Suppose that $Y_1, \ldots, Y_{n}$ are iid from a Bernoulli distribution with parameter $p$ and consider $H_0 : p = p_0\,.$ The test statistics are $$ T_W = \frac{n ({\widehat p} - p_0)^2}{{\widehat p} (1 - {\widehat p})}\,, \:\:\:\: T_S = \frac{n ({\widehat p} - p_0)^2}{p_0 (1 - p_0)} \,,$$ and $ T_{LR} = -2 [ X \log(p_0/{\widehat p}) + (n-X) \log[(1 - p_0)/(1 -\widehat p)] ], $ where $X = \sum_{i=1}^{n} Y_i\,.$ I want to Show that $T_W, T_S$ and $T_{LR}$ are asymptotically equivalent under $H_0$ by showing that their differences converge to $0$ in probability.
I showed that $T_W $ is asymptotically equivalent to $T_S$ as follows We have \begin{align*} T_W-T_S &=n ({\widehat p} - p_0)^2 [\frac{1}{{\widehat p} (1 - {\widehat p})}- \frac{1}{p_0 (1 - p_0)}]\\ &= p_0 (1 - p_0) \left( \frac{\sqrt{n} ({\widehat p} - p_0)}{\sqrt{p_0 (1 - p_0) }}\right)^{2}[\frac{1}{{\widehat p} (1 - {\widehat p})}- \frac{1}{p_0 (1 - p_0)}] \end{align*} From the asymptotic normality of the MLE, we have that
$$\left( \frac{\sqrt{n} ({\widehat p} - p_0)}{\sqrt{p_0 (1 - p_0) }}\right)^{2} \overset{d}{\rightarrow} \mathcal{N}(0,1)^{2}=\chi^{2}(1),$$ and from the consistency of the MLE we have that $$ {\widehat p} \overset{p}{\rightarrow} p_0. $$ From Slutsky’s theorem and the continuous mapping theorem (CMT) we have $$ [\frac{1}{{\widehat p} (1 - {\widehat p})}- \frac{1}{p_0 (1 - p_0)}] \overset{p}{\rightarrow} 0.$$ This implies that \begin{align*} T_W-T_S & \overset{d}{\rightarrow} 0\\ T_W-T_S & \overset{p}{\rightarrow} 0. \end{align*}
But I did not succeed at showing the same conclusion for $T_W$ and $T_{LR}$. Can someone help?
Use the Taylor expansion of $T_{LR}$ around $p_0$:
\begin{align} T_{LR}&=-2n\left[\hat p \ln\left(\frac{p_0}{\hat p}\right) + (1-\hat p) \ln\left(\frac{1-p_0}{1-\hat p}\right)\right] \\ &=\frac{n}{p_0(1-p_0)}(\hat p-p_0)^2+O_p(|\hat p-p_0|^3) \\[1em] &=T_S+o_p(1)=T_W+o_p(1). \end{align}