New answers tagged

1

I am gonna start by attempting to answer question 1, and I will re-use some of the arguments provided in this answer : Compute the price of a derivative which pays $\log(S_T)S_T$ in the Black Scholes world (but I will try to make it fit into the current question in a step-by-step easy to understand guide) I will add an answer to question 2 later on. Part 1: ...


2

As you said, the key is a numéraire change. We change from using the locally risk-free bank account, $B_t=e^{rt}$ as numéraire (with probability measure $\mathbb Q^0$) to the stock price, $S_t$ as numéraire. Thus, the new meaure, $\mathbb Q^1$, is defined via \begin{align*} \frac{\mathrm{d}\mathbb Q^1}{\mathrm d\mathbb Q^0} = \frac{S_T}{S_0}\frac{B_0}{B_T}=\...


0

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 ...


5

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 ...


1

Best-of + Worst-of = Call1 + Call2 The right hand side is independent of correlation (and you can check it in your model). Therefore if Best-of is short correlation, worst-of must be long correlation. Increasing correlation makes the two assets more similar and therefore makes the best-of more like a vanilla. This is why a best-of is short correlation. (Hope ...


3

Intuitively, they should both be short correlation, that is the less correlated the assets are the higher the value of the worst of/best of option. The best of option payoff is sandwiched by an exchange option payoff (plus other vanilla forward/option payoffs on single stock, insensitive to correlation): $$ X_T -K + (Y_T-X_T)^+ \leq \max(X_T - K ,Y_T - K,0) \...


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.


2

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

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 ...


5

Just a small addendum to @noob2's answer. The discrete shape of $\lambda$ is: $$\lambda \approx \frac{V_1 - V_0}{S_1 - S_0} \times \frac{S_0}{V_0} $$ which can be rewritten as $$ \lambda \approx \frac{\frac{V_1 - V_0}{V_0}}{\frac{S_1 - S_0}{S_0}} $$ which is as @noob2 said just the ratio of the relative returns for option and stock.


9

A better, clearer, answer is to compute Lambda (leverage) of the option (link) and see if it is bigger or smaller than 1. Lambda is $\Delta \frac{S}{V}$ so we test $$\Delta \frac{S}{V} \lessgtr 1$$ which is what Joshi is saying in words: compare $\frac{1}{S}$ (delta to price for the stock, the delta of the stock is 1) to $\frac{\Delta}{V}$ (delta to price ...


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 \...


1

You are right to be concerned about this. The liquidity of many options is not as good as many people imagine. I have personally been in situations where a fund needed to liquidate a large option position and it was difficult to do. Once market makers detect a liquidity-crisis situation they pull back and the situation worsens. The liquidity of option ...


2

As has been said in the comments, unless you are working with an asset class that has a second dimension, i.e, swaptions where you have not only the option expiry but also the underlying tenor, a surface would suffice. In the swaptions case you can either have a surface for ATM options or a given strike, or you will need a cube (3 dimensions): option expiry,...


0

Without doing any calcs, I would guess it's because of the 2 days passed and the decrease in vol. Besides the more commonly known greeks (delta, gamma, vega), you have some other interesting second order greeks and the one that will explain what you are observing is probably the delta decay (Charm). Intuitively think of it like this: your put is in the money ...


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 ...


2

(I assume that by "260 stocks" you mean 260 shares of same corporation that are also the underlying of the options.) Since the payoff of the options is non-linear, you can't get a meaningful VaR by multiplying the delta by the volatility of the underlying stock. You can't even get a meaningful VaR by including gamma or higher-order terms of a ...


0

Yes and no: diversification always decreases the effects of surprise events (unless they hit the whole portfolio equally) and as such the overall risk of extreme moves. That is true for option portfolios in the same way as it is for stock. But there is a key difference: stocks statistically increase in value so diversification in a stock portfolio decreases ...


1

Bid Ask spreads should reflect the willingness of parties to exchange at a certain price, where market makers are the sellers it represents the risks they are prepared to take in order to make the the market, but as most trades now are through secondary sellers, not market makers or in dark pools, it would be impossible to estimate the initial risk of a ...


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

Even if you assume null cokurtosis terms, your equality is still off: \begin{align} \operatorname{Kurt}[X+Y] = {1 \over \sigma_{X+Y}^4} \big( & \sigma_X^4\operatorname{Kurt}[X] + \sigma_Y^4\operatorname{Kurt}[Y] \big). \end{align} Note that you need $\sigma_{X+Y}^2$. You already have $\sigma_X^2$ and $\sigma_Y^2$ (computed in the paper). Full formula is:...


2

I’m not an expert on Inflation derivatives, so I will just give you an explanation on why your finder doesn’t yield any root. In the Black & Scholes framework, it holds for the price of European Put options: $$P_{B S}(\sigma=0, T, K, S)=\left(K e^{-r(T-t)}-S\right)^{+},$$ $$P_{B S}(\sigma=\infty, T, K, S)=K e^{-r(T-t)}.$$ Given the parameters you ...


0

At time $T$, standard floorlet pays: $$ N\tau \left[\kappa - (I(T)I(S)^{-1} -1)\right]^+$$ with $N$ notional, $\kappa$ strike, $S < T$, and $\tau$ day count fraction. Standard floor is simply a strip of floorlets sharing a common strike paying at each $T_i$, $i=1,...,M$: $$ N\tau_i \left[\kappa - (I(T_i)I(T_{i-1})^{-1} -1)\right]^+$$ Your payoff is for a ...


