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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 $I(x)$.

Payoff=

$I(x)$ 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 $J$?

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  • $\begingroup$ 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
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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: $$\rho_{IJ}=\frac{\sigma_{IJ}}{\sigma_I\sigma_J}$$

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  • $\begingroup$ 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 Singh Duggal 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 Singh Duggal Oct 13 '20 at 11:41
  • $\begingroup$ No worries, glad it helped! $\endgroup$ – Daneel Olivaw Oct 13 '20 at 11:42

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