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 .

  • $\begingroup$ 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

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

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


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

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 $j=1,\ldots,N-1$).

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.