# Why is Ito applied this way?

Given the price of a call option :

$$C = \mathbb{E}\left[ D_{0,T} (s-K)1_{s>K} |\mathcal{F_0}\right]$$

with $$D_{0,T}=e^{-\int_0^Tr(u)du}$$

I read somewhere that applying Itô gives :

$$dC = \mathbb{E} \left[d D_{0,T} (s-K)1_{s>K} |\mathcal{F_0}\right]$$

$$dC = \mathbb{E} \left[ \frac{dD_{0,T}}{dT} (S_T-K) 1_{S_T>K} dT+ D_{0,T} \frac{d}{ds} \left[ (s-K)1_{s>K} \right] \biggr\rvert_{s=S_T} dS_T + D_{0,T}\frac{1}{2}\frac{d^2}{ds^2} \left[ (s-K)1_{s>K} \right] \biggr\rvert_{s=S_T} dS_T dS_T |\mathcal{F_0}\right]$$

My questions are :

1) Why the $$d$$ can be placed inside the $$\mathbb{E}$$? I mean why does this hold ? :

$$d\int D_{0,T}(s-K)1_{s>K}\phi_{S_T}(s)ds=\int d \left[ D_{0,T}(s-K)1_{s>K}\phi_{S_T}(s)\right]ds$$

2) when I look at this term : $$D_{0,T} \frac{d}{ds} \left[ (s-K)1_{s>K} \right] \biggr\rvert_{s=S_T} dS_T$$ I don't know where it comes from, because for me when I do apply Itô I get :

$$\int d \left[ D_{0,T}(s-K)1_{s>K}\phi_{S_T}(s)\right]ds = \int \left[ (...)dT+\frac{d}{ds}\left[ D_{0,T}(s-K)1_{s>K}\phi_{S_T}(s) \right]dS_T+\frac{1}{2}(...)dS_TdS_T \right] ds$$

focusing on the $$\int \left[ \frac{d}{ds}\left[ D_{0,T}(s-K)1_{s>K}\phi_{S_T}(s) \right]dS_T \right] ds$$ part I get :

$$\int \left[ \frac{d}{ds}\left[ D_{0,T}(s-K)1_{s>K}\phi_{S_T}(s) \right]dS_T \right] ds =\int D_{0,T} \left[ \frac{d}{ds}\left[ (s-K)1_{s>K}\phi_{S_T}(s) \right]dS_T \right] ds \mathbf{\mathbin{\color{red}\neq}} \int D_{0,T} \phi_{S_T}(s) \left[ \frac{d}{ds}\left[ (s-K)1_{s>K} \right]dS_T \right] ds = \mathbb{E}\left[ D_{0,T}\frac{d}{ds}\left[ (s-K)1_{s>K} \right]dS_T \right]$$

any help on this ?

3) isn't $$T$$ constant? maturity of the call option. Why do we find $$dT$$ in Itô as if it were the current time $$t$$ ? if we applied Itô to $$C_t$$ or $$C(t,S_t)$$ we would certainly find a $$dt$$ term and not a $$dT$$ one !

• Can you please provide the source with the ito application you mentioned? – Gordon Oct 11 '18 at 17:16

Let me have a go at 1 and 2:

1)To understand this, recall that differentiation is a linear operation and interchanges with sum, and sum and integral are similar things. More broadly, if a function meets some technical conditions such as cont. derivatives and finiteness etc, then one can interchange differential and expectation, you will have to google the conditions. Now call option price is a nice function with continuous derivatives, so should meet those conditions. Dominated convergence theorem can help with the other condition.

2) This seems like just an application of Ito product rule. Letting $$f=(s-k) 1_{s>k}$$ and applying ito product rule $$d (D f)=f \,dD +D \,df+0$$ and substituting Ito Lemma $$df=f_S dS+\frac{1}{2}f_{SS} dS^2$$, gives 2. This is inside expectation so not sure if your point relates to some other steps further down the derivation chain.

3) See below comment from @Quantuple, which includes a reference as well.

Hope this helps.

• Agreed with @Magicisinthechain for 1) and 2). Itô's lemma allows to compute the differential of a time-dependent function of a stochastic process. Here, the time is $T$ and the random variable is $X_T = (S_T-K)1_{S_T>K}$ such that the function to which Itô is applied is $$f : (T,S_T) \to D_{0,T} X_T = D_{0,T} (S_T-K)1_{S_T>K}$$For 3) everything then unravels as if $T$ was the usual time variable $t$ (hence not constant). There is a similar development in Jim Gatheral's "Th volatility surface: A practioner's guide" on page 14. – Quantuple Oct 12 '18 at 7:54
• for 2) can you write the differentiation with the integral instead of the expectation? Cause doing that shows a derivative of the density function : $f=(s-K)1_{s>K}\phi(s)$ and thus $f_s=\phi(s) \frac{d}{ds}(s-K)1_{s>K} + (s-K)1_{s>K} \frac{d}{ds}\phi(s)$ and as you can see the derivative of $\phi(s)$ causes me a problem to recover the expectation operator – user30614 Oct 12 '18 at 8:41
• Is another way of saying this: how do I go from ‘interchange of differential and integral’ to ‘interchange of differential and expectation’? Reason for asking is I thought this was your way of checking 1, but not sure if this is from the source you alluded to. – Magic is in the chain Oct 12 '18 at 9:40
• no I'm just saying how would you workout the calculation using the integral. With integrals we get a derivative of the density function, where does the derivative go? it seems that it doesn't appear anywhere in the final result – user30614 Oct 12 '18 at 12:45
• The density is still implicit in E as the source has not applied the expectation to the components based on what we see here. Think of density as embedded in the dx that appears in ordinanry integrals. – Magic is in the chain Oct 12 '18 at 13:24