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The Skorokhod Embedding Problem is well known and has many solutions. Now let $B=(B_t)_{t\ge 0}$ be a standard Brownian motion and $\tau$ be an embedding to the centered distribution $\mu$, i.e. the stopped process $B=(B_{\tau\wedge t})_{t\ge 0}$ is uniformly integrable and $B_{\tau}$ has the law $\mu$. My question is the following: Given a set of real numbers $(K_i)_{1\le i\le m}$, for any $\varepsilon, \delta>0$, could we find some bounded stopping time $\sigma$ such that

$$P\left[|\tau-\sigma|~\ge~ \delta\right]~~\le~~ \varepsilon$$

and

$$E[(B_{\sigma}-K_i)^+]~~=~~E[(B_{\tau}-K_i)^+] \mbox{ for all } i=1,\ldots, m.$$

Here we assume that the filtration for the Brownian motion $B$ is sufficiently rich. A natural idea for me is to take first $\tau_n=\tau\wedge n := \min(\tau,n)$, then clearly one has

$$\lim_{n\to\infty}\tau_n~~=~~\tau~~\mbox{ and }~~\lim_{n\to\infty}E[(B_{\tau_n}-K_i)^+]~~=~~E[(B_{\tau}-K_i)^+]$$

However, I don't find a good way to modify properly $\tau_n$ to $\sigma$, such that introduce some random variable $G_1, G_2, \ldots$ that are independent of $B$ and $\tau$ and construct a functional $\sigma=\sigma(\tau_n, G_1, G_2, \ldots)$. If someone has a idea, I look forward to that! Many thanks!

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No, 4 values of $k_i$ will determine that the distribution is tightly concentrated around $\pm 1$. Set $x_a = E[(B_{\sigma}- a)^+] $. Then $x_{-1 - \epsilon} = 1 + \epsilon $ implies $B_{\sigma} > -1 - \epsilon $, $ x_{1-\epsilon} - x_{-1+\epsilon} = 2 - 2 \epsilon $ implies no mass between $-1 + \epsilon $ and $ 1 - \epsilon $, etc. These values are fine for $\tau = $ first hitting time of $\pm 1$ but together they imply $B_{\sigma}$ is similar, in particular, $\sigma > $ exit time from the interval $(-1 + \epsilon, 1 - \epsilon)$ and can't be bounded.

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  • $\begingroup$ Once we take $\tau$ as the first hitting time of $\pm 1$, then it follows that the $x_{-1-\varepsilon}=1+\varepsilon$, $x_{-1+\varepsilon}=1-\varepsilon/2$, $x_{1-\varepsilon}=\varepsilon/2$ and $x_{1+\varepsilon}=0$. How could we deduce that the support $B_{\sigma}$ does not have any intersection with $[-1+\varepsilon, 1-\varepsilon]$? $\endgroup$ – CodeGolf Nov 10 '15 at 15:43
  • $\begingroup$ look at the graph of $ f(x) = (x-(b-a))^+ - 2* (x-b)^+ + (x-(b_+ a))^+, a > 0$. It is positive on $( b-a, b+a) $ 0 elsewhere, and $Ef(X)$ depends only on the numbers you want to specify. If it is 0, then $X$ take on values in $( b-a, b+a) $ with probability 0. I may have messed up the condition in the post, but I think the idea is right. $\endgroup$ – Michael Nov 10 '15 at 17:00
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I claim that it is just an idea instead of a complete answer, as it is more convenient to put it here. First, take a positive random variable $G$ that is independent of $B$ and $\tau$ and takes values in a finite set $\{g_1, \ldots, g_M\}$. Denote $p_i=P[G=g_i]$. Nest, we shall choose suitable $M$, $(g_i)_{1\le i\le M}$ and $(p_i)_{1\le i\le M}$ to construct the required $\sigma$. Put

$$\sigma~~\equiv~~\sigma_n~~:=~~\tau_n+G.$$

Clearly, $\sigma$ is a bounded stopping time (for a filtration rich enough). Then let us compute

$$E\left[(B_{\sigma}-K_i)^+\right]~~=~~E\left[E\left[(B_{\tau_n+G}-K_i)^+\right]\big| \mathcal{F}_G\right]~~=~~\sum_{j=1}^ME\left[(B_{\tau_n+g_j}-K_i)^+\right]p_j~~=~~C_i,$$

where $C_j=E\left[(B_{\tau}-K_i)^+\right]$ for all $i=1,\ldots, m$. Thus, the construction consists in finding a solution $(g_j, p_j)_{1\le i\le m}$ of the following system:

$$\sum_{j=1}^Mc_i(g_j)p_j = C_j \mbox{ for all } i=1,\ldots, m,~~ \sum_{j=1}^Mp_j=1,~~ p_j\ge 0 \mbox{ for all } j=1,\ldots, M,$$

I'm not quite familiar about the related result in linear algebra, but I believe that there exist some results about the existence. Could someone completes this idea by choosing properly $n$, $M$, $(g_j)_{1\le j\le M}$ and $(p_j)_{1\le j\le M}$ ? Many thanks for the reply!

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