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1. Let me first reconcile the Black-Scholes pricing formula with the idea of prices being determined by supply-and-demand. Even if it is not explicitly said this way, from an equilibrium perspective, the Black-Scholes formula defines the unique price of risk that is consistent with the absence of arbitrage. In fact, you explicitly use this price when you ...

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If a structured product trader can directly hedge in the market, he will usually do it. Here your example is a little too simplified, many structured products have features that cannot be easily hedged in the market because they are path dependant (barriers), or illiquid (typically a 5 years 60% put on a single stock). Let's take your product, but imagine ...

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It is not a contradiction, we are looking at two different phenomena: The Vol Smile is about a comparison on two call options $C_1$ and $C_2$ at a point in time: S is the same for both options (and does not change!), but $C_1$ has strike $K_1$ and $C_2$ has strike $K_2$. To fix ideas let's say $K_2 > K_1$. Then: $$\Delta_2 < \Delta_1$$ and $$IV_2&... 3 What not to do What you are asking us, without knowing, is related to how to price a variance swap. Well, under a general diffusion process, variance swaps can be priced by forming a suitably weighted portfolio of options over a continuum of strike prices with the entire portfolio maturing on a given date. The intuition is that your exposure to volatility ... 3 You simply take limits. Recall that in the Black-Scholes world$$d_1=\frac{\ln\left(\frac{S_t}{K}\right)+\left(r-q+\frac{1}{2}\sigma^2\right)(T-t)}{\sigma\sqrt{T-t}}.$$As t\to T , we have d_1\to\begin{cases} \infty & \text{if } S_t> K \\ 0 & \text{if } S_t=K \\-\infty & \text{if } S_t<K \end{cases}. Thus, \Delta=\Phi(d_1)e^{-q(T-t)}... 3 The simplest long vol strategy is to be long an ATM straddle and delta hedge it, the problem is that when it is no longer ATM the exposure to vol weakens. You could then sell that straddle and enter another ATM one. Another solution is the vol swap or variance swap mentioned by Stephane below. It gives constant exposure no matter what the level of S&P. ... 2 Breezing through the referenced paper, the point of it seems to be to develop Ito calculus for non-anticipative functionals$$ F(t, X_t),$$where X_t := \left\{X(u)\mid 0\leq u\leq t \right\} and \left(X(u)\right)_{u\geq 0} is a stochastic process. For example, for$$ F(t,X_t)=t^{-1}\int_0^t X(u)du.$$I think that, in that introductive paragraph, the ... 2 This is a good question. See my answer to a question here The point is that under Black-Scholes (and also many SV models) not only European prices but also American options prices are homogeneous of degree 1 in strike and spot as the optimal exercise time does not affect the homogeneity property in strike and spot price. Hence also for American options ... 2 Your question makes perfect sense; one has to define volatility. Volatility can be used interchangeably for a number of different metrics. Realized volatility - the observed volatility of the underlying asset (and btw, there are many quite different ways of measuring it). Implied volatility - the number you get when you run your option pricer in reverse. ... 2 I do not mean to discourage you, but it sounds like you're a wee bit late for this round of volatility games, for two reasons: You are still trying to figure out how to implement a long vol strategy. The market has already priced the risk in, i.e. buying volatility is already expensive. However, never too late to learn and prepare for a next time. My ... 1 For derivatives, there usually exists some agreement between the counterparties on how to handle corporate actions. The same for exchange traded contracts. EUREX, for example, has some pieces on the treatment of corporate actions and also mergers, here: https://www.eurexchange.com/exchange-en/products/equ/corporate-actions-procedures 1 The only important thing with all the parameters you use for pricing is that they all refer to the same frequency. If in your data, you work with daily returns, daily interest rates and days to maturity, your estimates will reflect daily values. It's usually what people do. 1 Just to add to @KeSchn's answer: At T the option has price (S_T - K)_+. This is non-differentiable at S_T = K. Hence the delta is 1 when S_T > K, 0 when S_T <K and not defined when S_T = K. That is not a problem since the probability that S_T = K is 0 almost surely. The limit behaviour, i.e. when t \rightarrow T, is as in KeSchn's ... 1 If market makers are long options (either puts or calls) , the rebalancing trades that they do tend to limit the market movement, since they are selling when markets rise and buying when markets fall. If market makers are short options (either puts or calls ), then the rebalancing trades that they do tend to exacerbate market movements , since they are ... 1 Effectively, your example describes a market where your market makers are not ‚atomistic‘ anymore with respect to their influence on (underlying) prices. On the other hand, MM not only employ the spot market for hedging, but also the futures market, where an increase in a ‚position‘ does not, by itself, imply an upwards pressure on the spot markets. 1 I believe you are applying the cap formula to value the floor. From the link you sent, try this:$$floorlet = D [(K-F)N(-d) + \sigma \sqrt{T} n(d)] $$Where the d will be the same. 1 Rebonato called the whole process "the wrong number in the wrong formula to get the right price". We use Black-Scholes much the same way that we look at price-earnings ratio in equities. This is beneficial for traders. First, it translates a fast-moving price into a slow moving valuation metric. Second, it gives us an idea of value. Is this option rich or ... 1 OK, here is what I think. (But you should ask for advice from others in this forum or elsewhere). You computed \frac{dC}{dK} (the dual delta) by a discrete approximation. The result is negative and this is correct (it is negative for a Call and Positive for a Put). In the case of a European Call it is given by the formula -e^{-r T}N(d_2). (See here for ... 1 In the following, I am assuming the BS73 model and I assume that "ATM" means$$ S = Xe^{-r\tau} $$The pricing formula for a European call then becomes$$ \tag{1} O\propto N\left(+\frac{1}{2}\sigma\sqrt{\tau}\right)-N\left(-\frac{1}{2}\sigma\sqrt{\tau}\right) $$times some scaling factor which is irrelevant for our purpose. Clearly,$$ Vega\equiv\frac{\...

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Based on your computation, you can observe that the $N’$ term is always positive, between 0 and 0.4. As $\sigma$ is always positive, you can focus on the $-d_2$ term. When $d_2 > 0$, i.e. call is ITM, delta has a negative sensitivity to volatility ; conversely for OTM call. That is in line with your remark.

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For shorting a stock what you would do is to have a margin account with your broker. As an example, in the American jurisdiction, according to Regulation T from the Federal Reserve, you would provide a 150% of the value of your position as initial margin (50% of additional marging). And the daily margining would be done against your margin account (both ...

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Each path is evolved based on the vol and a random number. The higher the vol the more the paths will diverge. Paths will diverge if you increase time as well. The solution is to increase the number of paths as vol or time increases to get a standard deviation of terminal values that you are comfortable with.

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In the simplest case, he books the structured product as a sale of the call at 5, he then enters the market and buys a call at 3, so he pockets 2 on the trade.

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There are a few reasons for this: You can use grid methods for simple trades like barriers. But when you have trades which are not only dependent on the terminal distribution i.e. have payoffs which are path dependent like for instance fadeout, they can be only properly priced by an MC model because you would need to model vol of vol as well. When there are ...

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The simple explanation is that in the absence of calendar spread arbitrage, we should observe monotonic option prices with respect to maturity. And option prices are monotonic with respect to increase in volatility. Let $(X_t)_{t \geq 0}$ be a martingale, $L>0$ and $0\leq t_1, t_2$, then we have $$E[(X_{t_{2}} - L)^{+}] \geq E[(X_{t_{1}}-L)^{+}]$$ for ...

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