 Solution

Writing $S=\sum X_i$, the likelihood function is ${\theta }^S(1-\theta {)}^{n-S}$ and its supremum occurs at $\hat{\theta }=S/n$, including the limiting boundary cases zero and one. The likelihood ratio compares $2^{-n}$ under the null with that supremum. ThusUse $0\log 0=0$ at boundary outcomes. Wilks theorem gives asymptotic ${\chi }_1^2$ under the regular interior null, because one parameter is constrained. Directly, setting $\hat{\theta }=1/2+Z$ and expanding the bracket gives $2Z^2+O(Z^4)$, henceUnder the null, $2\sqrt{n}{Z}_n\stackrel{d}{\to }N(0,1)$ and $Z_n=O_P(n^{-1/2})$. The remainder vanishes in probability, proving the same chi-squared distribution limit. If an observed $Z_n$ is not small, the quadratic numerical approximation is unreliable and the exact statistic should be used; exact binomial calibration is available. Persistent non-small deviations under a fixed alternative instead make the statistic grow linearly with $n$, rather than having the null chi-squared distribution law.