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For any process with independent increments, by the very fact of statistical independence the variance of $x_{t3}-x_{t1}$ is going to be the sum of the variances of $x_{t2}-x_{t1}$ and $x_{t3}-x_{t2}$ for $t1\leq t2 \leq t3$. Many processes have independent increments, including ABM, GBM, Poisson, etc. Then if you add a homogeneity assumption (the ...

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I will attempt to summarise the content included in this book, which has a specific chapter dealing with carry and roll-down. There, two concepts are made completely separate. Costs-of-carry are defined as costs relating to holding a trade that are not directly related to market movements. For example, funding a margin requirement for an IRS facing a ...

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It turns out that the two things are the same, appropriately scaled. Proof: we can construct a 5 year swap using 3 month libor combined with a 3mo-4.75yr forward swap, weighted by the dv01s of each part. Thus, ignoring discounting, we have 5yr swap rate = (0.25*3mo libor + 4.75*forward rate)/5. This can be rewritten as 0.25*(5yr swap rate - ...

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If a european option value becomes lower than intrinsic value it gets negative time value. In this circumstance the theta becomes positive because as time approaches to expiry the option value has to converge to intrinsic value. For european options there are 2 circumstances that can lead to the option value being lower than intrinsic value deep ITM puts ...

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Theta on a European Put option on a non-dividend paying stock is: $$\Theta=-\frac{S_t \sigma}{2\sqrt{\tau}}N'(d_1)+rKe^{-r\tau}N(-d_2)$$ For deep in-the-money Puts, $d_1$ and $d_2$ go to negative infinity: consequently, the term $N'(d_1)$ goes to zero, whilst the term $N(-d_2)$ goes to 1. Therefore, deep ITM puts can have a positive Theta, with a limit ...

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With a long time to maturity, your options have a low theta because their time value decays quite slowly. If there are many months to go, the passage of one day does not change the exercise probabilities too much, whereas short life options with only a few days left have a much higher time value decay. Hence, the larger the time to maturity, the lower theta. ...

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if you have a portfolio of calls and puts with the same maturity then your portfolio is gamma neutral if and only if it is vega neutral. The reasons is that the BS gamma divided by the BS vega is a function of $S$ and $T$ that does not vary with $K.$ So if you construct a linear combination that has zero gamma then the vega is zero too, and vice versa.

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It's hard to be sure without seeing the inputs, but I'm guessing that the implied curve changes shape because the original curve does (which you can see from your output: except for the 1-year and 5-years points, the actual discounts are different). The reason the original curve changes is probably the different position of weekends or holidays (so that, ...

