# Expectation of N(d2)?

I am trying to find out the Pricing Equation for certain type of Options under Risk-Neutral pricing. This is the equation I am getting, but I am not sure if this can be solved or not. Any help is appreciated.

$$V = E[I\{S(T_0) \geq B\}N(d_2)]$$

$t_0 < T_0 < T_1$ This is a time line

$$I\{S(T_0) \geq B\} = \begin{cases} 1, & \text{if } S(T_0) \geq B \\ 0, & \text{otherwise} \end{cases}$$

where $S(t_0), S(T_0)$ is the stock price at different times.

$N(d_2)$ is the Black Scholes $N(d_2)$ but Stock Price used in $N(d_2)$ is $S(T_0)$, and time period is $T_1-T_0$. So $N(d_2)$ in itself is a Random Variable in this case.

I am trying to find the Expectation at time $= t_0$

$d_2 = [ln(S(T_0)/K)+(r-0.5vol^2)(T_1-T_0)] / (vol* sqrt (T_1-T_0))$

$S(T_0) = S(t_0) exp((r-0.5vol^2)(T_0-t_0) + vol * sqrt(T_0-t_0)Z)$

Z~N(0,1)

• You want the expectation of a digittal option?
– will
Apr 17 '17 at 11:09
• Not exactly. I want to compute the above expectation at time=t0. But the N(d2) that I have is based on a future time T0. So that is the Probability of S(T1)>K at time T1 standing at time T0 So my question is there anyway to compute this expectation? Apr 17 '17 at 11:15
• If it's jsut about the expectation, and you don't care about the value (i.e. you can ignore discounting), then why not just price the binary option at $T_0$ with no discounting?
– will
Apr 17 '17 at 11:18
• Aah i understand. You have a binary option, but it's multiplied by $N(d_2)$ on the future date? How are the other parameters of $d_2$ chosen on that date? Are they fixed in the terms?
– will
Apr 17 '17 at 11:20
• I see. So, normally the easiest way to find answers to these is to write it as an integral, and then check this page to see if you can match it to anything! As a side note -how are you choosing the single vol? The strike of the option? or the KI level? You could also take another shortcut - depending on $T0$ and $T1$ - you could price a KI option as a static portfolio at $T1$, with $B^\prime = B \frac{F(T_1)}{F(T_0)}$.
– will
Apr 17 '17 at 11:28

You are essentially interested in pricing a second order bond-binary option. In its most general form, this contract has a time $T_2$ payoff of

$$\mathcal{B}_{\xi_1, \xi_2}^{s_1, s_2} \left( S_{T_1}, S_{T_2}, T_2 \right) = \mathrm{1} \left\{ s_1 S_{T_1} > s_1 \xi_1 \right\} \mathrm{1} \left\{ s_2 S_{T_2} > s_2 \xi_2 \right\}.$$

In your case, we have $\xi_1 = B$, $s_1 = +1$, $\xi_2 = K$ and $s_2 = +1$. The time $T_1$ value of this option is equal to

$$\mathcal{B}_{\xi_1, \xi_2}^{s_1, s_2} \left( S_{T_1}, T_1 \right) = \mathrm{1} \left\{ s_1 S_{T_1} > s_1 \xi_1 \right\} e^{-r \left( T_2 - T_1 \right)} \mathbb{E} \left[ \left. \mathrm{1} \left\{ s_2 S_{T_2} > s_2 \xi_2 \right\} \right| S_{T_1} \right].$$

Apart from the additional discounting, this is the same expression as in your question.

This contract is a special case of the generalize multi-period and multi-asset $\mathbb{M}$-binary options analyzed by Skipper and Buchen (2003). Its time $0 \leq t < T_1$ value is given by

$$\mathcal{B}_{\xi_1, \xi_2}^{s_1, s_2} \left( S_t, t \right) = e^{-r \tau_2} \mathcal{N}_2 \left( \alpha_{0, 1}, \alpha_{0, 2}; \rho \right),$$

where $\tau_i = T_i - t$,

$$\alpha_{0, i} = \frac{s_i}{\sigma \sqrt{\tau_i}} \left( \ln \left( \frac{S}{\xi_i} \right) + \left( r - \frac{1}{2} \sigma^2 \right) \tau_i \right)$$

and

$$\rho = s_1 s_2 \sqrt{\frac{\tau_1}{\tau_2}}.$$

Here, $\mathcal{N}_2$ is the bivariate standard normal distribution function with the given correlation. See the original paper for a derivation of this result. You find a similar result in Chapter 2 of the Ph.D. thesis Veiga (2010). See also this related question and the corresponding answers.

References

Skipper, Max and Peter W. Buchen (2003) "The Quintessiential Option Pricing Formula", Working Paper, School of Mathematics and Statistics, University of Sydney, available online

Veiga, Carlos Manuel "Closed Formulas and Rating Schemes for Derivatives", Ph.D. Thesis, Frankfurt School of Finance & Management, available online