# 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

a. is correct, but you should derive it using appropriate logic, not just guessing the answer. Ie the drift of discounted stock should be 0. Define a bond dB = rBdt. d(S/B) should have no drift. This can help you find the correct mu. You can find the sde for S/B using two dimensional ito

b. don't really know about market price of risk.

c. In this case the pde is the same as the black scholes pde using your risk neutral process. Can you think of why this is? Does the type of call option change how the underlying changes? What are the other boundary conditions ie (for S = 0 and S = infinity). Take a look at dirichlet (also known as zero gamma condition) and other types of boundary conditions.

d. That is the right start, but what is the expectation? Lets define C = cash on payout. Then the payout(S) = C*I(S>K). Plug this into your formula. The expectation now looks like C*E(I(S>K)). The problem is that this expectation is in real probability space and you want it in your risk neutral space. You can use girsanov's theorem. Best proof (result to use) I found is (1) in http://math.ucsd.edu/~pfitz/downloads/courses/spring05/math280c/girsanov.pdf

e. In d you will basically find that E(I(S>K)) a function(t)*P(S>K) in your risk neutral space. You need to find P(S>k) this turns out to be N(d2). You can define a new variable (S-E(S))/std(S) = Normal(0,1) to transform P(S>k) into N(d2)

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As of writing this comment it seems as if there is a type in your first paragraph. Isn't it that S/B is a martingale under Q^B (that is, has drift zero). –  Christian Fries Jan 5 '13 at 20:23
Yes you are definitely right sorry about that. S_t/B_t = E_t(S_T/B_T) under risk neutral probability –  phubaba Jan 31 '13 at 0:49

For part (d), instead of using Girsanov's theorem as phubaba suggested, I believe that we can state directly that the price is $$V_t = e^{-r(T-t)} \mathbb{E}^Q \left[ u(S_T-K) \middle \vert \mathcal{F}_t \right],$$ where $u$ is the step function, $Q$ is the risk-neutral probability measure, and $\mathcal{F}_t$ is the filtration at time $t$, since the value of any European-style option with a payoff $f(S_T)$ is given by $V_t = e^{-r(T-t)} \mathbb{E}^Q \left[ f(S_T) \middle \vert \mathcal{F}_t \right]$.

For part (e), note that for a general payoff function $f(S_T)$, we can write $$V_t = e^{-r(T-t)} \int_{-\infty}^\infty f(S_0 e^x) \frac{1}{\sigma \sqrt{2 \pi (T-t)}} \exp \left\{-\frac{\left[x-(r-\sigma^2/2)(T-t)\right]^2}{2 \sigma^2 (T-t)} \right\} dx,$$ where $x \sim \mathcal{N}((r-\sigma^2/2)(T-t), \sigma^2(T-t))$. Plugging in $f(S_0 e^x) = u(S_0 e^x-K)$, I get $V_t = e^{-r(T-t)}N(d_2)$.

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