Replicating portfolio

I have a doubt about the replicating portfolio methodology.

Example - Consider an European Call with $$K=21$$ and underlying with current price $$S_0=20$$. We assume that, at the maturity, the underlying can be $$22$$ with probability $$p$$ or $$18$$ with probability $$1-p$$. Let $$r=0,12$$ be the risk-free rate.

The option price is $$f=e^{-rT}[pf_u+(1-p)f_d]=0,633$$ for $$p:=\frac{e^{rT}-d}{u-d}$$. So the seller has adopted a long position on $$\Delta=\frac{f_u-f_d}{S_0u-S_0d}=0,25$$ stocks in the face of short position on the option. It follow a risk-free portfolio because we have the same payoff irrespective of the future state of world.

In alternative we can say that, for the absence of arbitrages and the $$\mathbb{Q}$$-martingality guaranteed by the First Theorem of APT, the present value of portfolio is

$$V_0(\Theta)=\mathbb{E}^{Q}[e^{-\int_0^Tr_Sds}V_T(\Theta)|\Im_0]\overset{\operatorname{for}r\operatorname{constant}}{\overbrace{=}}e^{-rT}\mathbb{E}^{Q}[V_T(\Theta)]=4,367$$

which will coincide to the current value of portfolio less than the price of no-arbitrage for the option, that is

$$4,367=20\cdot 0,25-f\Rightarrow f=0,633$$.

Now my doubt.

Why does the Risk-Neutral Evaluation Theorem say that $$V_0(\Theta)=e^{-rT}\mathbb{E}^{\mathbb{Q}}[\varphi(S_T)]$$?

I know that $$\varphi(S_0)=e^{-rT}\mathbb{E}^{\mathbb{Q}}[p\varphi(S_T)^++(1-p)\varphi(S_T)^-]$$ and that $$V_0(\Theta)-\varphi(S_0)=e^{-rT}\mathbb{E}^{\mathbb{Q}}[V_T(\Theta)]$$, but if $$\Theta$$ is not an arbitrage we have $$V_0(\Theta)\neq 0$$. I thought that if the portfolio is replicating we have $$\varphi(S_T)=V_T(\Theta)$$, so maybe I can write that $$V_0(\Theta)=e^{-rT}\mathbb{E}^{\mathbb{Q}}[\overset{=0}{\overbrace{\varphi(S_T)+V_T(\Theta)}}]=0$$ but I'm not sure.

Could you please clarify me these formulas? Thanks you all in advance.

• I think that $V_0(\Theta)-\varphi(S_0)=e^{-rT}\mathbb{E}^{\mathbb{Q}}[V_T(\Theta)]$ is not correct: if $\varphi(S_T)$ is your pay-off function with $S_t$ being the underlying, then $V_0(\Theta)=e^{-rT}\mathbb{E}^{\mathbb{Q}}[V_T(\Theta)]$, whilst $\varphi(S_0)=e^{-rT}\mathbb{E}^{\mathbb{Q}}[\varphi(S_T)]=p\varphi(S_T)^++(1-p)\varphi(S_T)^-$ – Jan Stuller Oct 21 '20 at 10:29
• @JanStuller Right. And from this my doubt. Theorem says: "Let $\tilde{X}$ a replicable derivative in an arbitrage-free market. So for every replicating strategy $\alpha \in \Xi$ and for every martingale measure $\mathbb{Q}$ results that $\mathbb{E}^{\mathbb{Q}}[\tilde{X}|\Im_n]=\tilde{V}_n(\alpha)=:\tilde{H}_n,n=0,...,N$. The process $(\tilde{H}_n)$ is called discounted price of arbitrage of $X$ (or risk-neutral price)". – Francesco Totti Oct 21 '20 at 12:44
• @JanStuller Or from another text... "Let be $h \in \mathbb{L}^2(\Omega, \Im, \mathbb{Q})$. So exists one and only one self-financing portfolio such that $V_T=h$. In addition, for every $t \in [0,T]$ results that $V_t=\mathbb{E}^{\mathbb{Q}}[\frac{h}{e^{r(T-t)}}|\Im_t]$. In particular $V_0=\mathbb{E}^{\mathbb{Q}}[\frac{h}{e^{rT}}]$. – Francesco Totti Oct 21 '20 at 12:51
• @JanStuller This means that I should hade $V_0(\Theta)=\mathbb{E}^{\mathbb{Q}}[e^{-rT}\varphi(S_T)]$, but in the example I correctly obtain $V_0(\Theta)=4,367$ and $\mathbb{E}^{\mathbb{Q}}[e^{-rT}\varphi(S_T)]=0,633$ (for the formula that you also write). – Francesco Totti Oct 21 '20 at 12:55