# Risk Neutral and Real World Valuations using Monte Carlo

Assume I'm an investor that wants to sell exotic put options. No one else is selling my kind of put option, so I need to determine my own "Market Price" through Monte Carlo simulation. I know that by the law of one price, this should hold:

$$P_t = E^Q[P_t|\mathcal{F}_t] = E^P[P_t|\mathcal{F}_t]$$

In my risk neutral Monte Carlo valuation, I model my stock price as:

$$dS = rS_tdt + \sigma S_tdW_t$$

In my real world Monte Carlo valuation, I model my stock price as:

$$dS = \mu S_tdt + \sigma S_tdW_t$$

Just thinking about this intuitively though, the put option valued under my real world Monte Carlo simulation will be way cheaper than the put option under my risk neutral simulations, because the growth rate is so much higher. So what am I missing here? Am I wrong in my first statement, that expectation under the P and Q measures are equal, or am I formulating my second statements incorrectly?

• There is no contradiction only some confusion. Price at t given you are at t (condition on $\mathscr{F}_t$) is a number not a random variable. So your first equation holds trivially, it has nothing to do with the law of one price. Going to your second part, the growth rate in your simulations does not matter, if you condition on t. What you are really saying (at time t) is: The price I observe right now of my option is 10 (say). It is the same, no matter how I simulate the future.
– g g
Nov 26 '19 at 18:17
• The answers below are very good. I just want to add one observation, depending on how exotic your option is. If the underlying is illiquid and uncommonly traded then one might hesitate to apply arbitrage pricing theory at all for put options on the underlying. Nov 26 '19 at 19:41

There are at least two things going on here. First of all let $$\{Q_i \}$$ denote a set of equivalent probability measures, which includes your $$P$$ and $$Q$$ above.

1. Any $$F^i(t)$$ defined as $$F^i(t) = E_t^{Q_i} [P_T]$$ will be a martingale by application of the tower law.

2. With the definition above, it will not be the case that $$F^i(t) = F^j(t)$$. Instead, if $$dQ_i / dQ_j$$ denotes the measure change (technically called the Radon-Nikodym derivative), then

$$E_t^{Q_i} [P_T] = E_t^{Q_j} \left[ \frac{dQ_i}{dQ_j} P_T \right]$$

which is the correct form of the law of one price.

You probably wonder whether $$\mathbb{E}^\mathbb{P}[P_T\mid\mathcal{F}_t]= \mathbb{E}^\mathbb{Q}[P_T\mid\mathcal{F}_t]$$. Note the $$T$$ as index, i.e. the future unknown payoff and not the current price $$P_t$$.

Now, why should $$P_t$$ be a martingale under both, $$\mathbb{P}$$ and $$\mathbb{Q}$$? Most likely, it is not. Indeed, the reason why you use $$\mathbb{Q}$$ in the first place is because $$(P_t)$$ is not a martingale under $$\mathbb{P}$$. Instead, you define $$\mathbb{Q}$$ such that discounted basic assets (and hence derivatives) are martingales under $$\mathbb{Q}$$.

As you noted, the distribution of $$(S_t)$$ is quite different under $$\mathbb{P}$$ and $$\mathbb{Q}$$. Thus, the conditional expectations of $$P_T$$ differ too.