# price of a “Cash-or-nothing binary call option”

I'm stuck with one homework problem here:

Assume there is a geometric Brownian motion $$dS_t=\mu S_t dt + \sigma S_t dW_t$$ Assume the stock pays dividend, with the cont. compounded yield $q$.

a) Find the risk-neutral version of the process for $S_t$.

b) What is the market price of risk in this case?

c) Assume no yield anymore. Now, there is a derivative written on this stock paying one unit of cash if the stock price is above the strike price $K$ at maturity time $T$, and 0 else (cash-or-nothing binary call option). Find the PDE followed by the price of this derivative. Write the appropriate boundary conditions.

d) Write the expression for the price of this derivative at time $t<T$ as a risk-neutral expectation of the terminal payoff.

e) Writte the price of this option in terms on $N(d_2)$, where $d_2$ has the usual Black-Scholes value.

Here is what I came up with by now:

for a): This should become $dS_t'=(r-q)S_t'dt + \sigma S_t'dW_t^\mathbb{Q}$ (is this correct?)

for b): This would be $\zeta=\frac{\mu-(r-q)}{\sigma}$ (?)

for c): The boundary conditions should be: Price at $t=T$ is $0$ if $S<K, 1$ else; I have no idea what to write for the PDE.

for d): I can only think of $C(S_t,t)=e^{-r(T-t)}\mathbb{E}[C(S_t),T]$, where $C(S_t,T)$ is the value at time $T$, i.e. the payoff.

for e): I don't know how to start here.

Can anybody help me and solve this with me?

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I found that $\mathbb{Q}_t(S_T\geq K)=N(d_2)$, where $\mathbb{Q}$ denotes risk-neutral probability, which should solve part e): The present value is the discounted future payoff, which is just $p$ if $p$ is the probability that $S_T\geq K$. Hence, the current value is $e^{-r(T-t)}\mathbb{Q}_t(S_T\geq K)=e^{-r(T-t)} N(d_2)$ – Marie. P. Dec 20 '12 at 16:50
Not entirely correct. You did not convert correctly from real probability measure to risk neutral probability measure. See my answer. – phubaba Dec 26 '12 at 17:52