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Let $B$ be a standard Brownian motion and $\{t_i\}_{i=0}^n$ a partition of $[0,t]$.
Define $c_i= (1-c)t_{i+1}+ct_i$, for some $c \in [0,1]$.
Write $B_i$ for $B_{t_i}$ and $$ S_n=\sum_{i=0}^{n-1} B_{c_i} \left( B_{i+1}-B_i \right). $$

Calculate $$ \lim_{n \to \infty} E \left( S_n^2 \right). $$

The only way I know is to use orthogonality of increment of Brownian motion.
That is segmenting each variable in the summand in term of increment, should deliver a sum $S_n^2$ which telescopes to a simple result.
I tried this calculation over and over, but my resulting expression doesn't telescope.
Is there another way to do this?

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$S_n=\sum_{i=0}^{n-1} B_{c_i} \left( B_{i+1}-B_i \right) = \sum_{i=0}^{n-1} ( B_{c_i}- B_i) \left( B_{i+1}-B_i \right) +\sum_{i=0}^{n-1} B_i \left( B_{i+1}-B_i \right) = \sum_{i=0}^{n-1} ( B_{c_i}- B_i) \left( B_{c_i}-B_i \right)+ ( B_{c_i}- B_i) \left( B_{i+1}-B_{c_i} \right) +\sum_{i=0}^{n-1} B_i \left( B_{i+1}-B_i \right).$ last like $B^2 - t$, each of first 2 like quadratic variation terms, should be like $B^2 - t + ct$. –  mike Oct 23 '12 at 16:31
    
@mike After identifying the quadratic variation term, what is the next step? As $c_i$ is not the left-end point of $[t_i,t_{i+1}]$, I think itô isometry doesn't apply. –  Nicolas Essis-Breton Oct 23 '12 at 16:42

1 Answer 1

up vote 3 down vote accepted

It suffices to evaluate

$$ \Bbb{E} \big[ B_{c_i} \big( B_{t_{i+1}} - B_{t_{i}} \big) B_{c_j} \big( B_{t_{j+1}} - B_{t_{j}} \big) \big]. \tag{1} $$

To this end, we make the following simple observation:

Observation. For $w \leq v \leq u$,

$$\begin{align*} \Bbb{E} \big[ B_{v} \big( B_{u} - B_{w} \big) \big] &= \Bbb{E} \big[ B_{v} \Bbb{E} \big[ B_{u} - B_{w} \big\vert \mathcal{F}_{v} \big] \big] = \Bbb{E} \big[ B_{v} \big( B_{v} - B_{w} \big) \big] \\ &= \Bbb{E} \big[ \big( B_{v} - B_{w} \big)^2 + B_{w} \big( B_{v} - B_{w} \big) \big] \\ &= \Bbb{E} \big[ \big( B_{v} - B_{w} \big)^2 \big] + \Bbb{E} \big[ B_{w} \Bbb{E} \big[ B_{v} - B_{w} \big\vert \mathcal{F}_{w} \big] \big] \\ &= \vphantom{\big[} v - w. \end{align*}$$

Now assume first that $i \neq j$. We may assume $i < j$. Then $t_{i+1} \leq t_j$ and

$$ \begin{align*} &\Bbb{E} \big[ B_{c_i} \big( B_{t_{i+1}} - B_{t_{i}} \big) B_{c_j} \big( B_{t_{j+1}} - B_{t_{j}} \big) \big] \\ &= \Bbb{E} \big[ B_{c_i} \big( B_{t_{i+1}} - B_{t_{i}} \big) \Bbb{E} \big[ B_{c_j} \big( B_{t_{j+1}} - B_{t_{j}} \big) \big\vert \mathcal{F}_{t_j} \big] \big] \\ &= \Bbb{E} \big[ B_{c_i} \big( B_{t_{i+1}} - B_{t_{i}} \big) \Bbb{E} \big[ \big(B_{c_j} - B_{t_j} \big) \big( B_{t_{j+1}} - B_{t_{j}} \big) + B_{t_j} \big( B_{t_{j+1}} - B_{t_{j}} \big) \big\vert \mathcal{F}_{t_j} \big] \big] \\ &= \Bbb{E} \big[ B_{c_i} \big( B_{t_{i+1}} - B_{t_{i}} \big) \Bbb{E} \big[ \big(B_{c_j} - B_{t_j} \big) \big( B_{t_{j+1}} - B_{t_{j}} \big) \big\vert \mathcal{F}_{t_j} \big] \big] \\ &= \Bbb{E} \big[ B_{c_i} \big( B_{t_{i+1}} - B_{t_{i}} \big) \big] \Bbb{E} \big[ \big(B_{c_j} - B_{t_j} \big) \big( B_{t_{j+1}} - B_{t_{j}} \big) \big] \\ &= \big(c_i - t_i\big)\big(c_j - t_j\big) \\ &= (1-c)^2 \big(t_{i+1} - t_i\big)\big(t_{j+1} - t_j\big) \end{align*}$$

