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The asset-or-nothing European option pays at t = T the value of the stock when at time T that value exceeds or is equal to the exercise price E, and nothing if the value of the stock is below E. So, in mathematical terms:

$$V(S,T) = \left\{ \begin{array}{lr} S & \text{if}\quad S \ge E,\\ 0 & \text{if}\quad S < E. \end{array} \right. $$

The cash-or-nothing European option pays at t = T a fixed value B when at time T that value exceeds or is equal to the exercise price E, and nothing if the value of the stock is below E. So, in mathematical terms:

$$V(S,T) = \left\{ \begin{array}{lr} B & \text{if}\quad S \ge E,\\ 0 & \text{if}\quad S < E. \end{array} \right. $$

We know that the formulas for these options are the following: \begin{align} &\text{Cash-or-nothing call:}\quad c_{cn}=Be^{-rT}N(d_2),\\ &\text{Cash-or-nothing put:}\quad p_{cn}=Be^{-rT}N(-d_2),\\ &\text{Asset-or-nothing call:}\quad c_{an}=Se^{-qT}N(d_1),\\ &\text{Asset-or-nothing put:}\quad p_{an}=Se^{-qT}N(-d_1).\\ \end{align}

where $$ d_1=\dfrac{\ln(S/E)+(r-q+\sigma^2/2)(T-t)}{\sigma\sqrt{T-t}} $$ and $$ d_2=d_1-\sigma\sqrt{T-t}. $$

We also know that we are supposed to follow the derivation of Black-Scholes in order to derive these formulas but we are having trouble understanding how it differs from the derivation of Black-Scholes itself.

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You can derive these formulae by tweaking the black scholes derivation. If you are using PDE method, you will use different boundary conditions. If you are using integration over the risk neutral probability , you will use a different payoff function but the same risk neutral density.

Alternatively , you can observe that these payoffs are combinations of regular puts and calls. For example , the cash or nothing call is the limit of a [E, E+dE] call spread as dE tends to zero, so you can obtain it by differentiating the regular black scholes call price by E. Then, the asset or nothing call = the regular call option + the cash or nothing call, so you can derive that one as well.

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The value of an cash-or-nothing option is just the discounted expected payoff of the option. So the value of such a call should be $e^{-r (T - t)} N \mathbb{P} \left\{ S_T > K \right\}$, where $\mathbb{P} \left\{ S_T > K \right\} = \mathcal{N} \left( d_2 \right)$, and $N$ is the cash agreed to be paid.

The asset-or-nothing is a bit more complicated since it is $e^{-r (T - t)} \mathbb{E} \left[ \left. S_T \right| S_T > K \right]$. The last term is the expected value of stock price given that $S_T > K$. So you would need to use the lognormal stock price and integrate it with pdf of standard normal and "complete the square". You would end up with $e^{(r - q) (T - t)} S_0 \mathcal{N} \left( d_1 \right)$ for that, and the final formula would be $e^{-q (T - t)} S_0 \mathcal{N} \left( d_1 \right)$.

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    $\begingroup$ We already know the formulas (stated in the question), the OP was interested in how they were derived. In fact, it's the same idea as Black-Scholes. $\endgroup$ – SmallChess Nov 11 '15 at 7:07
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    $\begingroup$ About notations: asset-or-nothing call should be $$e^{-r(T-t)}\Bbb E[S_TI(S_T>K)\mid S_t]$$ instead. $\endgroup$ – Vim Jun 9 '17 at 4:47
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Write black and scholes equation, for asset or nothing put K = 0 And for cash or nothing put S = 0 And K = B and discount with time as e^-rT

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    $\begingroup$ Substituting $S_0 = 0$ or $K = 0$ is not correct since also the two terms $d_1$ and $d_2$ are functions of the strike and spot. I get what you intend to say. However, even this is in my opinion not a satisfactory answer to the question. Keep in mind that @Sertii knows the solution already and wants to know how to derive it. $\endgroup$ – LocalVolatility Dec 5 '16 at 15:17
  • $\begingroup$ For both the cases the d1 and d2 will be function of E not the strike price. Which can be shown during the derivation of black and scholes where we place limit as E and infinity not the strike price $\endgroup$ – user3692159 Dec 5 '16 at 15:45

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