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For an aperiodic positive Harris recurrent Markov chain with geometric ergodicity, an observable with stationary moment $\pi (|h{|}^{2+\delta })<\infty$ for some $\delta >0$ satisfies a central limit theorem. Its asymptotic variance is ${\gamma }_h(0)+2{\sum }_{k\ge 1}{\gamma }_h(k)$, with stationary covariances. Ergodicity alone does not guarantee this theorem.

For positive claims with finite variance and $c\le \lambda \mu$, the classical risk model has ruin probability one from any finite capital. At claim times, surplus increments are independent copies of $cT-X$ with mean $c/\lambda -\mu$. Negative mean sends their partial sums to minus infinity by the strong law of large numbers. At zero mean, the increments have finite nonzero variance. For every fixed $K$, the central limit theorem gives limiting probability $1/2$ of a partial sum below $-K$. The probability of unboundedness below is therefore at least $1/2$; as a tail event it has probability zero or one by the Kolmogorov zero-one law, and hence one.

For independent observations from the exponential distribution with rate $\theta$, the log-likelihood is $n\log \theta -\theta {\sum }_i{X}_i$, soThe central limit theorem and delta method give

With absolutely summable autocovariance, the asymptotic variance of the sample mean satisfiesThe sum is signed, rather than a sum of absolute values. It can be zero: the first difference of strong white noise has telescoping partial sums. A central limit theorem needs further dependence assumptions.

A normal approximation replaces a sampling distribution by a normal distribution, often through a central limit theorem. In a Wald test, approximate statistical power is determined by the standardized difference under the alternative. This yields Neyman allocation for fixed treatment-specific variances.