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Quiz by Peter Carr: Are You a Trader or a Quant?

Note: Received via email today. Thanks to Dr. Peter Carr who gave us first-to-publish permission on QuantNet

Quiz: Are You a Trader or a Quant?

Traders and quants both know that an option's premium and its gamma are intimately related. This quiz will help you learn whether you are a trader or a quant. Assume the Black Scholes model with zero rates and dividends. Let \(t_0\) be the valuation time and let \(S_0 > 0\) be the initial stock price.
Under risk neutral measure \(\mathbf{Q}\), the stock price process is:

\(dS_t = \sigma S_t dW_t, \qquad t \geq t_0,\)​

where \(\sigma > 0\) is the stock's (constant) volatility and \(W\) is a standard Brownian motion. Let:

\(C(S,t;K,T) \equiv E^{\mathbf{Q}} [(S_T - K)^+| S_t = S]\)​

be the call value function. Consider a call which is initially at or out of the money i.e. \(S_0 \leq K\)
The three equations below all relate the call's initial premium \(C(S_0,t_0;K,T)\) to the gamma it has over its life:


1. Initial call premium is profit from gamma scalping at the initial stock price:

\( C(S_0,t_0;K,T) = \int_{t_0}^T \frac{\sigma^2}{2} S_0^2 C_{ss}(S_0,t;K,T)dt. \)​
2. Initial call premium is initial expected value of the profit from gamma scalping at all future dates:

\( C(S_0,t_0;K,T) = E^{\mathbf{Q}}_0 \int_{t_0}^T \frac{\sigma^2}{2} S_t^2 C_{ss}(S_t,t;K,T)dt. \)​

3. Initial call premium is cost of starting the hedge, rehedging continuously over time, and winding up at expiry:

\( C(S_0,t_0;K,T) = S_0 C_s(S_0,t_0;K,T) + \int_{t_0}^T S_t C_{ss}(S_t,t;K,T) dS_t - K 1(S_T > K). \)​

Two of the 3 equations are correct and one of them is wrong. Which one is wrong?
 

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I don't quite understand how you decide whether you are a quant or a trader from this quiz...
Anyhow, the first one is correct and follows from the B-S PDE and the fact that an option is initially out-of-the-money.
Third one seems to be correct as well (with probability one) and represents the
delta hedging strategy.
Not sure about where the second one comes from though...
 
Heh. I came up with 2 and saw that the correct answers were 1 and 3. I tried to figure out why I got it wrong and then I realized, I picked the one that was wrong - you posted the ones that were right.

It seems I can do the harder things, but when it comes to adding 1 and 1, I try and take the eigenvalue. :\
 
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