To find a complete sufficient statistic for $\theta$ when $\mathbf{X}$ is a random sample from $N(0, \theta)$, we can use the definition of sufficiency and the factorization theorem.

Step-by-Step Solution:

1. Likelihood Function:

For a sample $\mathbf{X} = (X_1, X_2, \ldots, X_n)$ from $N(0, \theta)$, the probability density function for each observation $X_i$ is:

$f(x_i; \theta) = \frac{1}{\sqrt{2\pi\theta}} \exp\left(-\frac{x_i^2}{2\theta}\right)$

The likelihood function for the sample is:

$L(\theta; \mathbf{x}) = \prod_{i=1}^{n} \frac{1}{\sqrt{2\pi\theta}} \exp\left(-\frac{x_i^2}{2\theta}\right) = \left(\frac{1}{\sqrt{2\pi\theta}}\right)^n \exp\left(-\frac{1}{2\theta} \sum_{i=1}^{n} x_i^2\right)$

2. Apply the Factorization Theorem:

According to the factorization theorem, a statistic $S(\mathbf{X})$ is sufficient for $\theta$ if the likelihood function can be factored as:

$L(\theta; \mathbf{x}) = g(S(\mathbf{x}); \theta)h(\mathbf{x})$

In our case, $S(\mathbf{X}) = \sum_{i=1}^{n} X_i^2$.

3. Complete Sufficient Statistic:

A complete sufficient statistic for $\theta$ is one where no non-trivial function of the statistic follows a distribution independent of $\theta$.

For a normal distribution with known mean and unknown variance, the sum of squares, $\sum_{i=1}^{n} X_i^2$, is a complete sufficient statistic for $\theta$.

Conclusion:

Thus, $\sum_{i=1}^{n} X_i^2$ is a complete sufficient statistic for $\theta$ when $\mathbf{X}$ is a random sample from $N(0, \theta)$.