# Tag Info

3

I don't have much experience in the matter, but I've been doing some related literature research recently and I think these links can be helpful: A rather recent study from CME A (possible a bit biased) report by BlackRock A report by Lyxor (asset manager affialiated to Societe Generale)

1

I think one should start with the basic Asset Pricing papers such as: Mehra and Prescott 1985; Campbell and Cochrane 1999; Bansal and Yaron 2004; Most of the models are then variations of these in one way or another. Some other more complex but potentially interesting are on intermediation. Some authors provide codes. Take a look for instance ...

1

As always I recommend reading Rennie and Baxter for an introduction to option pricing that's not too technical and gives intuition about how it all works.

1

I read the question as follows: You have one stock $S_0$ and after one period it either goes up to $S^+$ where the option takes the value $f^+$ or it goes down to $S^-$ where the option takes the value $f^-$. The bond grows from $B_0$ to $B_1 = B_0 \exp(r)$. Then you need to solve $$a S^+ + b B_1 = f^+ \\ a S^- + b B_1 = f^-$$ for $a,b$ which are $2$ ...

1

The current contract value is roughly 30k euros. The bidask spread is 1 tick, which equals 10 euros. Lets say you buy the contract and roll 3 times a year and then liquidate your position at expiry. You will hence pay 1 full bidask spread + 3 rolls, which if done via spreads with market orders, are equal to 1 tick each, hence you will pay 40 euros on bidasks ...

1

A unique state price vector does not have to exist for there to be no arbitrage. It sounds like the state price vector in question has infinitely many solutions. Try to reduce the price matrix to row echelon form and show that at least one state price vector exists.

1

A variance swap can be replicated with vanilla European options. If you take derivative with respect to variance, you need to do the same thing on both sides. That is, you need also take derivative with respect to variance on those vanilla options. However, the resulting derivative is not the vega in the usual sense, which is the derivative with respect to ...

1

I think I have figured this out. The key to the understanding is to think of the options' vegas as "key-strike vegas" compared to the var swap/replication portfolio's vega, which is analogous to "key rate durations of a bond portfolio" to the total effective duration of the portfolio.

1

The variance swap's Vega that is equal to the variance notional refers to the realized variance. The Black-Scholes vega refers to the market implied volatility. Now if you want, you can estimate the realized variance at expiry from the volatility of the options (for instance taking the atm variance arbitrarily), and that's often what people do. But that's ...

Only top voted, non community-wiki answers of a minimum length are eligible