Context:
I'm given Black and Scholes model.
Problem:
For $T > 1$ the asset returns $$Y_T = \ln(S_T)\cdot 1_{S_T>1} $$
where $S_t$ is solution for Black and Scholes model:
$$S_t = S_0 \cdot exp[(\mu – \frac{\sigma^2}{2})t +\sigma W_t]$$
$W_t$ is Wiener process and $1_{S_T>1} = 1$ if $S_T>1$ and $1_{S_T>1} = 0$ otherwise.
It is given that today $t$ is equal to $1$.
Evaluate the current price $Y_1$.
Can you give some ideas?
Best Answer
If you choose the money-market instrument as the numeraire, you get
$$Y_t = e^{-r(T-t)}E\left[Y_T \vert \mathcal{F}_t\right]$$ where the expectation is taken under the risk-neutral measure. Under the risk neutral measure $S_t$ follows the SDE $$dS_t = rS_tdt + \sigma S_t dW_t$$ where $W$ is Brownian Motion. This is predicated of course on the Black-Scholes model.
Note that you can write $S_T$ as follows. $$S_T = S_t e^{\left(r-\frac{1}{2}\sigma^2\right)(T-t) + \sigma(W_T - W_t)}$$
$$E\left[Y_T \vert \mathcal{F}_t\right] = \left(\log{S_t} + \left(r-\frac{1}{2}\sigma^2\right)(T-t)\right)E[\mathbb{1}_{\log{S_T} > 0} \vert \mathcal{F}_t] + E[\sigma(W_T - W_t)\mathbb{1}_{\log{S_T} > 0} \vert \mathcal{F}_t]$$
Let's calculate the first conditional expectation. Define $d(t, S_t) := \log{S_t} + \left(r-\frac{1}{2}\sigma^2\right)(T-t)$. Since $W_T - W_t$ is independent of $\mathcal{F}_t$,
$$E[\mathbb{1}_{\log{S_T} > 0} \vert \mathcal{F}_t] = P\{d(t, S_t) + \sigma(W_T-W_t)\vert\mathcal{F}_t\} = \Phi\left(\frac{d(t, S_t)}{\sigma\sqrt{T-t}}\right)$$ where $\Phi$ is the distribution function of standard normal r.v.
The second conditional expectation is calculated similarly.Replace $W_T - W_t$ by $\sqrt{T-t}X$ where X is standard normal independent of $\mathcal{F_t}$. This is purely for notational convenience. You can then write $$E[\sigma(W_T - W_t)\mathbb{1}_{\log{S_T} > 0} \vert \mathcal{F}_t] = E[\sigma\sqrt{T-t}X\mathbb{1}_{d(t, S_t) + \sigma\sqrt{T-t}X > 0} \vert \mathcal{F}_t] = \sigma\sqrt{T-t}E[X\mathbb{1}_{X > -\frac{d(t, S_t)}{\sigma\sqrt{T-t}}} \vert \mathcal{F}_t]$$
We will now focus on $$E[X\mathbb{1}_{X > -\frac{d(t, S_t)}{\sigma\sqrt{T-t}}} \vert \mathcal{F}_t]$$
We are taking the conditional expectation of something that is a mixture of r.v.s that are either measurable wrt to the conditioning sigma algebra or independent of it. We can then treat the measurable ones as if they were constants and compute the conditional expectation of the independent r.v.s as if it is an unconditional expectation.
For a standard normal r.v. $X$ define the function $g$ as $$g(a) := E[X\mathbb{1}_{X>a}] = \frac{1}{\sqrt{2\pi}}e^{-\frac{a^2}{2}}$$
We can then write $$E[X\mathbb{1}_{X > -\frac{d(t, S_t)}{\sigma\sqrt{T-t}}} \vert \mathcal{F}_t] = g\left(-\frac{d(t,S_t)}{\sigma\sqrt{T-t}}\right)$$
Putting everything together
$$Y_1 = e^{-r(T-1)}d(S_1)\Phi\left(\frac{d(1, S_1)}{\sigma\sqrt{T-1}}\right) + \sigma\sqrt{T-1}g\left(-\frac{d(1,S_1)}{\sigma\sqrt{T-1}}\right)$$