# Differential of stochastic term [closed]

Question 1:

How does one come up with the equation in the red box below?

It looks like some kind product rule, but I'm not sure how to apply Ito's lemma here.

Bjork doesn't seem to explain it fully, and I can't find the Heath book. My prof gave another proof which I got.

Question 2:

Should the encircled s's be u instead?

## closed as off-topic by Did, Holo, Ethan Bolker, Nicolas FRANCOIS, TyroneAug 8 '18 at 17:11

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• Q2: Yes, $s$ doesn't make any sense. – saz Feb 23 '15 at 14:52

A less heuristic proof is the following. Define the function $Y(t,T,\mathcal{P})$ such that, for each partition $\mathcal{P}$ (of size $n$) of the interval $[t,T]$, we have

$$Y(t,T,\mathcal{P}) := -\sum\limits_{i=1}^{n} f(t,s_{i})(s_{i + 1} - s_{i}) = -\sum\limits_{i=1}^{n} f(t,s_{i})\Delta s_i\,.$$

Observe that, $$\begin{eqnarray*} \sum\limits_{j=1}^{n}\frac{\partial}{\partial f_{t, s_{j}}} Y(t,T,\mathcal{P}) ~\mathrm df(t,s_{j}) = -\sum\limits_{i=1}^{n} 1\cdot\mathrm df(t,s_{i})\Delta s_i\,.\tag{1}\newline \sum\limits_{j=1}^{n}\frac{1}{2}\frac{\partial^2}{\partial f^2_{t, s_{j}}} Y(t,T,\mathcal{P}) ~\mathrm d\langle f\rangle_{t,s_{j}} = -\sum\limits_{i=1}^{n} 0\cdot \mathrm d\langle f\rangle_{t,s_{t_i}}\Delta s_i = 0\,.\tag{2}\newline \sum\limits_{j<r=1}^{n}\frac{\partial^2}{\partial f_{t, s_{j}}\partial f_{t, s_{r}}} Y(t,T,\mathcal{P}) ~\mathrm d\langle f, f\rangle_{t,s_{j},s_{r}} = -\sum\limits_{i<r=1}^{n} 0\cdot ~\mathrm d\langle f, f\rangle_{t,s_{i},s_{r}}\Delta s_i = 0\,.\tag{3} \end{eqnarray*}$$

Therefore, by Ito's Lemma, $(1)$, $(2)$ and $(3)$ imply that $$\mathrm dY(t,T,\mathcal{P}) = \frac{\partial}{\partial t}Y(t,T,\mathcal{P})~\mathrm dt - \sum\limits_{i=1}^{n} \mathrm df(t,s_{t_i})\Delta s_i\,.$$

