Evaluate the price of asset in log Black and Scholes model 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?
 A: 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
$$Y_1 = e^{-r(T-1)}d(S_1)\Phi\left(\frac{d(1, S_1)}{\sigma\sqrt{T-1}}\right) + \sigma\sqrt{T-1}g\left(-\frac{d(1,S_1)}{\sigma\sqrt{T-1}}\right)$$
A: Ito's Lemma is the answer here. First note that $S_t$ is an Ito's Process:
$$ S_t:= S_0 + \int_{h=0}^{h=t} \left( \mu S_h \right)dh + \int_{h=0}^{h=t} \left( \sigma S_h \right)dW_h  $$
The solution to the above equation is indeed: $S_t=S_0e^{(\mu-0.5\sigma^2)t+\sigma W_t}$
Since $S_t$ is an Ito Process, Ito's Lemma states that for any ("well-behaved", twice differentiable) function $F()$ of $S_t$ and $t$, the function $F(S_t,t)$ follows the process:
$$F(S_t,t)= F(S_0,t_0) + \int_{h=0}^{h=t}\left(\frac{\partial F}{\partial t}+\frac{\partial F}{\partial S}a(S_h,h)_{=\mu S_h}+\frac{1}{2}\frac{\partial^2 F}{\partial S^2}b(S_h,h)^2_{=\sigma^2 S_h^2}\right)dh + \int_{h=0}^{h=t}\left(\frac{\partial F}{\partial S}b(S_h,h)_{=\sigma S_h}\right)dW_h $$
Taking $F(S_t,t)=ln(S_t)$, we get:
$$F(S_t,t)= ln(S_0) + \int_{h=0}^{h=t}\left(0+\frac{1}{S_h}a(S_h,h)_{=\mu S_h}-\frac{1}{2}\frac{1}{S_h^2}b(S_h,h)^2_{=\sigma^2 S_h^2}\right)dh + \int_{h=0}^{h=t}\left(\frac{1}{S_h}b(S_h,h)_{=\sigma S_h}\right)dW_h=\\=ln(S_0) + \int_{h=0}^{h=t}\left(\mu-\frac{1}{2}\sigma^2\right)dh + \int_{h=0}^{h=t}\left(\sigma \right)dW_h=\\=ln(S_0)+(\mu - 0.5 \sigma^2)t + \sigma W_t$$
(taking exp on both sides of the above will actually give the solution $S_t=S_0e^{(\mu-0.5\sigma^2)t+\sigma W_t}$ to the original GBM SDE for $S_t$.)
The equations above give valid price processes under the real world measure $\mathbb{P}$. If we want to price the claim under the risk-neutral measure $\mathbb{Q}$ (associated with the money market Numeraire), we'll need to perform a change of measure, using the following Radon-Nikodym derivative:
$$\frac{d\mathbb{Q}}{d\mathbb{P}}=exp\left\{0.5\left(\frac{\mu-r}{\sigma}\right)^2 t-W_t\left(\frac{\mu-r}{\sigma}\right)\right\}$$
Applying the above R-N to the stock price process under $\mathbb{P}$ will result in the change of the drift in the stock price process from $\mu$ to $r$ and will give the price process under $\mathbb{Q}$, which is what we need for pricing.
