# Compute the price of a derivative which pays $\log(S_T)S_T$ in the Black Scholes world

Compute the price of a derivative which has pays $$\log(S_T)S_T$$, you can assume that the Black Scholes model is valid.

Using the stock measure we can write the expectation as

$$D(0) = S_0 \mathbb{E}_S(\log S_T)$$

with the expectation in the stock measure. In this measure,

$$dS_t = (r + \sigma^2)S_t dt + \sigma S_t dW_t$$

How has this been derived?

and it follows from Ito's lemma that

$$d \log S_t = (r+0.5\sigma^2)dt + \sigma dW_t$$

Why are we using Ito's lemma here?

Following this answer, let $$\mathbb Q$$ be the probability measure associated to the risk-free bank account as numeraire and $$\mathbb Q^1$$ the probability measure associated to the stock as numeraire.

You know that the standard equation $$\mathrm{d}S_t=rS_t\mathrm{d}t+\sigma S_t\mathrm{d}W_t^\mathbb{Q}$$ can be written as $$\mathrm{d}S_t=(r+\sigma^2)S_t\mathrm{d}t+\sigma S_t\mathrm{d}W_t^{\mathbb{Q}^1}$$ under the stock measure by applying Girsanov's theorem (this is example 1 of section 3 of this answer). We simply use $$\mathrm{d}W_t^\mathbb{Q}=(\sigma\mathrm{d}t+\mathrm{d}W_t^{\mathbb{Q}^1})$$.

Similarly, applying Ito's Lemma to $$f(t,x)=\ln(x)$$, we have $$\mathrm{d}\ln(S_t)=\left(r-\frac{1}{2}\sigma^2\right)\mathrm{d}t+\sigma \mathrm{d}W_t^{\mathbb{Q}}$$ which translates to $$\mathrm{d}\ln(S_t)=\left(r+\frac{1}{2}\sigma^2\right)\mathrm{d}t+\sigma \mathrm{d}W_t^{\mathbb{Q}^1}$$ under the new measure. The latter equation is equivalent to $$\ln(S_t)= \ln(S_0)+\left(r+\frac{1}{2}\sigma^2\right)t+\sigma W_t^{\mathbb{Q}^1}.$$ Because $$W_t^{\mathbb{Q}^1}$$ is a standard Brownian motion under the stock measure $$\mathbb{Q}^1$$ (by construction) and thus has zero expectation, we have $$\mathbb{E}^{\mathbb{Q}^1}[\ln(S_t)]=\ln(S_0)+\left(r+\frac{1}{2}\sigma^2\right)t.$$

Turning now to the claim paying $$S_T\ln(S_T)$$, we can derive its price as follows \begin{align*} e^{-rT}\mathbb{E}^\mathbb{Q}[S_T\ln(S_T)] &= e^{-rT}\mathbb{E}^{\mathbb{Q}^1}\left[S_T\ln(S_T)\frac{\mathrm{d}\mathbb{Q}}{\mathrm{d}\mathbb{Q}^1}\right] \\ &= S_0 \mathbb{E}^{\mathbb{Q}^1}\left[\ln(S_T)\right] \\ &= S_0 \left(\ln(S_0)+\left(r+\frac{1}{2}\sigma^2\right)T\right). \end{align*} Here, I used $$\frac{\mathrm{d}\mathbb{Q}}{\mathrm{d}\mathbb{Q}^1}=\frac{S_0e^{rT}}{S_T}$$.

Of course, this value can be negative (just like the payoff this claim can be negative).

• lovely method!! Jul 12 '20 at 14:57

Part 1: deriving the drift of the stock price process under the stock Numeraire.

Under the risk-neutral measure, the process for $$S_t$$ is as follows:

$$S_t = S_0 + \int_{h=t_0}^{h=t}rS_h dh + \int_{h=t_0}^{h=t}\sigma S_h dW_h = \\ = S_0exp\left[ (r-0.5 \sigma^2)t+\sigma W(t) \right]$$

In the above model, the Numeraire is $$N(t)=e^{rt}$$ with $$N(t_0):=1$$. Specifically, $$W(t)$$ is a standard Brownian motion under the risk-neutral measure associated with the Numeraire $$N(t)$$.

The change of Numeraire formula is (I wanna change from $$N(t)$$ to some $$N_1(t)$$):

$$\frac{dN_1(t)}{dN(t)}= \frac{N(t_0)N_1(t)}{N(t)N_1(t_0)}$$

Using the stock as numeraire gives:

$$\frac{dN_{S}}{dN}(t) = \frac{1*S_t}{e^{rt}S_0}=\frac{S_0exp\left[ (r-0.5 \sigma^2)t+\sigma W(t) \right]}{e^{rt}S_0}=e^{-0.5\sigma^2t+\sigma W_t}$$

The radon-nikodym derivative above is directly applicable to $$W(t)$$ using the Cameron-Martin-Girsanov Theorem.

