New answers tagged

0

I'm not too sure if I interpret the question correctly, but I am inclined to say that as the call is long gamma, 'jumps' (second order moves) would always result in higher value in the delta hedged portfolio, and therefore should be built into the price of the option. So the call should be more expensive if it has jumps.


1

Following this answer, let $\mathbb Q$ be the probability measure associated to the risk-free bank account as numeraire and $\mathbb Q^1$ the probability measure associated to the stock as numeraire. You know that the standard equation $\mathrm{d}S_t=rS_t\mathrm{d}t+\sigma S_t\mathrm{d}W_t^\mathbb{Q}$ can be written as $\mathrm{d}S_t=(r+\sigma^2)S_t\mathrm{d}...


4

He circled the volume of the option in question. It was roughly 50,000 and one option contract is usually for 100 shares. Price was 0.13. $50k*100*0.13 = 650k$


0

An easy way to check if you've made a mistake during a longish calculation like your derivation of the skew for a straddle is to numerically evaluate the original integral. It would be a one-liner in Mathematica.


3

Monte Carlo is more natural to perform a forward induction (think TARNs), whereas trees are more natural to do a backward induction/dynamic programming (think Bermudans). Forward induction may be the way to go in case you have a trade that is path dependent (i.e. The price at a time depends on the past history of the process (in the sense that at a ...


2

Consider a vanilla european option. The optimal strategy in the risk neutral measure is to exercise if $S(T)>K$. (because in the risk neutral world, I value my payoff at its expectation, which at that point is the payoff itself. At time T, if $S(T)-K$ is positive, I will take it over not exercising, which is a payoff of 0. This is of course also optimal ...


2

Question 1 is answered in parts 1 through to 6: the idea is that each part slowly builds the tools required to derive the process equation for $S_t$ under the $S_t$ Numeraire. Question 2 & Question 3 are then answered in part 7. Part 1: Expectation of a function of a Random variable: Let $X(t)$ be some generic Random Variable with probability density ...


5

I provide a solution in three steps. The first step carefully outlines how to split up the expectation and what new measures are used. This first step does not require any special model assumption and holds in a very general framework. I derive a formula for the option price that resembles the standard Black-Scholes formula. In a second step, I assume that ...


2

Black scholes formula based on $S_t$ measure , theory, and formulas you mention are derived in detail in "Steven Shreve: Stochastic Calculus and Finance" draft pdf from 1997 , page 328 "stock price as numeraire".


4

the Ask Price of a Call Option is always higher than the difference between the Strike Price and the price of underlying stock [more precisely, the price of the underlying minus the strike]. This is definitely true for American call options, that can be exercised immediately (as well as all the way to maturity): roughly speaking, you can "indirectly&...


3

When the call is at or out of the money, the result is obvious: the call will have some value but the difference between stock price and strike is nonpositive. Consider the case that the option is in the money and its current price $C$ is lower than the difference between stock price $S$ and the strike $K$, in symbols: $C < S-K$. If I own the stock, I ...


6

The call for the stock that can jump downward will be more valuable due to put-call parity. Suppose you have two stocks, both with a price of $100 and the same diffusive volatility. Stock A does not jump, whereas stock B can at some random time jump (for example) to zero. Clearly a put on stock B will be worth more, but the call must therefore also be worth ...


2

You can check out those discussions in Merton paper when introducing jumps "Option pricing when underlying stock returns are discontinuous". In the very last part he discusses the influence of considering jumps compared to the usual Black Sholes model. From what i remember it'sall about considering your option is ATM or not , that will usually make ...


7

@Jan Stuller already pointed to Rho, an option's sensitivity to changes in the risk-free rate. This number is indeed very low indicating that a non-flat term structure may not dramatically misprice equity options. A somewhat dated (but worth reading) empirical analysis was conducted by Bakshi, Cao, and Chen (1997, JF). They investigate how the Black-Scholes ...


4

I think the main reason is that $\rho$ for Equity options is much less significant that the other Greeks: so going into the length of modelling stochastic rates for equity options isn't worth it. I would be keen to hear what other's have to say if there are other major reasons.


3

It's just Girsanov's theorem. I suppose that under the risk neutral measure Q $$dS_{t}= r S_{t} dt + \sigma S_{t}dW_{t},$$ $$S_{t} = S_{0}\exp\left((r-\frac{\sigma^{2}}{2})T + \sigma W_{T}\right)$$ By multiplying by $e^{-rT}$ I have $e^{-rT}S_{T}$ which is a martingale so that I can change my measure under $Q$ to some equivalent probabilty $Q_{1}$ under ...


3

Part 1: deriving the drift of the stock price process under the stock Numeraire. Under the risk-neutral measure, the process for $S_t$ is as follows: $$ S_t = S_0 + \int_{h=t_0}^{h=t}rS_h dh + \int_{h=t_0}^{h=t}\sigma S_h dW_h = \\ = S_0exp\left[ (r-0.5 \sigma^2)t+\sigma W(t) \right] $$ In the above model, the Numeraire is $N(t)=e^{rt}$ with $N(t_0):=1$. ...


3

I believe the other answers are nearly exhaustive; but here's a bit of intuition I'd like to add: Think of the decision (= equilibrium price) of a market as: Decision = f(probabilities, risk aversion) where probabilities are the chances of various events happening, and risk aversion is the taste preference of the market. Now it turns out that the 'iso-curve' ...


