Consider a (stochastic) linear index, say $I(t)$, in that it grows at the risk free rate (with some volatility of course). There exists a maturity date $T$ on which I receive $I(T)$; however there is another index $J(t)$ which on crossing a barrier $B$ between $[0,T]$, say at time $x$, I receive $I(x)$.


$I(x)$ if there exists $x$ in $[0,T]$ such that $J(x)>B$, paid at $x$.

$I(T)$ otherwise, paid at maturity $T$.

I don't understand why this product shows vega with respect to any index. Since (discounted) $I$ is a martingale, it really doesn't matter when I get paid a martingale since the expected discounted value is the same. Can you help me conceptually understand why this shows vega w.r.t index $J$?

  • $\begingroup$ Arshdeep let me know if there is anything else I can address, otherwise please feel free to mark as answered if you think I've addressed your concerns. $\endgroup$ – Daneel Olivaw Oct 16 '20 at 10:57

Let $\sigma_J$ be the volatility of the index $J$. Assume that $J(0)\leq B$. Consider the following 2 extreme cases:

  • $\sigma_J=0 \Rightarrow \forall x\in[0,T],J(x)=J(0)\leq B$: hence you will always be paid $I(T)$ at expiry.
  • $\sigma_J=\infty \Rightarrow \exists\epsilon>0, J(\epsilon)>B$: hence you will almost immediately be paid $I(0)\approx I(\epsilon)$.

Thus the payoff of your product depends on the volatility of the index $J$. More intuitively, the more volatile $J$ is, the more likely it is it will cross the barrier $B$, because the more the price varies along the life $[0,T]$.

Regarding the payment of $I$ itself, assuming a constant risk-free rate $r$, note that the value of your payoff can be written: $$V(0)=E^\mathcal{Q}\left(D(0,T)1_{\{\max_{0\leq x\leq T}J(x)\leq B\}}I(T)+D(0,\tau)1_{\{\max_{0\leq x\leq T}J(x)> B\}}I(\tau)\right)$$ where: $$\tau:=\min\{x:x\in[0,T],J(x)>B\}$$ If $I$ is deterministic, then $I(t)=1/D(0,t)$ and the value simplifies to: $$\begin{align} V(0)&= E^\mathcal{Q}\left(1_{\{\max_{0\leq x\leq T}J(x)\leq B\}}+1_{\{\max_{0\leq x\leq T}J(x)> B\}}\right) \end{align}$$ which is equal to $1$ because either $\max_{0\leq x\leq T}J(x)$ is above $B$ or it isn't, there are no more outcomes. On the other hand, if $I$ is risky, that is it has a stochastic term, and log-normally distributed, you would have something along: $$V(0)=I(0)E^\mathcal{Q}\left(1_{\{\max_{0\leq x\leq T}J(x)\leq B\}}e^{-\frac{\sigma_I^2}{2}T+\sigma_IW_I(T)}+1_{\{\max_{0\leq x\leq T}J(x)> B\}}e^{-\frac{\sigma_I^2}{2}\tau+\sigma_IW_I(\tau)}\right)$$ This is a more complex product, because it depends upon the covariance structure between $I$ and $J$:

  • If both are positively correlated, then if $J$ crosses $B$ (which it needs to do from below, given $J(0)\leq B$), there are more chances the value of $I$ will be high;
  • On the other hand, if $I$ and $J$ have negative correlation, then if $J$ crosses the barrier it means it has gone upwards, so there will chances that the value of $I$ will have gone downwards due to negative correlation.

Note that correlation $\rho$ and volatility are related $-$ where $\sigma_{IJ}$ is covariance: $$\rho_{IJ}=\frac{\sigma_{IJ}}{\sigma_I\sigma_J}$$

  • $\begingroup$ But $E[DF(0,T)I(T)]=I(0)$, so it's the same payoff. This is because I grows at a risk free rate $\endgroup$ – Arshdeep Oct 13 '20 at 10:55
  • $\begingroup$ Ah you modified your comment. You wrote "with some volatility of course": that makes a difference. See my edit. $\endgroup$ – Daneel Olivaw Oct 13 '20 at 11:21
  • $\begingroup$ Yes, I realized that conditional distribution may not of course represent that $I$ will conditionally grow at the risk free rate. Thank you. $\endgroup$ – Arshdeep Oct 13 '20 at 11:41
  • $\begingroup$ No worries, glad it helped! $\endgroup$ – Daneel Olivaw Oct 13 '20 at 11:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.