Diving into the detail of how changing probability measure actually works, let's consider the probability distribution of $$W(t)$$ under the risk-neutral measure:

$$\mathbb{P}^Q(W_t \leq k)=\int_{h=-\infty}^{h=k}\frac{1}{\sqrt{2\pi}}e^{\frac{-h^2}{2t}}dh$$

We can define some new probability measure $$\mathbb{P}^2$$ using the Radon-Nikodym derivative $$y(W_t,t):=e^{-0.5\sigma^2t+\sigma W_t}$$ as follows:

$$\mathbb{P}^2(W_t\leq k):=\mathbb{E}^Q[y(W_t,t)I_{W(t) \leq k}]$$

Evaluating the expectation gives:

$$\mathbb{E}^Q[y(W_t,t)I_{W(t) \leq k}] = \int_{h=-\infty}^{h=k}y(W_t,t) f_{W_t}(h)dh = \\ = \int_{h=-\infty}^{h=k}e^{-0.5\sigma^2t+\sigma h} \frac{1}{\sqrt{2\pi}}e^{\frac{-h^2}{2t}}dh= \\ =\int_{h=-\infty}^{h=k}\frac{1}{\sqrt{2\pi}}e^{\frac{-(h^2-\sigma t)}{2t}}dh$$

Therefore we can see that applying the Radon-Nikdym derivative adds the drift $$\sigma t$$ to $$W_t$$ under the probability meaure $$\mathbb{P}^2$$ (we can see that via the probability distribution of $$W_t$$ under $$\mathbb{P}^2$$).

So in our case, $$\mathbb{P}^2$$ is the probability measure defined by using $$S_t$$ as numeraire, we can call it $$\mathbb{P}^{S_t}$$. The final step is to figure out the process of $$S_t$$ under $$\mathbb{P}^{S_t}$$:

Let's use the following algebric "trick": I am going to define a new process under the original risk-neutral measure $$Q$$, called $$\tilde{W_t}$$ as follows: $$\tilde{W_t}:=W_t-\sigma t$$.

Therefore, under the original measure $$Q$$, the process $$\tilde{W_t}$$ has a "negative" drift equal to $$-\sigma t$$.

Let's now insert $$\tilde{W_t}$$ into the original process equation for $$S_t$$ using $$W_t = \tilde{W_t} + \sigma t$$:

$$S_t=S_0exp\left[ (r-0.5 \sigma^2)t+\sigma W(t) \right]= \\ = S_0exp\left[ (r-0.5 \sigma^2)t+\sigma (\tilde{W(t)}+\sigma t) \right] = \\ = S_0exp\left[ (r-0.5 \sigma^2)t+\sigma^2 t + \tilde{W(t)} \right] = \\ = S_0exp\left[ (r+0.5 \sigma^2)t+ \tilde{W(t)} \right]$$

We know that applying the radon-nikodym derivative from before (i.e $$e^{-0.5\sigma^2t+\sigma W_t}$$ ) adds drift $$\sigma t$$, and we defined $$\tilde{W_t}$$ to have drift $$-\sigma t$$. Therefore applying the radon-nikodym to $$\tilde{W_t}$$ will remove the drift from $$\tilde{W_t}$$ and the process $$\tilde{W_t}$$ will become a driftless Standard Brownian motion under $$\mathbb{P}^{S_t}$$.

So we have the process for $$S_t$$ under $$\mathbb{P}^{S_t}$$ as:

$$S_0exp\left[ (r+0.5 \sigma^2)t+ \tilde{W(t)} \right]$$

Wehere $$\tilde{W(t)}$$ is a Standard Brownian motion without a drift.

Part 2: Ito's lemma to derive the process for $$log(S_t)$$.

I assume you know how to apply Ito's lemma to solve the standard GBM model for a stock price, i.e. our starting eqution above. Then by inspection, one can see that applying Ito's lemma to $$ln(S_t)$$ under measure $$\mathbb{P}^{S_t}$$ will produce the same result, but with a different drift. Indeed under $$\mathbb{P}^{S_t}$$:

$$S_t=S_0exp\left[ (r+0.5 \sigma^2)t+\sigma \tilde{W(t)} \right]$$

Therefore:

$$ln \left( \frac{S_t}{S_0} \right)= (r+0.5 \sigma^2)t+\sigma \tilde{W(t)}$$

I.e. the probability measure does not affect the way that Ito's lemma can be applied.

• I dont understand the applying radon nikodym bit Jul 3 '20 at 17:16
• $S_0 + \int_{h=t_0}^{h=t}rS_h dh + \int_{h=t_0}^{h=t}\sigma S_h dW_h = \\ = S_0exp\left[ (r-0.5 \sigma^2)t+\sigma W(t) \right]$ how was this solved? How was the third term in the LHS evaluated? Jul 3 '20 at 18:57
• $S_t$ is an Ito Process (by definition of Ito Process). By Ito's Lemma for any twice-differentiable function $F(S_t,t)$, we have: $F(S_t,t)=F(S_0,t_0)+\int_{h=t_0}^{h=t}\left( \frac{\partial F}{\partial t}+ \frac{\partial F}{\partial S}\mu S_h + 0.5\frac{\partial^2 F }{\partial S^2} \sigma^2S_h^2 \right) dh + \int_{h=t_0}^{h=t}\frac{\partial F}{\partial S}\sigma S_h dW_h$. Take $F(S_t,t)=ln(S_t)$, compute the derivatives, and you get: $ln(S_t)=ln(S_0)+\int_{h=t_0}^{h=t}\left(\mu + 0.5\sigma^2 \right) dh + \int_{h=t_0}^{h=t}\sigma dW_h$. The two integrals then evaluate directly to the result. Jul 5 '20 at 12:50
• @Permian: Radon-Nikodym: look at Cameron-Martin-Girsanov theorem. Radon-Nikdodym "derivative" is a random variable that allows you to change probability measure from some $\mathbb{P^1}$ to some $\mathbb{P^2}$. Jul 5 '20 at 12:52
• i know these things individually, im just struggling to see how to fit them together Jul 5 '20 at 16:16