# Brownian bridge density and risk neutral density for derivative pricing

The book The Volatility Surface by Gatheral (2006) introduces the Brownian bridge like density $q(x_t,t;x_T,T)$ of $x_t$ conditional on $x_T = log(K)$. Can we use $q(x_t,t;x_T,T)$ as the risk neutral measure? Why $q(x_t,t;x_T,T)$ can easily exceed one, and sometimes it can be extremely large?

Here are some formulae and interpretation from the book (from page 29 to 30). For fixed strike $K$ and maturity $T$, define the Black-Scholes gamma

$$\Gamma_{BS}(S_t,\bar{\sigma}(t)):=\frac{\partial ^2}{\partial S_t^2}C_{BS}(S_t,K,\bar{\sigma}(t),T-t)$$

and the Black-Scholes forward implied variance function

$$v_{K,T}(t)=\frac{E\big[\sigma_t^2 \cdot S_t^2 \cdot \Gamma_{BS}(S_t,\bar{\sigma}(t))|\mathcal{F}_0\big]}{E\big[S_t^2 \cdot \Gamma_{BS}(S_t,\bar{\sigma}(t))|\mathcal{F}_0\big]}$$

where $\bar{\sigma}^2(t):=\frac{1}{T-t}\int_{t}^T v_{K,T}(u)du$

The Black-Scholes implied variance, by following Lee(2005),can be expressed as:

$$\sigma_{BS}(K,T)^2=\bar{\sigma}(0)^2=\frac{1}{T}\int_{0}^{T}E^{G_t}[\sigma_t^2]dt$$

so interpreting $v(t)$ as the expectation of $\sigma_t^2$ with respect to the probability measure $\mathbb{G}_t$ defined, relative to the pricing measure $\mathbb{P}$, by the Radon-Nikodym derivative

$$\frac{d\mathbb{G}_t}{d\mathbb{P}}:= \frac{S_t^2\Gamma_{BS}(S_t,\bar{\sigma}(t))}{\mathbb{E}\big[S_t^2 \Gamma_{BS}(S_t,\bar{\sigma}(t))|\mathcal{F}_0\big]}$$

We compute a risk-neutral expectation: $$\mathbb{E}^p[f(S_t)] = \int dS_t p(S_t,t;S_0)f(S_t)$$

we get the risk neutral pdf of the stock price at time $t$ by taking the second derivative of the market price of European options with respect to strike price

$$p(S_t,t;S_0) = \frac{\partial ^2 C(S_0,K,t)}{\partial K^2}\bigg|_{K=S_t}$$

Then, from equation of Black-Scholes implied variance $\sigma_{BS}(K,T)^2$ we have $$v_{K,T}(t) = \mathbb{E}^{G_t} \big[\sigma_t^2 \big] = \mathbb{E}^p \big[\sigma_t^2 \frac{d\mathbb{G}_t}{d\mathbb{P}} \big] =\int dS_t \cdot q(S_t;S_0,K,T) \cdot \mathbb{E}^p \big[ \sigma_t^2 \big| S_t \big] =\int dS_t \cdot q(S_t;S_0,K,T) \cdot v_L(S_t,t)$$

where further define $$q(S_t;S_0,K,T) := \frac{p(S_t,t;S_0) S_t^2 \Gamma_{BS}(S_t)}{\mathbb{E} \big[ S_t^2 \Gamma _{BS}(S_t) \big| \mathcal{F}_0}$$

and $v_L(S_t,t)= \mathbb{E}^p[\sigma_t^2|S_t]$ is the local variance. q(S_t;S_0,K,T) looks like a Brownian Bridge density for the stock price. Then, the equation of $v_{K,T}(t)$ can be rewritten in terms of $x_t :=log(S_t/S_0)$

$$v_{K,T}(t)=\int dx_t \cdot q(x_t;x_T,T) \cdot v_L(x_t,t)$$

My question is: from the last equation of $v_{K,T}(t)$, it seems that q(S_t;S_0,K,T) is the risk neutral measure. However, sometimes it can be extremely large, such as 8 or 12. Should it be some value between 0 and 1?

• If you can define $q$ more specifically and provide all the related information, it would be more beneficial for discussion. A suggestion is to make your question self-contained, if possible. – Gordon Nov 16 '16 at 14:35
• Are you saying that whenever you evaluate a pdf you should always obtain a number smaller than one? $q$ is not a density anyway, as you've mentioned, so why you would like to use it as such? – Quantuple Nov 16 '16 at 16:27
• @ Quantuple I mean the probability measure should be within 0 and 1. The book tells us that q is the density. – Smirk Nov 16 '16 at 16:31
• Ahem a density does not have to be between 0 and 1. The area under the density has to be one. For example a uniform density on -0.1,0.1 has a value (ordinate) of 5 in this interval. – noob2 Nov 16 '16 at 17:44