Combining Convergence in Distribution and Almost Surely

Question

Suppose I know that $\frac{1}{\sqrt{m}}X(mt)\xrightarrow[m\to\infty]{d} B(t)$ where $B$ is Brownian motion, and that $\alpha(mt)/m \xrightarrow{a.s.} 0$. Then I am trying to prove the following convergence: $$ \frac{1}{\sqrt{m}}X(mt+\alpha(mt))\xrightarrow[m\to\infty]{d} B(t) $$ My approach is to try and prove as follows. Fix $t,x,\epsilon$. So by the continuity of the density of $B(t)$, $\exists \delta>0$ such that $|s-t|<\delta \implies |\mathbb{P}(B(s)\leq x)-\mathbb{P}(B(t)\leq x)|<\epsilon$. Then since $\alpha(mt)/m \to 0$ almost surely, therefore: $$ \mathbb{P}\left(\frac{1}{\sqrt{m}}X\left(m(t+\alpha(mt)/m)\right)\leq x\right)=\mathbb{P}\left(\frac{1}{\sqrt{m}}X\left(m(t+\alpha(mt)/m)\right)\leq x, \quad \exists M\; s.t.\; m>M\implies |\alpha(mt)/m|<\delta\right) $$ And I want at this point to say that this converges to $B(s)$ in distribution for some $s$ such that $|s-t|<\delta$, so then since $\epsilon$ is arbitrary, this converges in distribution to $B(t)$. This seems intuitive enough, however it is this step when I somehow want to use the almost sure convergence within the $\mathbb{P}(.)$ which is the step I am unsure on. Do I need more justification than this? Any help is appreciated.

Answer 1

I posted this answer at MathSE also.

Here, use your symbols, a proof will be given under following condition (1): \begin{equation*} \lim_{m\to\infty}\mathsf{P}\Big(\sup_{0\le t\le 1}\Big| \frac{\alpha(mt)}{m} \Big| \ge\delta \Big)=0,\qquad \forall \delta>0. \tag{1} \end{equation*} Let \begin{align*} Y_m(t) &=\frac{1}{\sqrt{m}}X(mt), \tag{2} \\ Z_m(t) &=\frac{1}{\sqrt{m}}X\big(m(t+\tfrac{\alpha(mt)}{m})\big) =Y_m\big(t+\tfrac{\alpha(mt)}{m}\big), \tag{3}\\ U_m(t) &=Z_m(t)-Y_m(t)=Y_m\big(t+\tfrac{\alpha(mt)}{m}\big) - Y_m(t). \tag{4} \end{align*}

In the following, the concept of modulus of continuity of an arbitrary function $x(.)$ on $[0,1]$ also be used(cf. P. Billingsley, Convergence of Probability Measures, 2ed. $\S$7, pp 80-81). Let \begin{equation*} w(x,\delta)=\sup_{|s-t|\le \delta}|x(s)-x(t)|, \qquad 0<\delta \le 1. \end{equation*} For fixed $\delta>0$, since \begin{equation*} |w(x,\delta)-w(y,\delta)|\le 2\sup\limits_{0\le t\le 1}|x(t)-y(t)|, \end{equation*} $w(x,\delta)$ is continuous in $x$, using this and $Y_m\stackrel{d}{\longrightarrow}B$(i.e., the distributions of $Y_m$ converge to to the distribution of BM $B$ weakly on $C[0,1]$), get \begin{gather*} \lim_{m\to\infty}w(Y_m,\delta)\stackrel{d}{=} w(B,\delta), \qquad \forall \delta>0,\\ \varlimsup_{m\to\infty} \mathsf{P}(w(Y_m,\delta)\ge \epsilon)\le \mathsf{P}(w(B,\delta) \ge \epsilon)\quad \forall \epsilon,\delta>0, \\ \lim_{\delta\to0}\varlimsup_{m\to\infty} \mathsf{P}(w(Y_m,\delta) \ge \epsilon)=0, \quad \forall \epsilon>0.\tag{5} \end{gather*} Next, the following fact will be proved, \begin{equation*} \lim_{m\to\infty}\mathsf{P}(\sup_{0\le t\le 1}|U_m(t)|>\epsilon)=0, \quad \forall \epsilon>0. \tag{6} \end{equation*} Denote \begin{equation*} A_m(\delta)=\Big\{\omega: \sup_{0\le t\le 1}\Big| \frac{\alpha(mt)}{m} \Big| \ge \delta \Big\}, \end{equation*} then from (1) \begin{gather*} \lim_{m\to\infty}\mathsf{P}(A_m(\delta))=0, \qquad \forall \delta>0,\\ \lim_{m\to\infty}\mathsf{P}([\sup_t|U_m(t)|]I_{A_m(\delta)}>\epsilon)=0, \qquad \forall \delta, \epsilon>0. \tag{7} \end{gather*} On the other hand, \begin{gather*} |U_m(t)| I_{A_m^c(\delta)}\le w(Y_m,\delta), \qquad \forall t\in[0,1],\\ [\sup_{0\le t\le 1}|U_m(t)|]I_{A_m^c(\delta)}\le w(Y_m,\delta). \tag{8} \end{gather*} From (8),(5), get \begin{equation*} \lim_{\delta\to0}\varlimsup_{m\to\infty} \mathsf{P}([\sup_{0\le t\le 1}|U_m(t)|]I_{A_m^c(\delta)}>\epsilon)=0, \quad \forall \epsilon>0.\tag{9} \end{equation*} Hence, from (7)and (9), (6) is true. Furthermore, $Z_m\stackrel{d}{\longrightarrow}B$ comes from $Y_m\stackrel{d}{\longrightarrow}B$ and (4),(6).