# Tag Info

10

What you have to do is to show that the dollar gamma satisfies the Black-Scholes PDE. Using Feynman-Kac it then follows that the dollar gamma is an expectation of a "payoff", just like the Black-Scholes claim price is an expectation of a payoff. And if something is the expectation of a payoff then it's a martingale. I'll leave the above for you to carry out....

8

By delta hedging you are saying that you have a view on the path and the volatility of the option you are trading, but not on its direction; in your case, that being short delta. From a theoretical perspective, all options are priced fairly and not delta hedging simply increase the variance of your payouts. In your example, selling a call and delta ...

7

$C= S_0 N(d_1) - K e^{-rT} N(d_2)$ $C$, $S_0$ and $K$ have units of currency (e.g. USD). $N(d1)$ and $N(d_2)$ are unit-less (dimensionless), the formula is dimensionally correct. Considering, $d1 = \frac {ln{\frac {S_0} K} + r T + \frac {\sigma^2} {2} T} {\sigma \sqrt T }$ $r$ and $\sigma^2$ have units of "per year", as they are stated on an annualized ...

6

The conjecture is true when the interest rate is zero. Note that, from this question, under the Black-Scholes model, \begin{align*} \Gamma(t,S_t) &= \frac{N'(d_1(t))}{S_t \sigma \sqrt{T-t}}\\ Vega(t,S_t) &= S_tN'(d_1(t)) \sqrt{T-t}, \end{align*} where \begin{align*} d_1(t) = \frac{\ln \frac{S_t}{K} + \big(r+\frac{1}{2}\sigma^2\big)(T-t)}{\sigma \...

4

As @Alex C mentioned, this is how banks make money. They will sell/buy you an instrument with a markup and hedge their position. Over the life of the trade, they will adjust the hedge. If the costs of hedging over the life of the trade are less than the markup, they will profit. Of course the banks, as market makers, will have a portfolio of positions ...

3

@Gordon has already given the answer but here is a little more notes to it... At time time $T_2$ the holder receives $X=(S_{T_1}-K)^+$. According to Risk Neutral Valuation the value at time $t$ $(t<T_1<T_2)$ is $$V_t = e^{-r(T_2-t)}E_t[(S_{T_1}-K)^+] = \\ e^{-r(T_2-t+T_1-T_1)}E_t[(S_{T_1}-K)^+]=\\ e^{-r(T_2-T_1)}e^{-r(T_1-t)}E_t[(S_{T_1}-K)^+]$$ $e^... 2 The choice of hedging strategy cannot affect the expected p/l, because hedging just consists of doing at-market purchases or sales of the underlying, each of which have zero expected value at the time of transacting. 2 The theta PnL here is the option price paid (for the time-value of the option); it is just a greek word for it with an extra feature showing how the option premium continously declines with the passage of time. Say that you buy an out of the money option and then the market just dies. You then get noting but theta losses. They will add up to the premium you ... 1 As you write, a bear put spread is a combination of going long a European put with a higher strike (here, 175) and short a European put with a lower strike (here, 150). Given the remaining parameters in Black-Scholes, i.e. the current underlying asset price$S$, the risk-free rate$r$, the volatility$\sigma$and time to maturity$T\$, you can simply value ...

1

If you perfectly hedge (infinitesimal moves), theta will offset gamma but if you do periodic hedges for finite moves, you would have gamma slippage and then you end up in a distribution of Pnl around zero.

1

An important assumption in BS is that you have to do instantaneous hedging, i.e, an infinitesimal move. In reality, you can't. Therefore you won't recover option price -- instead your price pnl minus the option price will be +- around zero.

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