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Take two independent statistical samples from beta distributions with parameters $(\theta ,1)$ and $(\tau ,1)$, of positive sizes $n,m$. Put $A=-{\sum }_i\log X_i$, $B=-{\sum }_j\log Y_j$, and $T=A/(A+B)$. To test the null hypothesis $\theta =\tau$, divide the maximized common-rate likelihood function by the unrestricted maximum to obtain the displayed likelihood ratio. Indeed the unrestricted fitted rates are $n/A,m/B$, and the common fitted rate is $(n+m)/(A+B)$; substitution cancels the exponential factors.

Under the null hypothesis, $A$ and $B$ have independent gamma distributions with a common rate. Beta-gamma independence gives $T\sim \mathrm{Beta}(n,m)$. The derivative of $\log (T^n(1-T{)}^m)$ changes sign at $n/(n+m)$, so the exact critical region of size $\alpha$ is $T<k_1$ or $T>k_2$, with $k_1<n/(n+m)<k_2$. Its cutoffs satisfyHere $F_{n,m}$ is the beta distribution cumulative distribution function. Equal likelihood ratios at the cutoffs do not generally mean equal tail probabilities. Increasing the ratio threshold continuously adds positive null distribution mass on both sides, proving uniqueness for $0<\alpha <1$.

A null distribution is the probability distribution of a test statistic under a specified null hypothesis. It calibrates a hypothesis test's test size and p-value. For a simple hypothesis it is one distribution; for a composite null hypothesis it may depend on unknown nuisance parameters.

Let $S={\sum }_i{X}_i$. For the exponential distribution with mean $\theta$, the likelihood function is $L(\theta )={\theta }^{-n}{e}^{-S/\theta }$ on positive observations. Fix any $0<{\theta }_1<{\theta }_0$. The alternative-to-null likelihood ratio isSince $1/{\theta }_1-1/{\theta }_0>0$, this is strictly decreasing in $S$. The Neyman-Pearson lemma says to reject for small $S$. Under the null hypothesis, the given sum rule yields $S\sim \mathrm{Gamma}(n,1/{\theta }_0)$, where the second argument is a rate, and henceLet $q_{\alpha }$ be the lower $\alpha$ quantile of this chi-squared distribution. The required critical region isIts test size is exactly $\alpha$ because the null distribution is continuous. Its cutoff is the same for every ${\theta }_1<{\theta }_0$, so it is simultaneously most powerful against every such value: it is a uniformly most powerful test. Equivalently, reject if $S$ is below the lower $\alpha$ quantile of the specified gamma distribution.

A likelihood-ratio test rejects for small $\lambda$. The derivative of the natural logarithm of its variable factor isThus $t^n(1-t{)}^m$ increases up to $n/(n+m)$ and decreases thereafter, tending to zero at both endpoints. Every nontrivial lower level set of $\lambda$ therefore consists of two tails. For a prescribed test size $0<\alpha <1$, the exact critical region isWriting $F_{n,m}$ for the cumulative distribution function of the beta distribution with parameters $(n,m)$, the required two equations areThe first makes both cutoffs have the same likelihood ratio; the second calibrates the null hypothesis rejection probability. Equivalently, the second left-hand side isAs the common likelihood ratio threshold increases from zero to one, this probability increases continuously and strictly from zero to one; hence the equations determine unique cutoffs. The two tail probabilities need not each equal $\alpha /2$ when $n\ne m$. Exact calibration holds for every common rate because the given null distribution of $T$ is independent of that rate. Boundary equalities have zero probability.

True, with the test's rejection rule fixed. If its rejection region in the space of the test statistic is $C$, its statistical power against the specified simple hypothesis $H_1$ isThus the probability distribution of $T$ under the alternative hypothesis determines the statistical power of this test. The null distribution is used to choose a rule with the desired test size, but statistical power is evaluated under the alternative hypothesis. Different rejection rules can have different statistical power even for the same probability distribution of $T$.