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 ...


1

The bisection method, Brent's method, and other algorithms should work well. But here is a very recent paper that gives an explicit representation of IV in terms of call prices through (Dirac) delta sequences: Cui et al. (2020) - A closed-form model-free implied volatility formula through delta sequences


1

I tried to answer a similar question recently here: How can we estimate new stock price after a large purchase? Whilst it might not be exactly what you might be looking for, it might help you understand the problem in a practical context.


5

Actually the recent work by Koijen and Gabaix tries to tackle precisely a similar question. They do not have a paper yet. Here's the link: https://www.youtube.com/watch?v=apCxV8zxoOc They find that buying 1% of the market increases prices by 5-12%. I.e. markets are inelastic. So if this is true for the market, I am assuming for individual stocks would be ...


2

In terms of the settings, we know the current stock price, we have assumed that the stock price dynamics follow Geometric Brownian motion (GBM), we know the parameters of this process (volatility etc), and we know the characteristics of the options (option type, maturity). In practice, we know the current price of the option as well, but we pretend we don’t, ...


2

Hence how would we know the value of V(S,T)=max(S−T,0),for S∈R+? You know the value at time T as a function of S: it is simply the pay out, which is $\max(S-K,0)$, where $K$ is a strike. Moreover, why do we care to solve for V(S,t),for t<T if a European option may only be exercised at the maturity time t=T? No, we're not interested in value at time T. ...


0

How about using the ATM option expiring closest to your futures contract delivery date? As in : Stock price : price today Futures price : price at a future date, usually the nearest expiring contract. Option price : price ATM at a comparable future date.


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

Without the math, look at it this way: I give you a die to toss. You can toss it thrice and take the payoff as the number on the die. On each turn you can either accept the payoff or move on. At the last throw, you must accept the payoff. Your logic would state that this product has value equivalent to tossing a die once and only once (here, all 'Europeans'...


1

Ok as an example consider a 1yr-10yr 3pct Bermudan payer (the right to pay fixed at 3pct vs libor starting at any annual date from 1yr onwards with a maturity of 11yrs from today). For simplicity assume a flat yield curve. If rates are 1pct, the probability of exercise on the first date is low (a long way to cross the 3pct strike). If rates are 6pct, the ...


1

Given that liquidity of NIFTY options decreases rather quickly with moneyness, using the most liquid ATM option is your best bet (i.e. the least bid ask spread). Although, keep it consistent in that you use the same option (i.e. say the one that expiries 3M into the future) for different days. Also, better to use implied volatility as a performance measure ...


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.


3

With $V$ American option value, $H$ holding (aka continuation) value, and $B$ bank account value, we have: $$V_N(S_N) = (K-S_N)^+$$ and for $i$ backwards from $N-1$ down to 0, we have: $$ H_i(S_i) = \mathbf{E}\left[B_{i}B_{i+1}^{-1}V_{i+1}(S_{i+1})|S_i\right]$$ $$ V_i(S_i) = \max (K-S_i, H_i(S_i)) $$ The algorithm result is $V_0(S_0)$. The different ...


2

At the minimum, the difference is the hedging instrument's market cost, liquidity (especially for dynamic hedging), funding costs, and counterparty risk costs (e.g., exchange traded futures vs. OTC swaps). Options in particular can backfire possibly complicating the vega and gamma picture of the overall portfolio (main and hedge positions), if care is not ...


1

Definelty barrier shift is necessary when hedging PDI and CUI, thats because you always want to hedge your barrier as it was no hitted because you would experience enormous gamma in that region. So you shift the barrier and in this way you will never hit the barrier in your hedge


2

As the comments suggest, there is no "objectively" optimal strategy, and it depends on your goals and risk tolerance. I personally like to sell covered calls on a weekly schedule, for calls that expire the week after. Of course, the collected credit is lower than with expiration dates further in the future, but it gives me more flexibility. I can let them ...


2

As you point out in your d1 formula: $$d_1 = \frac{ln \left( \frac{S}{K} \right)+\left(r+0.5\sigma^2 \right)T}{\sigma \sqrt{T}} $$ Therefore, $N(d_1)$ (where $N(.)$ stands for the Standard Normal CDF) is only equal to half when $d_1$ is exactly zero. When an option is ATM, then $S=Ke^{-rT}$. So $N(d_1)$ won't be exactly 0.5, because: $$d_1 = 0.5\sigma\...


4

The delta you mentioned is the Black-Scholes delta. If you let $r=0$, $T=1$ and solve the equation $d_1=0$, you get what is in the article.


6

The Stock price after the single period can be 1.05 or 0.95. If the stock ends up at 1.05, the option pay-off is 0.05, if the stock price ends up at 0.95 the option pay-off is zero. We want to figure out the price of the option by replicating it with the underlying Stock and a Bond (rates are zero, so the bond price at time zero is 1 and after the single ...


2

Let $S_t$ be the exchange rate from one unit foreign currency FGN to units of domestic currency DOM. Note that, for maturity $T$ and strike $K$, \begin{align*} \max(S_T-K, \, 0) = S_TK\max\Big(\frac{1}{K}-\frac{1}{S_T}, \, 0\Big). \end{align*} Moreover, let $B^d_t$ and $B^f_t$ be the respective domestic and foreign money-market account values at time $t$, ...


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