European Call Option - Expectation of Normal CDF

I'm trying to understand a paper of Sergii Kuchuk-Iatsenko and Yuliya Mishura, Pricing the European Call Option in the Model with Stochastic Volatility Driven by Ornstein-Uhlenbeck Process, Exact Formula, as its has similarity of what I am researching right now. So we know that European call option at $t=0$ is $$V_0=SN(d_1)+Ke^{-r(T-t)}N(d_2)$$ where $$d_1=\frac{ln(S/K)+(r+\frac{1}{2}\sigma^2)T}{\sigma\sqrt{T}}$$ $$d_2=d_1-\sigma\sqrt{T}$$ $S$ is price of underlying asset, $N(.)$ is standard normal CDF, $K$ is strike price, $r$ is risk-free rate, and $T$ is time to expiration. In Sergii Kuchuk paper, $N(d_1)$ can be stated into below equation $$N(d_1)=\frac{1}{2}+\frac{1}{\sqrt{2\pi}}I_{d_1\gt0}\int_{0}^{d_1}e^{-s^2/2}ds-\frac{1}{\sqrt{2\pi}}I_{d_1\lt0}\int_{d_1}^{0}e^{-s^2/2}ds$$ I already understand until this part. Then the paper stated that $$\mathbf{E}(N(d_1))=\frac{1}{2}+\frac{1}{\sqrt{2\pi}}\int_{0}^{\infty}\mathbf{Q}(S\lt{d_1})e^{-s^2/2}ds-\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{0}\mathbf{Q}(S\gt{d_1})e^{-s^2/2}ds$$ where $\mathbf{Q}$ is probability measure which is equivalent to objective measure $\mathbf{P}$. Which means $$\mathbf{E}(\frac{1}{\sqrt{2\pi}}I_{d_1\gt0}\int_{0}^{d_1}e^{-s^2/2}ds)=\frac{1}{\sqrt{2\pi}}\int_{0}^{\infty}\mathbf{Q}(S\lt{d_1})e^{-s^2/2}ds$$ I don't have a clue on this part.

• I just glanced through the paper: in the usual black-scholes case, $d_1$ for $t=0$ is not a random variable but you have a different setting, the authors introduce a $\bar{\sigma}_t$ which actually is a random variable (also for the $t=0$ case), which makes $d_1$ itself a random variable – user190080 Jul 4 '18 at 11:17
• Ah I see, I missed that point, thank you very much – Ben Jul 4 '18 at 11:46

The only way I can make any sense of this is $d_1$ is a random variable taking values in $\mathbb{R}$ with PDF $f$ and CDF

$$Q(d_1 \leqslant x) = \int_{-\infty}^x f(t) \, dt,$$

so

$$E\left(I_{d_1 > 0}\int_0^{d_1} e^{-s^2/2} \, ds \right) = \int_0^\infty f(x)\left(\int_0^xe^{-s^2/2} \, ds\right) \, dx$$

Integrating by parts, we get

$$E\left(I_{d_1 > 0}\int_0^{d_1} e^{-s^2/2} \, ds \right) = \left.\int_0^xe^{-s^2/2} \, ds \int_{-\infty}^xf(t) \, dt\right|_0^\infty - \int_0^\infty \left(\int_{-\infty}^xf(t) \, dt\right) e^{-x^2/2}\, dx$$

Since $\int_{-\infty}^\infty f(t) \, dt = 1$ we have

$$E\left(I_{d_1 > 0}\int_0^{d_1} e^{-s^2/2} \, ds \right) = \int_0^\infty e^{-s^2/2} \, ds - \int_0^\infty \left(\int_{-\infty}^xf(t) \, dt\right) e^{-x^2/2}\, dx \\ = \int_0^\infty \underbrace{\left(\int_x^\infty f(t) \, dt\right)}_{Q(x < d_1)}e^{-x^2/2} \, dx$$

The second term can be handled in a similar way.

• But if we see the formula of $d_1$, is it possible that $d_1$ to be a variable random? – Ben Jul 4 '18 at 10:36
• For the integration by part, is it evaluated from 0 to $\infty$ because of the indicator variable? And if we suppose $u=\int_{0}^{x}e^{s^2/2}ds$ and $dv=f(t)dt$, then we get $du=(e^{x^2/2}-1)ds$ as we use fundamental theorem of calculus, am I missing something here? – Ben Jul 4 '18 at 11:08
• @Ben: The paper is about stochastic volatility. Since $d_1$ depends on $\sigma$ it is random. The expected value is an integral over $(-\infty,\infty)$ so the indicator reduces the integral to one over $[0,\infty)$ and the answer to the first question in the second comment is yes. For the last question $du = e^{-x^2/2} \, dx$. – RRL Jul 4 '18 at 14:31