Let $X$ be a single observation from $N(0,\theta).\ \ (\theta=\sigma^2)$
Writing joint density.
$L(x;\theta)=\dfrac{e^{-\frac{x^2}{2\theta}}}{\sqrt{2\pi\theta}}=\dfrac{1}{\sqrt{2\pi\theta}}\cdot e^{-\frac{x^2}{2\theta}}$
We see that $T(X)=X^2$ sufficient for $\theta$
$(a)$ Is $X$ is sufficient statistic?
Since the original sample is always sufficient we conclude that $X$ is sufficient.
$(b)$ Is $|X|$ is sufficient statistic?
$|X|=\sqrt{X^2}$
I am stuck in this one. So please guide me here.
$(c)$ Is $X^2$ is sufficient statistic?
Yes, that's what we get from Neymann Factorization theorem.
$(d)$ What is the maximum likelihood estimator of $\sqrt{\theta}$
When $X \sim N(\mu,\sigma^2)$ MLE of $\sigma^2$is $\dfrac{\sum_{i}(X_i-\bar{X})^2}{n}$
in our case, we have $\dfrac{X^2-0}{1}$
using invariance property MLE of $\sqrt{\theta}$ is $\sqrt{X^2}=X$
B part is where I am stuck and I want to know if I calculated everything right? Where I am going wrong? Any tips to keep in mind and please correct me.
Best Answer
(a) is wrong. You found with factorization theorem that $T=X^2$ is sufficient for $\sigma^2$. $X$ is not sufficient because it is NOT a monotonic function of $T$
(b)
$$T_1=|X|=\sqrt{X^2}$$
In this case, $|X|$ is a monotonic function of $T$ thus $T_1$ is sufficient too for $\sigma^2$
(c) this is wrong too...
Easy find that $\hat{\theta}=X^2$ (observe that MLE must be function of the sufficient estimator, if it exists)
Now using invariance property
$$\hat{\sigma}_{ML}=\sqrt{\hat{\theta}}=|X|$$