# How to express the conditional expected variance under the risk neutral measure?

The conditional expection of variance under risk neutral measure is

$$\mathbb{E}^Q[V_T |S_T=K]$$

where $S_T$ and $K$ represent the spot price at maturity and strike price, respectively.

Assume I know the risk neutral density $q(V,S,t)$, I want to calculate the inverse of conditional expectation of $V_T$. Should I use this formula
$$\frac{1}{\mathbb{E}^Q[V_T|S_T=K]} = \frac{1}{\int V_t \cdot q(v,s,t)\cdot dV}$$ or another formula

$$\frac{1}{\mathbb{E}^Q[V_T|S_T=K]} = \frac{\int q(v,s,t)\cdot dV}{\int V_t \cdot q(v,s,t)\cdot dV}$$

Further, if
$$g(v,s,t)\approx q(v,s,t)\cdot dv \cdot ds$$ I wonder if we can get
$$\Bbb{E}^\Bbb{Q}\left[ V_T \vert S_T = s \right] = \frac{\int_\Omega v\,q_{V_T,S_T}(v, s,T) dv}{\int_\Omega q_{V_T,S_T}(v, s,T) dv} \approx \frac{\sum v\cdot q_{V_T,S_T}(v,s,t) \cdot dv }{\sum q_{V_T,S_T}(v,s,t) \cdot dv}$$

$$= \frac{\sum v \cdot \frac{g(v,s,t)}{dv \cdot ds} \cdot dv }{\sum \frac{g(v,s,t)}{dv \cdot ds} \cdot dv} = \frac{\sum v \cdot g(v,s,t)}{\sum g(v,s,t)}$$

Thanks!

• What is the expectation of variance? What is the dynamic of your model? – user16651 Nov 24 '16 at 11:43

Assume that $q_{V_T,S_T}(v,s,T)$ represents the known joint pdf of 2 continuous random variables $V_T$ and $S_T$ under some probability measure $\Bbb{Q}$. Further assume that the definition domain of $V_T$ is $\Omega$.
By definition of the expectation operator + conditional probability density function $$\Bbb{E}^\Bbb{Q}\left[ V_T \vert S_T = s \right] = \int_\Omega \, q_{V_T \vert S_T}(v, s, T) dv$$
From the product rule of probability $$q_{V_T \vert S_T}(v, s, T) q_{S_T}(s,T) = q_{V_T,S_T}(v, s,T)$$ hence, going back to the expectation calculation $$\Bbb{E}^\Bbb{Q}\left[ V_T \vert S_T = s \right] = \frac{\int_\Omega v\, q_{V_T,S_T}(v, s,T) dv}{q_{S_T}(s,T)}$$ Finally using the sum rule to express the marginal $q_{S_T}$ from the joint pdf $q_{V_T,S_T}$ one gets: $$\Bbb{E}^\Bbb{Q}\left[ V_T \vert S_T = s \right] = \frac{\int_\Omega v\, q_{V_T,S_T}(v, s,T) dv}{\int_\Omega q_{V_T,S_T}(v, s,T) dv}$$ which is equivalent to the second formula you mention.