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This is possible if the option is long-dated and interest rates are high enough. For example, a five-year put struck at \$90 where the spot is \$100 (so it is in the money with respect to the spot price) with implied volatility 20% and interest rates 10% has a theta of \$0.19. 4 That is quite possible. You have negative time value and a positive theta if the option price is below the intrinsic value. Look at deep ITM put options, the stock price is basically so low, the chance of it rising is negligible and the option price is the discounted payoff. This has a positive theta since the longer the time of maturity, the lower the ... 4 Think of this in terms of Taylor series. Let's say the option price today is$C\left(S,t\right)$where S is the underlying price and t time. Let's say the underlying price changes by$\Delta S$in a time interval$\Delta t$, so your P/L will be:$\mathrm{P/L}=C\left(S+\Delta S,t+\Delta t\right)-C\left(S,t\right) $Use Taylor series to first order in t and ... 4 I can argue your case as follows, consider a portfolio such that The value of$\Pi$of a portfolio satisfies the differential equation given by: $$\frac{\delta \Pi}{\delta t}+rS\frac{\delta \Pi}{\delta S}+\frac{1}{2}\sigma^{2}S^{2}\frac{\delta^{2}\Pi}{\delta S^{2}}=r\Pi$$ From the differential equation, $$\Theta=\frac{\delta \Pi}{\delta t}$$ $$\Delta=\frac{... 4 It’s just the effect of interest. If you are long a deep ITM European put, it is worth the PV of K minus the stock price. But one day later the PV of K has grown a bit. That’s it. It’s the opposite for calls because you have to pay the K, so bringing the date closer costs you money. This is all assuming interest rates are positive. 3 The relationship between theta and gamma is the Black-Scholes PDE. Let's take normal B-S dynamics with r=0: dS_t = \sigma S_t dW_t The pricing PDE for a derivative g(S_T) is (with terminal condition g): \frac{dp}{dt} = \frac{1}{2}\sigma^2S^2 \frac{d^2p}{dS^2} Or \Theta = \frac{1}{2}\sigma^2S^2 \Gamma This PDE has a solution (Feynman-Kac Theorem): ... 3 Well it all depends how theta is calculated in the first place. Depending on your pricing scheme those could be very different things. Anyways assuming that you are dealing with european vanilla then the BS theta is an instantaneous quantity that assumes that volatility does not change so you definetely don’t get any carry effect from this quantity. Now ... 3 If both options are out of the money, your higher strike put (of which you are short) has a higher theta than your lower strike put (of which you are long). Thus earn more theta than you lose. 3 No Because the P&L it generates is in O(dt^2). Ito's lemma tells you that you can ignore this P&L.$$PnL = \frac{\partial^2 V}{\partial t^2}dt^2 = 0$$3 I think your FX theta is probably not the same as the theta in black scholes sense. I think it may mean time pnl, which is applicable to all products, i.e., the PnL of time passing 1 day, but keeping all the market data the same. Note that here you have two markets, market 1 (original market) and market 2 (the original market's but moving 1 day forward) ... 3 The reason is that in many common models including geometric Brownian motion, the variance of the logarithmic returns is proportional to time. Thus, their standard deviation/volatility is proportional to the square root of time. Consider for example the class of Levy models where X is is the logarithmic stock price process such that S_t = S_0 e^{X_t}. ... 3 The value of an option is based on its intrinsic value plus its time value. Intrinsic value is simply based on, for example for a plain option, the strike price of the option and the underlying instrument’s spot price. Intrinsic value remains unchanged as the maturity is approached as long as the underlying instrument’s price remains unchanged. Time value, ... 2 To answer that question you first have to define what "no change other than the passage of time" means. So you could make one of the following "no change" assumptions. the shape of the term structure will remain unchanged. assumption of realized forwards. assumption that some other hypothesized scenario will realize. Based on one of those assumptions you ... 2 The value of a call option that is near ATM can be approximated as C(S,T)≈ 0.4 \sigma \sqrt T. Therefore, under the unrealistic assumption that S does not change very much (i.e. the option stays near the money) the value decays as the square root of the remaining time. In words, yes it does speed up considerably as you get close to expiration. 2 I don't understand why you think the numbers dont match up. In my opinion it all works out. Perhaps best if you first convert all numbers to percentages and for 1 underlying instead of 100 multiplier. From OVML you have multiplier = 1 troy ounce S = 1075 K = 1075 r = 0.0033 T = 2/12 sig = 0.12 Convert all into percentages: S = 100 K = 100 r = 0.0033 T ... 2 Disclaimer: I did not check your example, i.e., that "theta * 1 day" will predict a negative option price. Theta is the derivative with respect to time-to-maturity. It is the change of the option price with respect to an infinitessimal change in time and not with respect to a change of one day - even if the derivative is "scaled" towards a time scale having ... 2 There are two ways you can lose money: The actual volatility of the stock is less than the IV you assumed. For example (extreme case) let's say that the stock price does not move at all: you make no money at all on the gamma scalping and you lose on your theta. The gamma scalping counterbalances the theta only if the stock moves "enough". The stock price ... 2 If you are long gamma, your delta is increasing at an increasing rate. In order to delta hedge this position, you will be selling stock as the stock price goes up and buying stock as the stock price falls. An explanation of gamma trading can be found in my response to this question: What really is Gamma scalping? If you are long gamma, you are long ... 2 I dont think that people would usually use one as the substitute for the other, as: \theta/\Gamma=-\frac{S^{2}\sigma^{2}}{2} which is arrived upon by neglecting the terms of the formula for \theta, which are preceded by the interest rate r. I think the background to your question stems from the fact that option market practitioners will consider ... 1 The theta for puts and calls at the same strike should be the same, so it seems the SPX theta is somehow wrong. Edit: thanks @maxim, I see now what the issue is. I think the difference is coming from the fact that the options on the e-mini futures are using the Black formula where the futures price is held constant when calculating the theta. However ... 1 Say the value of your option on day t is V(S_t, \sigma_t, \tau) where S_t is the spot price, \sigma_t is the implied volatility and \tau is the number of days to expiry (it also depends on the strike price, interest rate etc but I've ignored these for simplicity). Your P&L from t to t+1 is$$ V(S_{t+1}, \sigma_{t+1}, \tau-1) - V(S_t,\... 1 First and foremost I do not agree with you Closed Form value. I get$\Theta=-8.963$. There are various of BS calculator you can use the check your results and in general you should do that. Here is one: https://goodcalculators.com/black-scholes-calculator/ Have in mind that maturity T is fixed then your forward FD problem should look like this:$\$ \Theta(T-...

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