Statistic converging to a consistent statistic is convergent--why is this wrong? I was doing a problem in statistics and I was trying to show that a certain estimator is consistent. Basically, we have an estimator $\hat{\theta}$ which is known to be consistent for some parameter $\theta$,
$\hat{\theta} = \frac{X}{X + 2}$
We also have a statistic, for some $a, b > 0$,
$\tilde{\theta} = \frac{a + nX}{a + nX + 2n + b}$.
I wanted to show that $\tilde{\theta}$ is also consistent. I have an argument but I doubt whether or not it makes sense--can someone point out the error? Let $\epsilon > 0$;
$$ 
\begin{align}
\boldsymbol{P}(|\tilde{\theta} - \theta|>2\epsilon) &= \boldsymbol{P}(|\tilde{\theta} - \hat{\theta} + \hat{\theta} -  \theta|>2\epsilon)\\
& \leq \boldsymbol{P}(|\tilde{\theta} - \hat{\theta}| + |\hat{\theta} -  \theta|>2\epsilon)\qquad(*)
\end{align}
 $$
since if $\omega$ in the sample space is such that $|\tilde{\theta}(\omega) - \hat{\theta}(\omega) + \hat{\theta}(\omega) -  \theta|>2\epsilon$, then it must be (?) that $\omega$ is also such that
$ |\tilde{\theta}(\omega) - \hat{\theta}(\omega) |+| \hat{\theta}(\omega) -  \theta|>2\epsilon $.
I also thought that
$$ 
\begin{align}
(*) \leq \boldsymbol{P}(\{|\tilde{\theta} - \hat{\theta}|>\epsilon\} \cup \{|\hat{\theta} -  \theta|>\epsilon\})
\end{align}
$$
since either  of $|\tilde{\theta} - \hat{\theta}|$ or $|\hat{\theta} -  \theta|$ must be greater than $\epsilon$. I thought this should hold since if $\omega$ in the sample space is such that
$$\omega \in \{ \omega' | \ \ |\tilde{\theta}(\omega') - \hat{\theta}(\omega') |+| \hat{\theta}(\omega') -  \theta(\omega')|>2\epsilon \}$$ then $\omega$ should be such that
$$
\omega \in \{\omega' \ \ |\tilde{\theta}(\omega') - \hat{\theta}(\omega')|>\epsilon\} \cup \{|\hat{\theta}(\omega') -  \theta|>\epsilon\}.
$$
It would then follow that
$$
\boldsymbol{P}(\{|\tilde{\theta} - \hat{\theta}|>\epsilon\} \cup \{|\hat{\theta} -  \theta|>\epsilon\}) \leq \boldsymbol{P}(|\tilde{\theta} - \hat{\theta}|>\epsilon) + \boldsymbol{P}(|\hat{\theta} -  \theta|>\epsilon).
$$
Now, I figure since $\tilde{\theta}$ converges to $\hat{\theta}$, and since $\hat{\theta}$ is consistent for $\theta$, (i.e. the probability that $\theta$ and $\hat{\theta}$ are different goes to zero as $n \rightarrow \infty$), the expression we started with tends to zero as $n \rightarrow \infty$, hence $\tilde{\theta}$ should be consistent for $\theta$.
I assume this must be wrong because I never see arguments like this, and I probably have my inequalities pointing the wrong way. I'm sorry if I have made really elementary mistakes. Can someone please tell my why this is wrong?
Thank you!
 A: It's not wrong and is a common proof found on the Internet.
If $\hat\theta\rightarrow^p \theta$ and $\hat\theta-\tilde\theta\rightarrow^p0$ then $\tilde \theta\rightarrow ^p\theta$. If an estimator is consistent for a parameter and another estimator converges in probability to that estimator, then it must be that the second estimator also converges in probability to the parameter/is consistent.
First for any arbitrarily small $\epsilon>0$, $\lim_{n\rightarrow\infty}\Pr(|\hat\theta-\tilde\theta-0|>\epsilon)=\Pr\left(\big|\frac{X}{X+2}-\frac{X}{X+2}\big|>\epsilon\right)=\Pr(0>\epsilon)=0$ so $\hat\theta-\tilde\theta$ converges in probability to $0$.
Choose a small positive $\epsilon$. From the triangle inequality and the fact that $A+B>2\epsilon \subseteq A>\epsilon\cup B>\epsilon$ (if $A+B>2\epsilon$, then $A$ is at least $\epsilon$ or $B$ is at least $\epsilon$, but there are some cases where $A>\epsilon$ or $B>\epsilon$ but $A+B\not\gt2\epsilon$), we have that
$$\begin{split}\Pr(|\tilde \theta-\theta|>2\epsilon)&=\Pr(|\tilde \theta-\hat\theta+\hat\theta-\theta|>2\epsilon)\\
&\le\Pr(|\tilde\theta-\hat\theta|+|\hat\theta-\theta|>2\epsilon)\\
&\le\Pr(|\tilde\theta-\hat\theta|>\epsilon\cup|\hat\theta-\theta|>\epsilon)\\
&\le\Pr(|\tilde\theta-\hat\theta-0|>\epsilon)+\Pr(|\hat\theta-\theta|>\epsilon)\end{split}$$
so that $$\lim_{n\rightarrow\infty}\Pr(|\tilde \theta-\theta|>2\epsilon)\le0$$
In fact probabilities are nonnegative so
$$\lim_{n\rightarrow\infty}\Pr(|\tilde \theta-\theta|>2\epsilon)=0$$
Since $\epsilon$ was arbitrary, $\tilde\theta\rightarrow^p\theta$.
