Question

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be iid \(N(0, \theta), 0<\theta<\infty\). Show that \(\sum_{1}^{n} X_{i}^{2}\) is a sufficient statistic for \(\theta\).

Short answer

Through the application of the Neyman-Fisher factorization theorem, we showed that the joint pdf of \(X_{1}, X_{2}, ..., X_{n}\) can be factorised into a function of \(\sum_{i=1}^{n}x_{i}^{2}\) and \(\theta\), and a function of X alone. Therefore, \(\sum_{i=1}^{n}x_{i}^{2}\) is indeed a sufficient statistic for \(\theta\).

Step-by-step solution

Expression of the Joint Probability Density Function

The first step is to express the joint probability density function (pdf) of \(X_{1}, X_{2}, ..., X_{n}\) since they are independent and identically distributed (iid) as \(N(0, \theta)\). The joint pdf can be written as the product of their individual pdfs, that is \(f(x_{1}, x_{2}, ..., x_{n}| \theta) = \prod_{i=1}^{n}f(x_{i}|\theta)\). Substituting the normal distribution pdf into this equation gives us \(f(x_{1}, x_{2}, ..., x_{n}| \theta) = \prod_{i=1}^{n}\frac{1}{\sqrt{2\pi\theta}}e^{-\frac{x_{i}^{2}}{2\theta}}\)

Simplifying the Joint pdf

We can simplify the joint pdf by collecting similar terms together to get \(f(x_{1}, x_{2}, ..., x_{n}| \theta) = \frac{1}{(2\pi\theta)^{\frac{n}{2}}}e^{-\frac{\sum_{i=1}^{n}x_{i}^{2}}{2\theta}}\). It is noticed that the exponential portion of this function involves the statistic \(\sum_{i=1}^{n}x_{i}^{2}\)

Application of Neyman-Fisher Factorization Theorem

The Neyman-Fisher factorization theorem states that a statistic T(X) is sufficient for \(\theta\) if and only if the joint pdf or pmf can be factorized as a function of T(X) and \(\theta\), and a function of X alone. Looking at our simplified joint pdf, we can express it as \(f(x_{1}, x_{2}, ..., x_{n}| \theta) = h(x_{1}, x_{2}, ..., x_{n})g(T(x_{1}, x_{2}, ..., x_{n}), \theta)\) where \(h(x_{1}, x_{2}, ..., x_{n}) = 1\), \(g(T(x_{1}, x_{2}, ..., x_{n}), \theta) = \frac{1}{(2\pi\theta)^{\frac{n}{2}}}e^{-\frac{T(x_{1}, x_{2}, ..., x_{n})}{2\theta}}\) and \(T(x_{1}, x_{2}, ..., x_{n}) = \sum_{i=1}^{n}x_{i}^{2}\)

Conclusion

Given the factorizable form of the joint pdf and applying the Neyman-Fisher factorization theorem, we can conclude that \(\sum_{i=1}^{n}x_{i}^{2}\) is a sufficient statistic for \(\theta\).