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!


1 Answer 1



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.

  • $\begingroup$ A couple questions. (1) What exactly is meant by 'tradeable' here? I had a suspicion $S_t^2$ was not a legit numeraire but couldn't articulate why. Also, why IS $E_t(S_T^2)$ a tradeable? (2) Is the Expectation above in your answer under what measure? I assume I can compute it under the risk-neutral measure in which case I get that $E_t(S_T^2) = S_t^2e^{(2r+\sigma^2)(T-t)}$. Adjusting method 3 with this, it seems I get the factor $e^{(2r+\sigma^2)T}$ instead of $e^{(r+\sigma^2)T}$ as in Method 1 and 2. Where am I still off? $\endgroup$
    – jmac
    Commented Feb 2 at 8:41
  • $\begingroup$ @jmac Apologies - I forgot the discount factor in the claim (which is under the risk neutral measure) and edited my answer / hint. If you now write down the SDE for the claim you'll see that it has the form $d E[\cdot] = r E[\cdot] dt + 2\sigma E[\cdot] dW$ which has a rate of return equal to $r$, just like the money market and the stock itself. $\endgroup$
    – Frido
    Commented Feb 2 at 9:15
  • $\begingroup$ Thank you so much for that correction. I went through it all and now Method 3 gives the same answer as the first two. One last quick question/inquiry. Why is the numeraire you suggested tradeable? I guess I'm not seeing this in a real-world sense. Like I can get $S_t^2$ is not tradeable because you can only trade the stock not its square. But why is the discounted expected value of the terminal squared stock price all of a sudden 'tradeable'? Like how would we even hedge changes in the derivative security due to fluctuations in the numeraire? $\endgroup$
    – jmac
    Commented Feb 2 at 20:30
  • $\begingroup$ @jmac Because by the Martingale Representation Theorem we can write any function of $S_T$ as follows: $$ f(S_T) = E_0 [ f(S_T) ] + \int_0^T \phi (S_t) dW_t $$ where the first term on the right hand side is the price of the claim and the second term is the delta hedge strategy. You can see this by noting that $\phi (S_t) dW_t = \frac{ \phi(S_t) }{\sigma S_t} dS_t$. See also: en.wikipedia.org/wiki/Martingale_representation_theorem $\endgroup$
    – Frido
    Commented Feb 3 at 10:09
  • $\begingroup$ @jmac Again I forgot to include discount factor in my comment above (as I'm used to working under forward measure or just setting r=0), but it's not hard to see how MRT looks like with a discount factor. Summary: the claim is tradable because it can be replicated by (delta) hedging it. $\endgroup$
    – Frido
    Commented Feb 3 at 10:18

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