Evaluating a Truncated Stock Price Expectation in Black–Scholes

# How do I find this Expectation?


# How do I find this Expectation?












I have an expectation given as: $\mathbb{E}\left(S_{T}\mathbb{1}_{S_{T}\geq K} \right)$

where $K$ is just an arbitrary number (i.e. the strike price, but that's unimportant) and $S$ can be modelled by the equation $S_{t} = \exp((r-\frac{1}{2})t + \sigma W_{t})$. Also, the expectation is under the $\mathbb{P}$-measure, not the $\mathbb{Q}$-measure, so effectively this expectation is $\mathbb{E}^{\mathbb{P}}\left(S_{T}\mathbb{1}_{S_{T}\geq K} \right)$

Now, when trying to evaluate $\mathbb{E}\left(\mathbb{1}_{S_{T}\geq K} \right)$ under the $\mathbb{Q}$-measure, then solving this expectation is fairly easy, since you can integrate the SDE and take the $\log$ of $S$ to get $\log(S_{T}) = \log(S_{0}) + (r-\frac{1}{2})T + \sigma\sqrt{T}N(0,1)$ (since $W_{t}\approx \sqrt{T}N(0,1)$) and thus we have for $\mathbb{E}\left(\mathbb{1}_{S_{T}\geq K} \right) = \mathbb{P}({S_{T}\geq K})$:

$\log(S_{T}) = (r-\frac{1}{2})T + \sigma\sqrt{T}N(0,1) > \log(K)$ and thus rearranging this equation gives

$N(0,1) > \frac{\log(K/S_{0}) - (r-\frac{1}{2}\sigma^{2})T}{\sigma\sqrt{T}}$

and through a little bit more rearrangement (i.e. knowing that $N(0,1)<-x = 1 - N(x)$ we get finally $d_{2} = \frac{(r-\frac{1}{2}\sigma^{2})T + \log(S_{0}/K)}{\sigma\sqrt{T}}$ and thus $\mathbb{E}\left(\mathbb{1}_{S_{T}\geq K} \right) = N(d_{2})$.

The problem I have now however is that I'm a bit unsure how to find $\mathbb{E}^{\mathbb{P}}\left(S_{T}\mathbb{1}_{S_{T}\geq K} \right)$ (i.e. under the $\mathbb{P}$-measure) since I've never done any measure theory before, so if someone could help me out I'd really appreciate it. Thanks in advance.

## Answer by LocalVolatility (score 1)

https://quant.stackexchange.com/a/30102

In the context of deriving the European plain vanilla price in the Black and Scholes (1973) model, you also evaluate the expectation

\begin{equation} \mathbb{E}^{\mathbb{Q}} \left[ S_T \mathrm{1} \left\{ S_T > K \right\} \right] \end{equation}

under the $\mathbb{Q}$ (risk-neutral) measure. Remember that the value of a contract is its discounted expected value under $\mathbb{Q}$.

I think what you might confuse here is that the above expectation is often solved by changing to an auxiliary measure under which the stock price is the numeraire. This measure is often denoted $\mathbb{S}$ but is different from the physical measure $\mathbb{P}$.

You do not need to change the measure however to solve the expectation, though it makes things much easier once you feel comfortable with the corresponding machinery. Alternatively you can just solve the corresponding integral. Let $\phi(x)$ be the standard normal density function. Then

\begin{eqnarray} \mathbb{E}^{\mathbb{Q}} \left[ S_T \mathrm{1} \left\{ S_T > K \right\} \right] & = & \int_{-d_-}^\infty S_0 \exp \left\{ \left( r - \frac{1}{2} \sigma^2 \right) T - \sigma \sqrt{T} x \right\} \phi(x) \mathrm{d}x. \end{eqnarray}

Here we defined $d_-$ through

\begin{eqnarray} S_T > K & \qquad \Leftrightarrow \qquad & S_0 \exp \left\{ \left( r - \frac{1}{2} \sigma^2 \right) T - \sigma \sqrt{T} x \right\} > K\\ & & x > \frac{\ln \left( K / S_0 \right) - \left( r - \frac{1}{2} \sigma^2 \right) T}{\sigma \sqrt{T}} =: -d_-. \end{eqnarray}

Substituting for $\phi(x)$ and collecting terms in $x$ in the integral yields

\begin{eqnarray} \ldots & = & \int_{-d_-}^\infty S_0 \exp \left\{ \left( r - \frac{1}{2} \sigma^2 \right) T \right\} \frac{1}{\sqrt{2 \pi}} \exp \left\{ -\frac{x^2}{2} + \sigma \sqrt{T} x \right\} \mathrm{d}x\\ & = & \int_{-d_-}^\infty S_0 \exp \left\{ \left( r - \frac{1}{2} \sigma^2 \right) T \right\} \frac{1}{\sqrt{2 \pi}} \exp \left\{ -\frac{x^2 - 2 \sigma \sqrt{T} x \pm \sigma^2 T}{2} \right\} \mathrm{d}x\\ & = & S_0 e^{r T} \int_{-d_-}^\infty \frac{1}{\sqrt{2 \pi}} \exp \left\{ -\frac{(x - \sigma \sqrt{T})^2}{2} \right\} \mathrm{d}x. \end{eqnarray}

Here we completed the square in the second and third steps. Apply a change of variables by setting $y = x - \sigma \sqrt{T}$ and you get

\begin{eqnarray} \ldots & = & S_0 e^{r T} \int_{- \left( d_- + \sigma \sqrt{T} \right)}^\infty \phi(y) \mathrm{d}y\\ & = & S_0 e^{r T} \mathcal{N} \left( d_+ \right), \end{eqnarray}

where we defined $d_+ = d_- + \sigma \sqrt{T}$.