5
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Computing the integral $\int \Phi(x_t,t)dx_t$ writing the equation in the form we can write the integral as the mean square limit

$$\int \Phi(x_t,t)dx_t=\lim_{\Delta \to 0} \sum^{j=1}_{N-1} [\Phi((\frac {x(t_j)+x(t_{j+1})}{2},t_j)][x(t_j+1)-x(t_j)]\ \ \ \ \ \ .\ (3)$$

and in Ito's form as $$\int \Phi(x_t,t)dx_t=\lim_{\Delta \to 0} \sum^{j=1}_{N-1} [\Phi x(t_j),t_j][x(t_j+1)-x(t_j)] \ \ \ \ \ \ \ \ \ \ .\ (4)$$ Let us prove the existence of the limit in $(3)$ and find the formula relating the two indicated integrals. To do this we select the $\Delta$ partitioning and consider the difference between the limit expressions on the right-hand sides of $(3)$ and $(4)$. Making use of the differentiability with respect to $x$ of the function $\Phi(x_t, t)$ we get

$$D_{\Delta}=\sum^{j=1}_{N-1} [\Phi((\frac {x(t_j)+x(t_{j+1})}{2}),t_j-\Phi(x(t_j),t_j][x(t_{j+1})-x(t_j)] \ \ \ \ \ \ .(5)$$ $$= \frac{1}{2} \sum ^{N-1}_{j=1} \frac {\partial \Phi}{\partial x}[(1-\theta)x(t_j)+\theta x(t_{j+1}),t_j)][x(t_{j+1})-x(t_j)]^2 ,0\le \theta \le 1/2,t_j=t_j^{\Delta}. $$ my question is how to proceed after $(5)$, how this final equation comes .

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  • $\begingroup$ So your question is how $(5)$ implies the identity in the next line, right? Or do you want to know how to proceed using the identity in the last line? $\endgroup$
    – saz
    Feb 19 '16 at 17:47
  • $\begingroup$ yes how 5 implies the identity in next ? @saz $\endgroup$
    – Bogorovich
    Feb 19 '16 at 18:41
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+50
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If $x \mapsto \Phi(x,t)$ is differentiable, it follows from Taylor's formula that

$$\Phi(x,t) = \Phi(y,t) + (x-y) \frac{\partial}{\partial x} \Phi(\zeta,t)$$

for some intermediate point $\zeta$ between $x$ and $y$ (i.e we can find $\lambda \in (0,1)$ such that $\zeta = \lambda x+ (1-\lambda) y$). Using this identity for

$$x :=\frac{x(t_j)+x(t_{j+1})}{2} \qquad y := x(t_j) \qquad t = t_j$$

we find

$$\Phi \left( \frac{x(t_j)+x(t_{j+1})}{2}, t_j \right) = \Phi(x(t_j),t_j)+ \frac{x(t_{j+1})-x(t_j)}{2} \frac{\partial}{\partial x} \Phi(\zeta,t_j) \tag{1}$$

with

$$\zeta = \lambda x(t_j)+ (1-\lambda) \frac{x(t_j)+x(t_{j-1})}{2} \tag{2} $$

for some $\lambda \in (0,1)$. Note that $(2)$ is equivalent to

$$\begin{align*} \zeta &= x(t_j) \left[ \frac{2\lambda}{2} + \frac{(1-\lambda)}{2} \right] + \underbrace{\frac{1-\lambda}{2}}_{=:\theta} x(t_{j+1}) \\ &= x(t_j)(1-\theta) + \theta x(t_{j+1}) \end{align*}$$

for some $\theta \in (0,1/2)$. Hence, by $(1)$,

$$\Phi \left( \frac{x(t_j)+x(t_{j+1})}{2}, t_j \right) -\Phi(x(t_j),t_j)= \frac{x(t_{j+1})-x(t_j)}{2} \frac{\partial}{\partial x} \Phi ( x(t_j)(1-\theta) + \theta x(t_{j+1}), t_j).$$

Multiplying this expression with $x(t_{j+1})-x(t_j)$ and summing over $j=1,\ldots,N-1$ yields the identity you are looking for. (Mind that $\theta = \theta(j)$; we cannot expect to find one $\theta$ which works for all $j=1,\ldots,N-1$).

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  • $\begingroup$ isn't the taylor formula $f(x)=\sum_{0}^{\infty} (x-a)^n f^n(a)/n!$ @saz .you wrote $\Phi(x,t) = \Phi(y,t) + (y-x) \frac{\partial}{\partial x} \Phi(\zeta,t)$,why it is not $(x-y)$ instead of $(y-x)$ $\endgroup$
    – Bogorovich
    Feb 20 '16 at 7:45
  • $\begingroup$ @Bogorovich Yeah, you are right; it should read $x-y$. $\endgroup$
    – saz
    Feb 20 '16 at 8:26

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