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Let $S_t$ be the Geometric Brownian Motion, we know that $$dS_t=rS_tdt+\sigma S_tdW_t, t\in [0,T], S_0>0, r>0,\sigma>0$$ and the distribution of $S_t$ is known explicitly. Please see the link above for the density function of $S_t$. Let $h(.)$ be a bounded continuous function (as nice as you want), and consider the discretely monitored $S_{t_1},S_{t_2},\ldots, S_{t_n}$ with log-normal distribution of $S_t$. Assume that we have $$S^{k}_{t_i}\Longrightarrow S_{t_i},\quad\text{weakly as}\quad k\to\infty,$$ where $S_{t_i}$ is log-normally distributed.

My question here : Do we have the following weak convergence $$ \displaystyle\sum_{i=1}^{n}h(S^{k}_{t_i}) \Longrightarrow \displaystyle\sum_{i=1}^{n}h(S_{t_i}),\quad\text{as}\quad k\to\infty? $$ In other words, do we have $$(S_{t_1}^k,\ldots,S_{t_n}^k)\Longrightarrow (S_{t_1},\ldots,S_{t_n}) \quad\text{as}\quad k\to\infty?$$ We note the above is not true in general but I am not sure for the log-normal random variable case. It would be nice if some one could give me some ideas or hints. Thank you so much for your time. I truly appreciate it.

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    $\begingroup$ If the $S_t^k$ are all generated like $S_t$ as the solutions to a stochastic SDE, you will have the joint distributions as well, but if they are arbitrary subject only to each converging weakly to a log normal, no. $\endgroup$
    – user83457
    Commented Apr 3, 2017 at 7:37
  • $\begingroup$ Thanks mike, could you point me to a reference ? I could not find any $\endgroup$
    – KNN
    Commented Apr 3, 2017 at 15:45
  • $\begingroup$ Any book on browinan motion will have the joint distribution of the $log(S_{t_i})$. They are joint normal, and covariances are determined by the condition that the increments be independent. It is easy and standard to show that joint normals converge in distribution iff their means and covariances converge. I am not sure where to send you for this result, although I recall that the one dimensional version is in Chung's book. It is proven just by inspection of the characteristic function. $\endgroup$
    – user83457
    Commented Apr 3, 2017 at 15:52
  • $\begingroup$ And maybe I should add that once n-tuples $(S_{t_1}^k,...,S_{t_1}^k)$ are converging in distribution, you the expectation of a continuous function of r.v.'s which are converging in distribution. $\endgroup$
    – user83457
    Commented Apr 3, 2017 at 15:57
  • $\begingroup$ As stated, the $S^k_{t_1},…,S^k_{t_n}$ need not be defined in the same probability space, so the sum $∑_{i=1}{n}h(S^k_{t_i})$ is not well defined. Under the assumption that all variables are defined on the same space, still, the hypothesis $S^k_{t_i}⟹S_{t_i}$ weakly as $k→∞$ does not say anything about the joint distribution of $(S^k_{t_1},…,S^k_{t_n})$ $\endgroup$
    – VictorZurkowski
    Commented Apr 4, 2017 at 6:14

1 Answer 1

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For each i, $S_{t_i}$ is distributed as $exp( (r - \sigma^2/2)t_i + Z \sqrt{t_i})$, where $Z$ is a standard normal. If we take $S^k_{t_i} = exp( (r - \sigma^2/2)t_i + Z \sqrt{t_i})$ for all k and i, then, trivially, $S^{k}_{t_i}\Longrightarrow S_{t_i},\quad\text{weakly as}\quad k\to\infty$. Take $h = ln$, then $\sum_{i=1}^{n}h(S^{k}_{t_i}) = (\sum_{i=1}^{n}(r - \sigma^2/2)t_i) + \sigma (\sum_{i=1}^{n} \sqrt{t_i}) Z$, but $\sum_{i=1}^{n}h(S_{t_i}) = (\sum_{i=1}^{n}(r - \sigma^2/2)t_i) + \sigma (\sum_{i=1}^{n} W_{t_i}) $, both normal, with same mean, but different variances, so, in general, $\sum_{i=1}^{n}h(S^{k}_{t_i})$ does not converge weakly to $\sum_{i=1}^{n}h(S_{t_i})$.

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  • $\begingroup$ Dear Victor: thank you ! Can we assume that $S_{t_i}^k=\exp\left((r-\sigma^2/2)t_i+\sigma Z\sqrt{t_i} \right) $? I am not sure if we can assume this. $\endgroup$
    – KNN
    Commented Apr 4, 2017 at 13:44
  • $\begingroup$ This is a counterexample to show that the assumtion of weak convergence for each $t_i$ is not enough. One needs some assumption on the join weak-convergence of the n-tuple. I am not assuming the form of $S^k$, I am saying: look at these random variables. They satisfy the weak convergence assumption for each $t_i$ but tghe n-tuple does not converge to the corresponding n-tuple of the sample of $S_t$. $\endgroup$
    – VictorZurkowski
    Commented Apr 4, 2017 at 20:32

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