Applying Itô’s Formula to a Discounted Call Option Expectation

# Why is Ito applied this way?


# 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 !

## Answer by Magic is in the chain (score 1)

https://quant.stackexchange.com/a/42161

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.