Now the actual problem (everything below is under the measure $\mathbb{Q}$, I assume we had applied the R-N derivative as above to the price process of the Stock):
$$Y_T = \ln(S_T) \mathbb{I}_{S_T>1}$$
Given some initial state $S_0$, the above can be re-written using what we derived above for $ln(S_t)$:
$$\ln(S_T) \mathbb{I}_{S_T>1}=\left(ln(S_0)+(r - 0.5 \sigma^2)T + \sigma W_T\right)\mathbb{I}_{S_T>1}$$
Given some initial state at time $t_1=1$, instead of $t_0 = 0$, we can rewrite the above simply as:
$$\ln(S_T) \mathbb{I}_{S_T>1}=\left(ln(S_1)+(r - 0.5 \sigma^2)(T-1) + \sigma W(T-1)\right)\mathbb{I}_{S_T>1}$$
We are interested in the (risk-neutral) price of this claim (given the initial state at $t=1$); the price at time $t_1$ is (by definition) computed as the discounted, risk-neutral expectation under the measure $\mathbb{Q}$ of the pay-off at maturity, as follows:
$$ Y(t_1):=e^{-r(T-1)}\mathbb{E}^{\mathbb{Q}}\left[\ln(S_T) \mathbb{I}_{S_T>1} | \mathcal{F}_{t_1} \right]$$
Substituting in for $ln(S_T)$, we get:
$$e^{-r(T-1)}\mathbb{E}^{\mathbb{Q}}\left[\left(ln(S_1)+(r - 0.5 \sigma^2)(T-1) + \sigma W(T-1)\right)\mathbb{I}_{S_T>1} | \mathcal{F}_{t_1} \right] =\\= e^{-r(T-1)}\mathbb{E}_1^{\mathbb{Q}}[ln(S_1)\mathbb{I}_{S_T>1}]+\mathbb{E}_1^{\mathbb{Q}}[(r - 0.5 \sigma^2)(T-1)\mathbb{I}_{S_T>1}] + \mathbb{E}_1^{\mathbb{Q}}[\sigma W(T-1)\mathbb{I}_{S_T>1}]$$
Now term by term:
$$ \mathbb{E}_1^{\mathbb{Q}}[ln(S_1)\mathbb{I}_{S_T>1}]=ln(S_1)\mathbb{P}^{\mathbb{Q}}(S_T>1)=ln(S_1)\mathbb{P}^{\mathbb{Q}}(S_1e^{(r-0.5 \sigma^2)(T-1)+\sigma W(T-1)}>1)=\\= ln(S_1)\mathbb{P}^{\mathbb{Q}}\left(Z>\frac{ln\left(\frac{1}{S_1}\right)-r(T-1)+0.5 \sigma^2 (T-1)}{\sigma \sqrt{T-1}}\right)=\\=ln(S_1)\mathbb{P}^{\mathbb{Q}}\left(Z<\frac{ln(S_1)+r(T-1)-0.5 \sigma^2 (T-1)}{\sigma \sqrt{T-1}}\right):=ln(S_1)\Phi(d)$$
Where: $d:=\frac{ln(S_1)+r(T-1)-0.5 \sigma^2 (T-1)}{\sigma \sqrt{T-1}}$ and $\Phi$ is the standard Normal CDF.
The second term then evaluates to:
$$\mathbb{E}_1^{\mathbb{Q}}[(r - 0.5 \sigma^2)(T-1)\mathbb{I}_{S_T>1}]=(r - 0.5 \sigma^2)(T-1)\Phi(d)$$
For the last term, I will use the fact that $$\mathbb{P}^{\mathbb{Q}}(S_T>1)=\Phi(d)$$ and  the fact that:
$$\mathbb{E}_1^{\mathbb{Q}}[\sigma W(T-1)\mathbb{I}_{S_T>1}]=\sigma \left(\sqrt{T-1}\right) \mathbb{E}_1^{\mathbb{Q}}[Z \mathbb{I}_{Z<d}]=\\=\sigma \left(\sqrt{T-1}\right)  \int_{-\infty}^{d}h\frac{1}{\sqrt{2\pi}}exp\left\{\frac{-h^2}{2} \right\}dh:=\sigma \left(\sqrt{T-1}\right) f(d) $$
The final answer should then be:
$$Y(t_1)=e^{-r(T-1)}\left(ln(S_1)\Phi(d)+(r - 0.5 \sigma^2)(T-1)\Phi(d)+\sigma \left(\sqrt{T-1}\right) f(d)\right)$$
