The question asks to identify which of the given statistics is NOT a sufficient statistic for the parameter $\theta$ of an exponential distribution, given a random sample \(X_1, X_2, \dots, X_n\).

Exponential Distribution and Likelihood

An exponential distribution with mean \(\theta\) has the probability density function (PDF):

\begin{equation*} f(x; \theta) = \frac{1}{\theta} e^{-x/\theta}, \quad \text{for } x \ge 0, \theta > 0 \end{equation*}

Given a random sample \(X_1, X_2, \dots, X_n\), the likelihood function is the joint PDF of the sample, treated as a function of the parameter \(\theta\):

\begin{align*} L(\theta; x_1, \dots, x_n) &= \prod_{i=1}^n f(x_i; \theta) \\ &= \prod_{i=1}^n \left( \frac{1}{\theta} e^{-x_i/\theta} \right) \\ &= \left(\frac{1}{\theta}\right)^n e^{-\frac{1}{\theta} \sum_{i=1}^n x_i} \\ &= \theta^{-n} e^{-\frac{1}{\theta} \sum_{i=1}^n x_i}\end{align*}

Let \(S = \sum_{i=1}^n X_i = X_1 + X_2 + \dots + X_n\). The likelihood function can be written as:

\begin{equation*} L(\theta; x_1, \dots, x_n) = \theta^{-n} e^{-\frac{1}{\theta} S} \end{equation*}

Sufficient Statistics Definition

A statistic \(T(X_1, \dots, X_n)\) is called a sufficient statistic for a parameter \(\theta\) if it contains all the information about \(\theta\) that is present in the sample. Informally, knowing the value of a sufficient statistic is as good as knowing the entire sample for the purpose of estimating the parameter.

The Factorization Theorem provides a formal way to check for sufficiency. A statistic \(T(X_1, \dots, X_n)\) is sufficient for \(\theta\) if and only if the likelihood function can be factored in the following way:

\begin{equation*} L(\theta; x_1, \dots, x_n) = g(T(x_1, \dots, x_n); \theta) \cdot h(x_1, \dots, x_n) \end{equation*}

where \(g\) is a function that depends on \(\theta\) and the statistic \(T\), and \(h\) is a function that depends only on the sample data \((x_1, \dots, x_n)\) but NOT on \(\theta\).

Analyzing Statistic Options

From the likelihood function \(L(\theta; x_1, \dots, x_n) = \theta^{-n} e^{-\frac{1}{\theta} \sum x_i}\), we see that it depends on the sample data only through the sum \(S = \sum_{i=1}^n X_i\). Applying the Factorization Theorem, we can set \(T(x_1, \dots, x_n) = \sum_{i=1}^n x_i = S\). Then \(g(S; \theta) = \theta^{-n} e^{-\frac{1}{\theta} S}\) and \(h(x_1, \dots, x_n) = 1\). Here \(g\) depends on \(S\) and \(\theta\), and \(h\) does not depend on \(\theta\). Therefore, \(S = \sum_{i=1}^n X_i\) is a sufficient statistic for \(\theta\).

• Option 2: \(X_1 + X_2 + \dots + X_n\). This is exactly \(S\), which we have shown is a sufficient statistic for \(\theta\).
• Option 1: \(\frac{1}{X_1 + X_2 + \dots + X_n} = \frac{1}{S}\). If \(T\) is a sufficient statistic for \(\theta\), then any one-to-one function of \(T\) is also a sufficient statistic for \(\theta\). Since \(S = \sum X_i > 0\) (for samples from exponential distribution where \(X_i \ge 0\) and at least one \(X_i > 0\)), the function \(f(s) = 1/s\) is a one-to-one function for \(s > 0\). Thus, \(\frac{1}{S}\) is also a sufficient statistic for \(\theta\).
• Option 4: \((X_n, X_1 + X_2 + \dots + X_{n-1})\). Let \(T = (X_n, S_{n-1})\) where \(S_{n-1} = \sum_{i=1}^{n-1} X_i\). The likelihood function is \(L(\theta; x_1, \dots, x_n) = \theta^{-n} e^{-\frac{1}{\theta} (x_n + S_{n-1})}\). We can choose \(g((x_n, S_{n-1}); \theta) = \theta^{-n} e^{-\frac{1}{\theta} (x_n + S_{n-1})}\) and \(h(x_1, \dots, x_n) = 1\). Here \(g\) depends on the statistic \((X_n, S_{n-1})\) and \(\theta\), and \(h\) does not depend on \(\theta\). By the Factorization Theorem, the pair \((X_n, X_1 + X_2 + \dots + X_{n-1})\) is a sufficient statistic for \(\theta\). (Alternatively, since \(S = X_n + S_{n-1}\) is a sufficient statistic and \(S\) can be calculated from the pair \((X_n, S_{n-1})\), this pair is also sufficient).
• Option 3: \(\frac{X_n}{X_1 + X_2 + \dots + X_{n-1}} = \frac{X_n}{S_{n-1}}\). Let's check if this statistic is sufficient. If \(T = \frac{X_n}{S_{n-1}}\) is sufficient, then the ratio of likelihoods for any two samples \(\mathbf{x}_1\) and \(\mathbf{x}_2\) such that \(T(\mathbf{x}_1) = T(\mathbf{x}_2)\) must be independent of \(\theta\). Consider the case with \(n=2\). The statistic is \(T = X_2/X_1\). Let \(\mathbf{x}_1 = (1, 2)\) and \(\mathbf{x}_2 = (2, 4)\). \(T(\mathbf{x}_1) = 2/1 = 2\). \(T(\mathbf{x}_2) = 4/2 = 2\). So \(T(\mathbf{x}_1) = T(\mathbf{x}_2)\). The likelihoods are: \(L(\theta; \mathbf{x}_1) = \theta^{-2} e^{-\frac{1}{\theta}(1+2)} = \theta^{-2} e^{-3/\theta}\) \(L(\theta; \mathbf{x}_2) = \theta^{-2} e^{-\frac{1}{\theta}(2+4)} = \theta^{-2} e^{-6/\theta}\) The ratio of the likelihoods is: \(\frac{L(\theta; \mathbf{x}_1)}{L(\theta; \mathbf{x}_2)} = \frac{\theta^{-2} e^{-3/\theta}}{\theta^{-2} e^{-6/\theta}} = e^{-3/\theta - (-6/\theta)} = e^{3/\theta}\) Since the ratio \(e^{3/\theta}\) depends on \(\theta\), the statistic \(T = X_2/X_1\) is NOT a sufficient statistic for \(\theta\). This reasoning extends to the general case \(\frac{X_n}{S_{n-1}}\).

Conclusion

Based on the analysis using the Factorization Theorem and checking the properties of sufficient statistics, the statistic \(\frac{X_n}{X_1 + X_2 + \dots + X_{n-1}}\) is NOT a sufficient statistic for the parameter \(\theta\) of the exponential distribution.