_i^2$ as well." Do you? Sure about that? // Here is a suggestion: read the WP page, at the moment your understanding of sufficient statistics seems well below this introduction.
– Did
Commented Apr 21, 2014 at 15:33
• Only when you pointed out that did I realize that knowing the value of $\sum X_i$ doesn't mean I know $\sum X_i^2$.I know how to extend this to show that $\bar x$ is sufficient.I was wondering why I couldn't stop it at this stage.Now I understand part a Commented Apr 21, 2014 at 15:37
• Did you intend to write "unknown" where you wrote "known"? Commented Jan 2, 2018 at 1:38

$$\sum_{i=1}^n (x_i - \theta)^2 = \left( \sum_{i=1}^n x_i^2 \right) -2\theta \left( \sum_{i=1}^n x_i \right) + n\theta^2$$ Therefore $$\exp \left( \frac {-1} 2 \sum_{i=1}^n (x_i-\theta)^2 \right) = \underbrace{ e^{-n\theta^2/2}\cdot \exp\left( \theta\sum_{i=1}^n x_i \right)}_{\large\text{first factor}} \cdot \underbrace{ \exp\left( \frac{-1} 2\sum_{i=1}^n x_i^2 \right) }_{ \large \text{second factor}}$$

The first factor depends on $(x_1,\ldots,x_n)$ only through $\displaystyle\sum_{i=1}^n x_i.$ The second factor does not depend on $\theta.$

Therefore by Fisher's factorization theorem, $\displaystyle\sum_{i=1}^n x_i$ is sufficient for $\theta.$

(As your question now stands, it says "known mean", but "$N(\theta,1)$" means the mean is unknown and the variance is known.)

• Should't there be a minus sign in the second factor? exp$(-\sum x_i^2)$ Commented Jun 29, 2018 at 13:08
• @JohnCataldo : Actually the whole factor of $-1/2$ was missing. Now I've put it there. $\qquad$ Commented Jun 29, 2018 at 16:10

(a) Taking your joint probability density of $\frac{1}{(2\pi)^{n/2}}e^{{-1 \over 2}\sum(x_i-\theta)^2}$, you can expand this into $$\left(\frac{1}{(2\pi)^{n/2}}e^{-\sum x_i^2 /2}\right)\left(e^{-n\theta^2/2+\theta \sum x_i }\right)$$ where the left part does not depend on $\theta$ and the right part is a function of $\theta$ and $\sum x_i$, implying by Fisher's factorisation theorem that $\sum x_i$ is a sufficient statistic for $\theta$

(b) $S_n^2$ (you do not say, but presumably the sample variance, or possibly the sample second moment about $0$ or perhaps $\sum x_i^2$) is not a sufficient statistic for $\theta$. One way of seeing this is that multiplying all the $x_i$ observations by $-1$ would not change $S_n^2$, and so it cannot give any information to distinguish between the population mean of the original normal distribution being $\theta$ or being $-\theta$