Risk-Neutral and Real-World Expectations in Option Valuation

# Risk Neutral and Real World Valuations using Monte Carlo


# 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?

## Answer by user34971 (score 6, accepted)

https://quant.stackexchange.com/a/49935

Just to add to the answer by @Kevin :

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.

- 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.

- 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.

## Answer by Kevin (score 6)

https://quant.stackexchange.com/a/49934

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.