# Why does an autocall on a linear payoff have vega?

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 $$J$$?

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

• 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 – Arshdeep Oct 13 '20 at 10:55
• Ah you modified your comment. You wrote "with some volatility of course": that makes a difference. See my edit. – Daneel Olivaw Oct 13 '20 at 11:21
• Yes, I realized that conditional distribution may not of course represent that $I$ will conditionally grow at the risk free rate. Thank you. – Arshdeep Oct 13 '20 at 11:41
• No worries, glad it helped! – Daneel Olivaw Oct 13 '20 at 11:42