Girsanov Weighting Versus Simulating Risk-Neutral Paths

# Girsanov Implementation Question


# Girsanov Implementation Question












Motivated by a recent question here, I would like to understand better the difference between two possible implementations of a Monte Carlo model with a single underlying with zero dividend, zero interest rates. (A) start by modeling the underlying with a Brownian with drift $\mu$, then convert to risk neutral zero drift by multiplying sample path probabilities by the Girsanov factor. And (B) apply a constant negative drift across the same sample paths such that the net drift becomes zero.

The question is, will I price anything differently using A versus B ?

## Answer by Rylan (score 1, accepted)

https://quant.stackexchange.com/a/84030

I'm assuming the recent question is this one: Girsanov Theorem: how to implement it with code? -- linking it mostly beacuse I will use the same notation in this answer as I did in that one.

Apologies if I've misunderstood your question.

If you want to price some option with payoff $f(S_T)$, for example $f(S_T) = (S_T - K)^+$ and $S_T$ = $S_0e^{ (\alpha- 0.5 \sigma^2)T + \sigma W_T} = S_0e^{ (r- 0.5 \sigma^2)T + \sigma \tilde W_T}$, and you want to use Monte Carlo, here are two ways you can do it:

- Generate $\tilde W_T$ directly in Monte Carlo, plug it into the form of $S_T$ that takes $\tilde W_T$, plug that into $f$, and compute the (unweighted) mean of your payoffs. This is a direct computation of $E^\mathbb{Q}[D(T)f(S_T)]$, the classic "risk neutral pricing formula"

- Generate $W_T$, plug it into the form of $S_T$ that takes $W_T$, plug that into $f$, and compute the weighted mean of your payoffs -- the weights are given by $Z_T$, so we can think of this as the unweighted mean of $Z_tf(S_T)$ where $S_T$ is generated using $W_T$, not $\tilde W_T$. This implements the risk-neutral pricing formula as $E^{\mathbb{Q}}[D(T)S(T)] = E^{\mathbb{P}}[D(T)Z_Tf(S_T)$], which is fine as this expectation is in fact how $\mathbb{Q}$ is defined.

But I don't think we can take a path of $W_t + \alpha_t$, multiply it by some "probability correction" like the RN derivative, and call that a Brownian motion under some other measure (and therefore plug it into some pricing formula). This is because a path of $Z_t \tilde W_t$ is not a Brownian motion path. To see this, we can apply Ito to it and note that the quadratic variation will not generally be $t$. However, a path of $\tilde W_t$ is a Brownian motion (without drift) in the measure $\mathbb{Q}$ that $Z_t$ defines.

$Z_t \tilde W_t$ is in fact a martingale, so it may produce correct prices for some special cases... if I had to guess, affine functions $f$ work but other functions will run afoul of Jensen's inequality. However in general I think you can expect it will not work.