3

If you put some numbers into down-in/out barrier call option formulae that can be found in many books, you will see that the down-in curve is not symmetric. It just looks like it in that plot. Below the barrier, the prices are obviously just Black-Scholes values, as the spot price goes higher the chance of it going below the barrier is obvious becoming ...


2

Intuitively, underlying call keeps losing value as the spot goes down, but the barrier option value (which starts at almost nothing for high spot) keeps growing as the spot approaches the barrier level (the chance to get something, even if it's an out-of-money call, is growing). When the spot hits the barrier level, the value of the call is still ok (unless ...


4

There's a lot left unspecified in this question, since it is stated without precision, but the effective idea of the answer given here is that those jumps introduce extra variation into the forward distribution of the underlying. And such variation is the bread-and-butter of option value. That said, the ambiguity in the question leaves room for other ...


1

If $X$ is FOR-DOM exchange rate (asset always on the left, numeraire on the right), then its dynamics in the QUANTO currency measure (currency different from FOR and DOM; let $Y$ be the DOM-QUANTO exchange rate) is: $$ dX/X = \left(r_{\rm DOM}-r_{\rm FOR}-\rho_{XY}\sigma_X \sigma_Y \right) dt + \sigma_X dW $$ Terminal distribution: $$ \ln (X_T/X_0) \sim \...


2

As long as you are able to generate a joint terminal distribution, any model will do the job. Copula is only one such approach. Now, in theory, you cannot completely statically replicate this payoff in general. To see this, know that all you have is vanillas, and the most you can do is imply the marginal distribution from them (the usual risk neutral density)...


3

The three ways to manufacture pseudo-implied vols I know of are: Find a related underlying and, even if only few options trade on it, 'borrow' its implied vols. Compute statistical vol from historical underlying prices (not strike dependent, still useful to know). Compute breakeven vol, still based on historical underlying prices, strike dependent, by ...


4

Great answer given by KeSchn above. I would like to add an additional perspective. My experience with and my understanding of the Risk Neutral measure is entirely based on "no arbitrage" and "replication / hedging" arguments. The way I would like to explain this view is via the following three-step construction: (i) First, I want to build ...


17

Life Without a Risk-Neutral Measure How would we price assets without the measure $\mathbb Q$? Well, we would start with some version of the Euler equation $P_t=\mathbb{E}_t[M_{t+1}P_{t+1}]$, where $M$ is the stochastic discount factor (SDF). This equation holds under very weak assumptions (law of one price) and uses real-world probabilities. So, we take the ...


2

Below, Gamma is denoted by $\Gamma$, and $IV=\sigma$: $$ \Gamma = \left(\frac{1}{S_0*\sigma*\sqrt{T}}\right)*\left(\frac{1}{\sqrt{2\pi}}e^{\frac{-d_1^2}{2}} \right) $$ The expression in the first bracket is inversely proportional to $\sigma$, so isolating just this first expression: higher $\sigma$ will trivially lower this first expression. The second ...


2

I'm not sure I understand the question, but I'll give it a try anyway. The mean and variance specified for the terminal distribution $S_T$ are dependent on current asset price, $S_0$, and implied volatility, $\sigma_i$ (which needs to come from the market via hopefully same pricer that one uses). The expectation of a payoff, function $f(S_T)$, is hence a ...


4

Give QuantLib a try: import QuantLib as ql today = ql.Settings.instance().evaluationDate averageType = ql.Average.Geometric option_type = ql.Option.Call strike = 120.0 exerciseDate = ql.TARGET().advance(today, 90, ql.Days) payoff = ql.PlainVanillaPayoff(option_type, strike) exercise = ql.EuropeanExercise(exerciseDate) option = ql....


4

First, the error is because you should input the cds_vol as a quote. So instead of cds_col use ql.QuoteHandle(ql.SimpleQuote(cds_vol)) Apart from that the .setPricingEngine() method will affect the cds_option object directly, so you should use it as: cds_option.setPricingEngine((ql.BlackCdsOptionEngine(probability, recovery_rate, risk_free_rate, ql....


0

A model 'understands' the price of risks that are assumed to exist. For example, the Black-Scholes model undertands the cost of delta-hedging, but not of vega-hedging. Hence we have stochastic volatility models: these understand the cost of delta-hedging and volatility hedging. However, none of these models take into account transaction costs. Hence you ...


2

In a put back-spread, you are short one put and long two puts with a lower strike. As far as Vega (the spread's price sensitivity to IV) is concerned, it depends on where you enter the spread. Usually, you'd enter the spread where the short put is at the money or near being at the money: this ATM put will have a high Vega, whist the two puts with lower ...


0

The price of an American option is the Bermuda option in the pointwise limit in $S$ as the maximal exercising interval approaches zero. See the proof in this answer. The Bermuda option at any exercising time can be evaluated inductively via the dynamic programming principle as the maximum of the payoff and the risk-neutral expected value, i.e, the European ...


3

For Fourier methods, you always need the characteristic function of the log-asset price $\ln(S_t)$. In the Black-Scholes model, $\ln(S_t)\sim N\left(\ln(S_0)+\left(r-\delta-\frac{1}{2}\sigma^2\right)t,\sigma^2t\right)$. It is well-known that the characteristic function of $X\sim N(m,s^2)$ is given by $$\phi_X(u)=\exp\left(imu-\frac{1}{2}s^2u^2\right).$$ You ...


1

If the underlying is driftless (think futures) and the value of the option is not discounted (think future style options with daily bilateral variation margin or CSA's with zero collateral interest rates) then the value of an american put and a european put would be the same by Jensen's inequality.


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