# Joint convergence of stochastic processes

Suppose I have processes $X_n(t)\overset{d}{\longrightarrow} X(t)$, and $Y_n(t)\overset{p}{\longrightarrow} ct$ for some constant $c$. Then, can I conclude like in Slutsky's theorem that $(X_n(t),Y_n(t))\overset{d}{\longrightarrow} (X(t),ct)$?

I tried to adapt the proof for normal random variables, so proving that for $\phi$ bounded and continuous, $\mathbb{E}[\phi(X_n(t),Y_n(t))]\to\mathbb{E}[\phi(X(t),ct)]$. First, we see that $\phi(X_n(t),ct)$ is also a bounded, continuous function, and by the convergence of $X_n(t)$, therefore $\mathbb{E}[\phi(X_n(t),ct)]\to\mathbb{E}[\phi(X(t),ct)]$, and thus $(X_n(t),ct)\overset{d}{\longrightarrow} (X(t),ct)$.

Also, $|(X_n(t),Y_n(t))-(X_n(t),ct)|\overset{p}{\longrightarrow}0$. Then, as here (Convergence in probability to a sequence converging in distribution implies convergence to the same distribution), we can show that $(X_n(t),Y_n(t))\overset{d}{\longrightarrow}(X(t),ct)$.

This all seems to go very well, but I am not completely sure whether my arguments actually hold for processes instead of random variables. So do you think this argument is correct? Or does an equivalent of Slutsky's theorem not hold for stochastic processes?

• some clarification are needed about your notations $X_n(t)\overset{d}{\longrightarrow} X(t)$ are you claiming that this is true for every $t$ or that convergence of X_n to X holds on the probability space supporting the trajectories of X and X_n ? Best regards. Feb 8, 2016 at 22:47
• I mean convergence with respect to the Skorokhod topology.
– ckoe
Feb 9, 2016 at 8:02

Let $S$ and $T$ be complete separable metric spaces, and let $(X_n)$, $X$ (resp. $(Y_n)$, $Y$) be random elements of $S$ (resp. $T$) such that $X_n\overset{d}{\longrightarrow} X$ and $Y_n\overset{P}{\longrightarrow} Y$. Crucially, suppose that $Y$ is deterministic. Let $f:S\times T\to\Bbb R$ be bounded and continuous. Write the difference $\Bbb E[f(X_n,Y_n)]-\Bbb E[f(X,Y)]$ as the sum of $\Bbb E[f(X_n,Y_n)]-\Bbb E[f(X_n,Y)]$ and $\Bbb E[f(X_n,Y)]-\Bbb E[f(X,Y)]$. The latter difference tends to $0$ as $n\to\infty$ because $X_n\overset{P}{\longrightarrow} X$ and $Y$ is deterministic.
To deal with the former difference, let $M$ be a bound for $f$: $|f(x,y)|\le M$ for all $(x,y)\in S\times T$. Fix $\epsilon>0$. By tightness of the sequences $(X_n)$ and $(Y_n)$ there are compact sets $K_S\subset S$ and $K_T\subset T$ such that $$\Bbb P[X_n\notin K_S]+\Bbb P[X\notin K_S]+\Bbb P[Y_n\notin K_T]<\epsilon$$ for all $n$. The restriction of $f$ to $K_S\times K_T$ is uniformly continuous, so there exists $\delta>0$ such that if $(x,y)$ and $(x',y')$ are elements of $K_S\times K_T$ with $d_S(x,x')+d_T(y,y')<\delta$ then $|f(x,y)-f(x',y')|<\epsilon$. Thus \eqalign{ |\Bbb E[f(X_n,Y_n)]-\Bbb E[f(X_n,Y)]| &\le 2M\Bbb P[d_T(Y_n,Y)>\delta]+2M\epsilon\cr &\phantom{bbbbb}+\Bbb E[|f(X_n,Y_n)-f(X_n,Y)|;X_n\in K_S,Y_n\in K_T,d_T(Y_n,Y)<\delta)]\cr &\le 2M\Bbb P[d_T(Y_n,Y)>\delta]+2M\epsilon+\epsilon.\cr } Because $Y_n\overset{P}{\longrightarrow} Y$, $\lim_n\Bbb P[d_T(Y_n,Y)>\delta]=0$. Therefore $$\limsup_n |\Bbb E[f(X_n,Y_n)]-\Bbb E[f(X_n,Y)]|\le (2M+1)\epsilon,$$ for all $\epsilon>0$. This shows that $\lim_n |\Bbb E[f(X_n,Y_n)]-\Bbb E[f(X_n,Y)]|=0$.
• Yes, the test function $f$ is to be continuous with respect to the product topology on $S\times T$. Feb 13, 2016 at 16:33