Computing Derivative Security with Change of Numeraire

Question

Under Black-Scholes, price a contract worth $S_T^{2}log(S_T)$ at expiration.

This is a question from Joshi's Quant Book (an extension question).

Ok, so I solved this with 3 different methods to make sure I understood the concepts. Unfortunately 2 of the 3 methods give the same answer but the 3rd one does not (although it is only off by a factor). I'm curious where I am going wrong and which solution is correct (provided one of them is indeed correct).

Let $V_0$ be the value of our derivative security.

Method 1: Pricing under the Risk-Neutral Measure

$V_0 = e^{-rT}\tilde{E}(S_T^{2}log(S_T))$ where expectation is taken under the risk-neutral measure.

Here, I just compute the expectation directly with the help of the moment generating function as follows:

Under the risk-neutral measure,

$S(t) = e^{ln(S_0) + (r-0.5\sigma^2)t + \sigma(W_t - W_0)}$

Define a random variable $X \sim N(ln(S_0) + (r-0.5\sigma^2)T, \sigma^2T)$.

Setting $log(S_T) = X$ and $S_T = e^X$, we have:

Then, $V_0 = e^{-rT}\tilde{E}(Xe^{2X})$.

Now, we use the mgf of X to compute this expectation.

$M_X(s) = \tilde{E}(e^{sX})$ and so by differentiating with respect to s and evaluating at $s=2$, we obtain:

$M_X'(2) = \tilde{E}(Xe^{2X}) = (ln(S_0) + (r + 1.5\sigma^2)T)S_0^2e^{(2r+\sigma^2)T}$

(Here, I use that for $X \sim N(\nu, \lambda^2)$, we have $M_X(s) = e^{s\nu + 0.5s^2\lambda^2}$ and thus $M_X'(2) = (\nu + 2\lambda^2)e^{2\nu + 2\lambda^2})$

Hence, $V_0 = e^{-rT}\tilde{E}(Xe^{2X}) = (ln(S_0) + (r + 1.5\sigma^2)T)S_0^2e^{(r+\sigma^2)T}$.

Method 2: Using the stock as numeraire.

Here, $V_0 = S_0 \hat{E}(S_Tlog(S_T))$ where expectation is taken under the stock measure.

Under the stock measure,

$S(t) = e^{ln(S_0) + (r+0.5\sigma^2)t + \sigma(W_t - W_0)}$

Define a random variable $X \sim N(ln(S_0) + (r+0.5\sigma^2)T, \sigma^2T)$.

Setting $log(S_T) = X$ and $S_T = e^X$, we have:

Then, $V_0 = S_0\hat{E}(Xe^{X})$.

Using the same process as in Method 1, I get the exact same answer.

(the only difference in the process is that I evaluate the mgf at 1 for the first derivative instead of 2 and multiply by $S_0$ at the end instead of $e^{-rT}$).

Method 3: Using the squared stock as numeraire.

Here, $V_0 = S_0^2 E^\star(log(S_T))$ where expectation is taken under the associated measure.

Under this change of measure with Radon-Nikodym Derivate $Z(t) = d\hat{P}/d\tilde{P} = e^{-rt}S_t^2/S_0^2$, I determined that the drift term is $(r+2\sigma^2)$ for the stock process $S_t$. Since the diffusion term is unaffected by change in measure, we have:

$dS_t = (r+2\sigma^2)S_tdt + \sigma S_tdW^\star_t$

Thus, $S(t) = e^{ln(S_0) + (r+1.5\sigma^2)t + \sigma(W_t - W_0)}$

Recall we wish to compute $V_0 = S_0^2 E^\star(log(S_T))$.

Define a random variable $X \sim N(ln(S_0) + (r+1.5\sigma^2)T, \sigma^2T)$.

Setting $log(S_T) = X$ we have:

Then, $V_0 = S_0^2\hat{E}(X) = S_0^2(ln(S_0) + (r + 1.5\sigma^2)T)$

So, here I am missing that $e^{(r+\sigma^2)T}$ term.

So, what am I doing wrong? And which is correct?

Thank you in advance!

Answer 1

Hint:

I think you need to use a tradable as numeraire. So the money market and the stock price are tradables. But $S_t^2$ is not a tradable. How to solve this:

Notice that for $t\in[0,T]$ the claim $E_t \left[ e^{-r(T-t)}S_T^2 \right]$ is a tradable, and in particular $S_T^2 = E_T \left[ S_T^2 \right]$.

So, you need to write down the SDE for $E_t \left[ e^{-r(T-t)} S_T^2 \right]$ and use that as numeraire.