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In Wilmott's third volume, on p. 857, he tries giving an insight into the market price of risk by showing what it is for traded assets. For this he constructs a portfolio of two different options: long one option worth $V$ and short $\Delta$ units of another option worth $V_1$ giving a portfolio value of $$ \Pi = V - \Delta V_1. $$ Following the derivation he gave above for stochastic volatility, he gives the PDE $$ \frac{\partial V}{\partial t} + \frac{1}{2}\sigma^2S^2\frac{\partial^2 V}{\partial S^2} + (\mu - \lambda_S \sigma)S\frac{\partial V}{\partial S} - rV = 0. $$ So far, so good. But he then states,

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My questions:

  1. Why does the stock being tradeable mean its price must satisfy this PDE?

  2. Assuming that $V=S$ is a solution of the PDE, where did the $\frac{\partial S}{\partial t}$ go? Is it that, by $S$ he really means the initial stock price, i.e., $S = S(0)$? We would then have the time derivative vanishing.

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$V_t $ is the price of a tradeable. Because we can delta hedge it, $V_t =v(t,S_t) $ where $v$ is a solution of the PDE on some domain whose boundary corresponds to the exercise of the option. For a European option with payoff $g(S_T)$ at time $T$, the price function $v$ is the solution $$ \frac{\partial v}{\partial t} + \frac{1}{2}\sigma^2S^2\frac{\partial^2 v}{\partial S^2} + (\mu - \lambda_S \sigma)S\frac{\partial v}{\partial S} - rv = 0. $$ on $[0,T] \times \mathbb{R}^*_+$ with boundary $$ v(T,S) = g(S) $$ Consider the special case of a derivative that delivers the stock at time $T$: $g(S) = S$. Obviously you can replicate this by buying the stock at time $0$ and holding it until $T$ so the price is $V_t = S_t$. The corresponding price function is thus $v(t,S) = S$. Its partial derivative with respect to $t$ is $0$ (no Theta), its partial derivative with respect to the variable $S$ (Delta) is $1$ and its second partial derivative with respect to the variable $S$ (Gamma) is 0. Since it is the price of a European option, it satisfies the PDE which reduces to
$$ (\mu - \lambda_S \sigma)S - rS = 0. $$ This allows to deduce the market price of risk.

Note: I think to avoid confusion you need to distinguish between the partial and total derivative.

The value of a portfolio is stochastic. Its "total derivative" is a formal notation $$ dV_t = \mu^V_t dt + \sigma^V_t dW_t $$ for an Ito integral. Intuitively, this is your P&L over a small time period $[t,t+dt]$.

On the other hand the price function $v(t,S)$ is deterministic. Its partial derivative are the Greeks: $\Theta = \partial_t v$, $\Delta = \partial_S v$, $\Gamma = \partial^2_S v$. Here $S$ is nothing but a letter to identify the second variable of $v$. We could write $\partial_2 v$ just as well.

The two are related when you plug the stochastic price process in the price function: $$V_t = v(t,S_t)$$ And Ito's lemma tells you how to compute the "total derivative" in terms of the partial derivatives of $v$: $$ dV_t = \partial_tv(t,S_t) dt + \partial_Sv(t,S_t) dS_t + \frac{1}{2}\partial^2_Sv(t,S_t) d\langle S,S\rangle_t $$ which leads to the BS PDE.

In the case $v(t,S) = S$ clearly the partial derivative is $\partial_t v(t,S) = 0$ for all $(t,S) \in [0,T]\times \mathbb{R}^*_+$ but the total derivative $dv(t,S_t) = dS_t = \mu S_t dt + \sigma S_t dW_t$ corresponds to a non-constant stochastic process.

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  • $\begingroup$ Okay, thanks...is it okay to say that any tradable in this market (of one stock and a money money account) must satisfy this PDE? Also, do you think my understanding of my second question is correct, that $\frac{\partial S}{\partial t} = 0$ since $S$ is actually $S = S(0)$ here? $\endgroup$
    – bcf
    Commented Jun 5, 2015 at 14:06
  • $\begingroup$ Yes any tradeable will satisfy this equation in some domain whose boundary corresponds to exercising the option. $\endgroup$
    – AFK
    Commented Jun 5, 2015 at 17:29
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    $\begingroup$ No S is just a variable here. The corresponding portfolio is just holding the stock so $v(t,S_t) = S_t$. No S_0 here. $\endgroup$
    – AFK
    Commented Jun 5, 2015 at 17:36
  • $\begingroup$ Okay, I wonder how Wilmott got the time derivative term to disappear? $\endgroup$
    – bcf
    Commented Jun 5, 2015 at 17:44
  • $\begingroup$ @AFK Did Dr Wilmott assumpted that the derivative of the stock to time is zero? $\endgroup$
    – SmallChess
    Commented Jun 6, 2015 at 2:38

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