This means that, for each partition $\mathcal{P}^{'}$ (of size $m$) of the interval $[0,t]$, we have $$\begin{array}{rcl} \displaystyle \sum\limits_{k=0}^{m} \Delta Y_{k}(s_k, T, \mathcal{P}) &=& \displaystyle \sum\limits_{k=0}^{m}\left(\frac{\partial}{\partial t}Y(s_k, T, \mathcal{P})\right)\Delta s_k - \sum\limits_{i=1}^{n} \left(\sum\limits_{k=0}^{m} \Delta f_k(s_k,s_{i})\right)\Delta s_i\,. \\ &&\\ \mbox{So, }\displaystyle \,\,\lim\limits_{\|\mathcal{P}\|\rightarrow 0}\sum\limits_{k=0}^{m} \Delta Y_{k} &=&\displaystyle \lim\limits_{\|\mathcal{P}\|\rightarrow 0} \sum\limits_{k=0}^{m}\left(\frac{\partial}{\partial t} Y(s_k, T, \mathcal{P})\right)\Delta s_k - \lim\limits_{\|\mathcal{P}\|\rightarrow 0}\sum\limits_{i=1}^{n} \left(\sum\limits_{k=0}^{m} \Delta f_k(s_k,s_{i})\right)\Delta s_i \\ &=&\displaystyle \sum\limits_{k=0}^{m}\frac{\partial}{\partial t} \left(\lim\limits_{\|\mathcal{P}\|\rightarrow 0} Y(s_k, T, \mathcal{P})\right)\Delta s_k - \sum\limits_{k=0}^{m} \Big(\int\limits_{t}^{T}\Delta f_k(s_k,s)~\mathrm ds\Big)_k \\ &=&\displaystyle \sum\limits_{k=0}^{m}\left(\frac{\partial}{\partial t} Y(s_k, T)\right)\Delta s_k - \int\limits_{t}^{T}\sum\limits_{k=0}^{m} \Big(\Delta f_k(s_k,s)\Big)_k~\mathrm ds \\ &=&\displaystyle \sum\limits_{k=0}^{m} \left(\frac{\partial}{\partial t} Y(s_k, T)\right)\Delta s_k - \int\limits_{t}^{T}\Big( f(t, s) -f(0, s) \Big)~\mathrm ds\,. \\ \therefore\,\, \sum\limits_{k=0}^{m} \Delta Y_{k}(s_k, T) &=&\displaystyle \sum\limits_{k=0}^{m} \left(\frac{\partial}{\partial t} Y(s_k, T)\right)\Delta s_k - \int\limits_{t}^{T}\Big( f(t, s) -f(0, s) \Big)~\mathrm ds \\ &&\\ \mbox{Consequently, }\quad\quad&& \\ Y(t,T) -Y(0,T)&=&\displaystyle \lim\limits_{\|\mathcal{P^{'}}\|\rightarrow 0}\sum\limits_{k=0}^{m} \Delta Y_{k}(s_k, T) \\ &=&\displaystyle \lim\limits_{\|\mathcal{P^{'}}\|\rightarrow 0}\sum\limits_{k=0}^{m}\left(\frac{\partial}{\partial t} Y(s_k, T)\right)\Delta s_k - \int\limits_{t}^{T}\Big( f(t, s) -f(0, s) \Big)~\mathrm ds \\ &=&\displaystyle \int\limits_{0}^{t} \left(\frac{\partial}{\partial t}Y(s,T)\right)\mathrm ds - \int\limits_{t}^{T}\Big( f(t, s) -f(0, s) \Big)~\mathrm ds\,. \end{array}$$ Or, if you prefer the SDE form,

$$\mathrm dY(t,T) = \Big(\frac{\partial}{\partial t} Y(t,T)\Big)\mathrm dt - \int\limits_{t}^{T} \Big(\mathrm df(t, s)\Big) ~\mathrm ds\,.$$

Admittedly, this is not as "air tight" as it can be. For instance, above, I assumed that as a differentiable function of $t$ and $s$ (not a process), $f(t,s)$ is sufficiently smooth to allow the interchange of the limiting operations "$\frac{\partial}{\partial t}$" and "$\|{\mathcal{P}}\|\rightarrow 0$", and that this is sufficient for the pertinent Ito-integrals to be well-defined and agree.

• No worries about last part. It's part of assumptions that we can switch. Thanks ^_^ – BCLC Feb 23 '15 at 17:09
• @BCLC, you are welcome. Also, just added a missing sum of covariations that necessarily form part of the application of Ito's lemma. – ki3i Feb 23 '15 at 17:21
• Fantastic answer! – AXH Mar 1 '15 at 0:34

Note that Bjork says the fundamental theorem of integral calculus. ki3i proves it rigorously, but we can also guess based on analogy with integral calculus specifically the general form of the Leibniz integral rule which can be derived using the fundamental theorem of calculus:

In product rule, we differentiate each factor holding the others constant and then add them all together. Here, we differentiate each 'part' holding the other 'parts' constant and then add. For example:

$$\frac{d}{dt} \int_0^t 5ts ds$$

$$= [\frac{d}{dt} \int_0^t 5xs ds]|_{x=t} + [\frac{d}{dt} \int_0^y 5ts ds]_{y=t}$$

$$= [5xt]|_{x=t} + [\int_0^y \frac{d}{dt} 5ts ds]_{y=t}$$

$$= 5t^2 + [\int_0^y 5s ds]_{y=t}$$

$$= 5t^2 + [2.5y^2]_{y=t}$$

$$= 5t^2 + 2.5t^2$$

$$= 7.5t^2$$

This is the same as

$$\frac{d}{dt} \int_0^t 5ts ds$$

$$= \frac{d}{dt} t\int_0^t 5s ds$$

$$= \frac{d}{dt} t(2.5t^2)$$

So, it's easy to check Leibniz integral rule if t and s are independent s.t. $f(t,s)=g(t)h(s)$. Luckily, Leibniz integral rule holds even if not independent such as $f(t,s)=\sin(t-s)$ (I suppose you could use Taylor or something, but that's not the point!).