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Let $(X^n_t)_{t \geq 0}$ be a sequence of cadlag stochastic processes, that is $X^n$ is a random element in the Skorokhod space $D([0, \infty), \mathbb{R}$) for each $n \in \mathbb{N}$. Also for each $n$ let $T^n_t$ be a cadlag stochastic process in $D([0, \infty), [0, \infty))$ which is non-decreasing.

Assume that

  1. $X^n$ converges in finite-dimensional distributions to a cadlag process $X$ (but not necessarily in distribution) and
  2. $T^n$ converges in distribution to the identity $\text{id}$.

Is it true that the composition $X^n \circ T^n$ (i.e. the process $(X^n_{T^n_t})_{t \geq 0}$) also converges in finite-dimensional distributions to $X$?

I know that this result is true when $X^n$ converges in distribution since

  • the composition has some nice continuity properties (see Ward Whitt, "Stochastic Process Limits") ... the limit $\text{id}$ of $T^n$ is continuous and strictly increasing
  • $T^n$ converges to a non-random process and thus we have joint convergence in distribution $(X^n, T^n)$ and
  • we can apply the Continuous Mapping Theorem.

But when I weaken the convergence of $X^n$ to only convergence in finite-dimensional distributions and on the other hand seek also only for convergence in finite-dimensional distributions of the composed process, then I don't know whether this type of convergence still holds.

yada
  • 3,535

1 Answers1

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The conjecture is wrong. Here is a counterexample.

For each $n$ consider a continuous-time Markov chain $X^n_t$ over two states 1 and 2 with 1 as initial state and only one transition from 1 to 2 with rate $n$. Denote by $P^n(t)$ the semigroup of $X^n$. Let $X_t$ be a Markov chain with semi group $P(0) = I$ and $P(t) = \begin {pmatrix} 0 & 1 \\ 0 & 1 \end{pmatrix}$ for $t > 0$ and same initial distribution. Note that $X$ is not continuous in probability. It holds that $P^n \to P$ pointwise and thus $X^n \to X$ in finite dimensional distributions but not in distribution. It also holds that $X^n_0 = 1 \to X_0 = 1$. Set $T^n$ as the first entrance time to state 2. Then $X^n_{T^n} = 2 \to 2 \neq 1 = X_0$.

yada
  • 3,535