By interchanging the role of $i$ and $j$, we find that this result holds for all $i \neq j$. Thus it remains to evaluate $(1)$ for the case $i = j$. But

$$\begin{align*} &\Bbb{E} \big[ B_{c_i}^2 \big( B_{t_{i+1}} - B_{t_{i}} \big)^2 \big] \\ &= \Bbb{E} \big[ B_{c_i}^2 \big\{ \big(B_{t_{i+1}} - B_{c_{i}} \big)^2 + 2\big(B_{t_{i+1}} - B_{c_{i}} \big)\big(B_{c_{i}} - B_{t_{i}} \big) + \big(B_{c_{i}} - B_{t_{i}} \big)^2 \big\} \big] \\ &= \Bbb{E} \big[ B_{c_i}^2 \big] \Bbb{E} \big[ \big(B_{t_{i+1}} - B_{c_{i}} \big)^2 \big] + 2 \Bbb{E} \big[ B_{c_i}^2 \big(B_{c_{i}} - B_{t_{i}} \big) \Bbb{E} \big[ \big(B_{t_{i+1}} - B_{c_{i}} \big) \big\vert \mathcal{F}_{c_i} \big] \big] \\ &\qquad + \Bbb{E} \big[ B_{c_i}^2 \big(B_{c_{i}} - B_{t_{i}} \big)^2 \big] \\ &= c_i \big( t_{i+1} - c_i \big) + 0 + \Bbb{E} \big[ \big\{ \big( B_{c_i} - B_{t_i} \big)^2 + 2B_{t_i} \big( B_{c_i} - B_{t_i} \big) + B_{t_i}^2 \big\} \big(B_{c_{i}} - B_{t_{i}} \big)^2 \big] \\ &= c_i \big( t_{i+1} - c_i \big) + \Bbb{E}\big[ \big( B_{c_i} - B_{t_i} \big)^4 \big] + \Bbb{E} \big[ 2B_{t_i} \big] \Bbb{E} \big[ \big(B_{c_{i}} - B_{t_{i}} \big)^3 \big] + \Bbb{E} \big[ B_{t_i}^2 \big] \Bbb{E} \big[ \big(B_{c_{i}} - B_{t_{i}} \big)^2 \big] \\ &= c_i \big( t_{i+1} - c_i \big) + 3\big(c_i - t_i\big)^2 + t_i \big( c_i - t_i \big) \\ &= c_i \big( t_{i+1} - t_i \big) + 2(1 - c)^2\big(t_{i+1} - t_i\big)^2. \end{align*}$$

Combining these results together, we have

$$ \begin{align*} \Bbb{E} \big[ S_n^2 \big] &= \sum_{i,j=0}^{n-1} \Bbb{E} \big[ B_{c_i} \big( B_{t_{i+1}} - B_{t_{i}} \big) B_{c_j} \big( B_{t_{j+1}} - B_{t_{j}} \big) \big] \\ &= \sum_{i,j=0}^{n-1} (1-c)^2 \big(t_{i+1} - t_i\big)\big(t_{j+1} - t_j\big) + \sum_{i=0}^{n-1} \big\{ c_i \big( t_{i+1} - t_i \big) + (1 - c)^2\big(t_{i+1} - t_i\big)^2 \big\} \\ &= (1-c)^2 t^2 + \sum_{i=0}^{n-1} c_i \big( t_{i+1} - t_i \big) + O\Big( t \max_{0 \leq i \leq n-1} \big( t_{i+1} - t_{i} \big) \Big) \\ &\xrightarrow[n\to\infty]{} (1-c)^2 t^2 + \int_{0}^{t} s \, ds = \left( \frac{1}{2} + (1 - c)^2 \right) t^2. \end{align*}$$

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The details consciousness you put into this proof really helps to understand it. Thank you. –  Nicolas Essis-Breton Oct 23 '12 at 19:30

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