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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?

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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).

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  • 2
    $\begingroup$ lovely method!! $\endgroup$ – Permian Jul 12 at 14:57
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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.

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  • $\begingroup$ I dont understand the applying radon nikodym bit $\endgroup$ – Permian Jul 3 at 17:16
  • $\begingroup$ $S_0 + \int_{h=t_0}^{h=t}rS_h dh + \int_{h=t_0}^{h=t}\sigma S_h dW_h = \\ = S_0exp\left[ (r-0.5 \sigma^2)t+\sigma W(t) \right]$ how was this solved? How was the third term in the LHS evaluated? $\endgroup$ – Permian Jul 3 at 18:57
  • $\begingroup$ $S_t$ 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. $\endgroup$ – Jan Stuller Jul 5 at 12:50
  • $\begingroup$ @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 $\mathbb{P^2}$. $\endgroup$ – Jan Stuller Jul 5 at 12:52
  • $\begingroup$ i know these things individually, im just struggling to see how to fit them together $\endgroup$ – Permian Jul 5 at 16:16

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