1

In the proof of Dupire equation we end up with an identity involving the Dirac delta function.

How to prove that $$\dfrac{E[\sigma_T^2\delta(S_T-K)]}{E[\delta(S_T-K)]}=E[\sigma_T^2|S_T = K].$$

where $\delta(x)$ is the Dirac delta function. $S_T$ is a random variable, and $\sigma_T$ also.

Quantuple
  • 14,622
  • 1
  • 33
  • 69
A.Oreo
  • 1,243
  • 15
  • 27
  • 2
    This is the definition of a conditional expectation. You could write the RHS in integral form and use Bayes to convince yourself of the equality though. – Quantuple Oct 26 '17 at 07:46
  • @Quantuple but it's not the indicator function, the problem is how to deal with the $\delta(0) = \infty$ in the integral. – A.Oreo Oct 26 '17 at 08:05
  • 1
    You should see this as follows, similarly to $\Bbb{E}\left[ 1{S_T \geq K} \right] = P(S_T \geq K)$, $\Bbb{E}\left[ \delta(S_T-K) \right] = p(S_T=K)$, or in integral form $\int_0^\infty \delta(x-K) p(x) dx = p(K)$. – Quantuple Oct 26 '17 at 08:22

1 Answers1

3

Slightly abusing notations \begin{align} \Bbb{E}\left[ \sigma^2_T \vert S_T = K \right] &= \int_{0}^{+\infty} \sigma^2_T \, p( \sigma^2_T \vert S_T = K) d\sigma^2_T \\ &= \int_{0}^{+\infty} \sigma^2_T \frac{p(\sigma^2_T, S_T=K)}{p(S_T=K)} d\sigma^2_T \\ &= \frac{\int_{0}^{+\infty} \sigma^2_T p(\sigma^2_T, S_T=K) d\sigma^2_T }{p(S_T=K)} \\ &= \frac{\int_{0}^{\infty} \int_{0}^{+\infty} \sigma^2_T \delta(S_T-K) p(\sigma^2_T, S_T) d\sigma^2_T dS_T }{\int_{0}^{+\infty} \delta(S_T-K) p(S_T) dS_T} \\ &= \frac{\Bbb{E}\left[ \sigma^2 \delta(S_T-K) \right]}{\Bbb{E}\left[ \delta(S_T-K) \right]} \end{align}

Quantuple
  • 14,622
  • 1
  • 33
  • 69