# Tradeable => Satisfies pricing equation?

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,

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.

$V_t$ 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.

• 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?
– bcf
Commented Jun 5, 2015 at 14:06
• Yes any tradeable will satisfy this equation in some domain whose boundary corresponds to exercising the option.
– AFK
Commented Jun 5, 2015 at 17:29
• 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.
– AFK
Commented Jun 5, 2015 at 17:36
• Okay, I wonder how Wilmott got the time derivative term to disappear?
– bcf
Commented Jun 5, 2015 at 17:44
• @AFK Did Dr Wilmott assumpted that the derivative of the stock to time is zero? Commented Jun 6, 2015 at 2:38