Answer to Question #227781 in Statistics and Probability for Jerry

Question

The probability density function (pdf) of the normal distribution is given by 1 V2πσ2 fx (x)= 0 (-(x - 1)

Answer 1

$f_x(x; \sigma^2)= \frac{1}{\sqrt{2 \pi \sigma^2}} e^{-\frac{x^2}{2 \sigma^2}}$

By factorization theorem , we can write it as

$=g(\sum_i^n x_i^2 \sigma^2)*h(x)$

Therefore

$T=\sum_1^n x_i^2$ is sufficient statistic for $\sigma^2$ where $h(x)=1$