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I'm told that the holder of a basket call is long correlation.

I understand that an increase in correlation leads to an variance of a portfolio.

But with one "degree of freedom" (high positive correlation), if one asset falls, so do the others. And with multiple degrees of freedom (correlation near zero), if one falls, the others may rise and bring the average into the money.

Is this reasoning wrong?

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  • $\begingroup$ Yes, it is wrong. "the others may rise": they may, but it is unlikely right? Since they are moving independently of each other most likely about half will move up and half will move down, offsetting each other to a large extent. The average will not move much. $\endgroup$ – Alex C Apr 26 '18 at 13:29
  • $\begingroup$ Alternatively, consider a particular case where the basket is made up of only $2$ components with correlation $-1$ such that their moves cancel each other. In that case the value of the basket will remain constant and therefore the value of the optionality is null. $\endgroup$ – Daneel Olivaw Apr 26 '18 at 14:59
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If I have an option on one stock then I am long implied volatility -> we can easily show this.

For a portfolios of stocks (the basket) the underlying turns out to be the sum of lognormals in the Black-Scholes setting. We don't know the true distribution. But we can write down the variance of the sum which is a function that increases if volatilies increase or if the correlation increases.

Thus we are long correlation for similar reasons as in the one stock call sense. This is not a proof but at least it makes the statement plausible.

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  • $\begingroup$ To add a little bit of math. $Cov(\alpha X, (1-\alpha) Y) = \alpha^2 Var(X) + (1-\alpha)^2 Var(Y) + 2 \alpha (1-\alpha) Cov(X, Y)$ Clearly $\alpha (1-\alpha)$ is positive. $\endgroup$ – Will Gu Apr 27 '18 at 6:19
  • $\begingroup$ @WillGu you are right. But it is just not that easy with baskets that's why I did not put math's there. But right as a start one can always put the formula of the variance of a sum (with weights). The variance of the sum of lognormals is different to the variance of a sum of normals. But of course speaking of second moments in general the formula is just correct. $\endgroup$ – Richard Apr 27 '18 at 6:27
  • $\begingroup$ oops typo alert: left side should be $Cov(αX + (1−α)Y)$. $\endgroup$ – Will Gu Apr 27 '18 at 6:30

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