Given a sequence defined by
$$ X_{n,\varepsilon}(\theta) = \int_0^1 \varepsilon f(Y_{n,s},\theta)\chi_{\{s \leq \tau\}} dW_s $$
where $W_s$ is a standard Brownian motion and $\tau$ is the time such that the process exceeds some $M >0$ and $Y_n \to Y$ as $n \to \infty, \varepsilon \to 0$, what $Y$ is specifically isn't important to the question other than it is a deterministic process.
Additionally, $f$ has a linear growth condition.
So for fixed $M$ can show point-wise convergence to $0$ as $n \to \infty$ and $\varepsilon \to 0$. Proof uses Chebyshev's inquality.
Now, consider
$$ \mathbb{P}(|X_{n,\varepsilon}(\theta)|\ge a) = \mathbb{P}(|X_{n,\varepsilon}(\theta)|^2\ge a^2) \leq \frac{1}{a^2}\mathbb{E}(|X_{n,\varepsilon}(\theta)|^2) $$ So for sufficiently large $a$ can make this small, say $a >> M$.
This means I can conclude the sequence $(X_{n,\varepsilon}(\theta)$ is Uniformly Tight, correct?
Where uniformly tight means For every $\delta >0$, there exists $K$ such that $$ \sup_{n,\varepsilon}\mathbb{P}(|X_{n,\varepsilon}(\theta)|\ge K) \leq \delta. $$
This seems correct to me but I want to insure I am indeed correct.