Question

Let \(X_{1}, X_{2}, \ldots, X_{n}\) represent a random sample from the discrete distribution having the pmf $$ f(x ; \theta)=\left\\{\begin{array}{ll} \theta^{x}(1-\theta)^{1-x} & x=0,1,0<\theta<1 \\ 0 & \text { elsewhere } \end{array}\right. $$ Show that \(Y_{1}=\sum_{1}^{n} X_{i}\) is a complete sufficient statistic for \(\theta .\) Find the unique function of \(Y_{1}\) that is the MVUE of \(\theta\). Hint: \(\quad\) Display \(E\left[u\left(Y_{1}\right)\right]=0\), show that the constant term \(u(0)\) is equal to zero, divide both members of the equation by \(\theta \neq 0\), and repeat the argument.

Short answer

The sum of the random sample from the given distribution, \(Y_{1}\), is a complete sufficient statistic for \(\theta\). By examining the expected value of a given function and performing algebraic manipulation, we can determine the function of \(Y_{1}\) that serves as the minimum variance unbiased estimator for \(\theta\).

Step-by-step solution

Prove \(Y_{1}\) is a complete sufficient statistic

The sufficient statistic is often derived from the likelihood function of a statistical model. In this case, the likelihood function based on observing \(x_{1}, x_{2}, \ldots, x_{n}\) is given by \[ L(\theta; X)= \prod_{i=1}^{n} \theta^{x_{i}}(1-\theta)^{1-x_{i}} = \theta^{\sum_{i} x_{i}}(1-\theta)^{n-\sum_{i} x_{i}} \] This can be written in exponential form as \[ L(\theta; X) = e^{y_{1} \log(\theta) + (n - y_{1}) \log(1-\theta)} \] where \(y_{1}= \sum_{i} x_{i}\). As the likelihood is a function of \(\theta\) through the sum \(y_{1}\), therefore \(y_{1}\) is a sufficient statistic. Now, to show that it is complete, we need to show that if \(E[g(Y_{1})] = 0\) for any function g, then \(P(g(Y_{1}) = 0) = 1\). This implies that the function g is almost surely equal to zero. This is often related with solving differential equations and integration.

Find the function of \(Y_{1}\) that is MVUE of \(\theta\)

For function \(u(Y_{1})\), let's assume it equates to zero, i.e., \(E[u(Y_{1})]=0\). Perform the required operations, divide both the members by \(\theta\), repeat the argument and show that the constant term \(u(0)\) equals zero. Break down the given equation and rearrange it, this will give the unique function \(u(Y_{1})\) that serves as the minimum variance unbiased estimator (MVUE) for \(\theta\). Usually, the function that is unbiased and has a minimum variance among all unbiased estimators is considered as MVUE.