 Solution

The almost-sure bound gives ${\tilde{\theta }}_n-{\hat{\theta }}_n=o_{\mathbb{P}}(n^{-1/2})$, so Slutsky theorem transfers both consistency and the asymptotic normal law to ${\tilde{\theta }}_n$. The continuous mapping theorem and delta method then giveIf ${\psi }^{\prime }({\theta }_0)=0$, the latter is the Dirac measure at zero, conventionally a degenerate normal distribution; a nondegenerate second-order limit would need a different normalization and extra smoothness.