# Tag Info

8

This is an interesting and not so easy question. Here's my 2 cents: First, you should distinguish between mathematical models for the dynamics of an underlying asset (Black-Scholes, Merton, Heston etc.) and numerical methods designed to calculate financial instruments' prices under given modelling assumptions (lattices, Fourier inversion techniques etc.). ...

6

the problem is that the pay-off has discontinuous first derivative. Try a contract with pay-off that is twice differentiable and it will probably work. The problem is that all the value comes from the tiny number of paths within $\Delta S$ of the strike, and these paths have huge value. This is a well-known problem. As the bump size goes to zero, the ...

5

Since the volatility is not changing, we can assume that the only change is the underlying asset price $S$. Then \begin{align*} C(S+\Delta) &\approx C(S) + Delta \times\Delta +\frac{1}{2} Gamma \times \Delta^2 \\ &=11.50 + 0.58 \times 0.5 + \frac{1}{2}\times 2 \times (0.5)^2\\ &=12.04. \end{align*}

4

Delta is a linear approximation of the change in price due to a small move of the relevant interest rate. Typically a parallel move of the whole interest curve is assumed here. This applies to all kind of fixed income instruments, in particular IRS. Interest rates can be given as coupon rates (these are the so called par rates, based on prices observable in ...

4

Since we are dealing with the change in the value of the call option in relation to the change in the value of the stock price, we are looking at the simple slope of the function. We do not need to know how much the stock price will change in order to estimate the slope. In the absence of gamma, the stock price could change 2% or 1% and this wouldn't ...

3

IMHO the 'definition' you mention is not a mathematical definition per se, but rather an approximation used by some practitioners. Mathematically, it is $N(d_2)$ in the BS formula which figures the conditional probability that the terminal asset price $S_T$ will finish above the strike level $X$ given the information we possess today (represented by $S_t$), ...

3

If you're asking what the FX Outright for 1M EUR/PLN is, given that table, then yes the answer is just outright = spot + fwd points, which is 3.4550 + 0.0079 = 3.4629 (you had the wrong column for your 1M value). Usually fwd points are quoted directly (i.e. not as an outright), using a divisor set by market convention. I expect EUR/PLN divisor to be 10,000, ...

3

Note that, at time $t$, \begin{align*} d_1(t) = \frac{\ln \frac{S_t}{K} + (r+\frac{\sigma^2}{2}) (T-t)}{\sigma \sqrt{T-t}}, \end{align*} which is a function of $S_t$, and then, it is a random quantity. Consequently, the delta $N(d_1(t))$ is also a random quantity. Note that, at time $t$, $N(d_1(t))$ is known. However, $\{N(d_1(t)) \mid t \ge 0\}$ is a ...

3

Options on interest rates futures in the listed markets are always traded 1-yield (100-yield) just like the futures which are traded 1-yield. So negative rates aren't an issue and its always black volatility. In the OTC market, both normal and black volatility are quoted, but the common practice is to use black volatility is what is way more frequently used....

3

It is false. here is an example. Let $$dS_t = rS_t dt + \sigma f(S_0) S_t dW_t.$$ $$dB_t = r dt.$$ The price is then the Black--Scholes price with volatility $f(S_0).$ The delta is the BS delta plus $$f'(S_0) \times \operatorname{BS Vega}.$$ Picking $f$ appropriately, we can make the Delta as big as we like. Note that example is highly artificial in ...

3

Given that by delta means that if the price goes up by 0.01% i.e. one basis point, you gain 15 and vice versa if the price goes down by one basis point. You know that the daily standard deviation is 2.2%, than again you know that $220*15 = 3300$ is the standard deviation of your portfolio. So, since we are using a normal distribution you can look at a table ...

3

Options have an asymmetric payoff profile: The payoffs are zero for almost all cases and positive else (as we well know). If the option is OTM, most of its payoffs are zero. A rise in volatility will hence increase the likelihood for instead positive payoffs from a change in the underlying price (i.e. delta increases). If the option is already ITM, most(...

3

The time to expiry is required, but it's included in the inputs: the two discounts $e^{-rT}$ and $e^{-qT}$ and the standard deviation $\sigma\sqrt{T}$. You might argue it could be documented more clearly, and I might agree with you.

2

Basically there are three steps to accomplish this. 1 - collect time series of options for several expirations and strikes. 2 - calculate implied volatility surface for every time period, and use model-based or model-free interpolation to create continuum of strikes / expirations. 3 - from the continuous surfaces you can calculate series of any specific ...

2

The strangle vol defined in your formula \begin{align*} Strangle(∆) = 0.5[Call Vol(∆) + Put Vol(∆)] - ATM Vol \end{align*} is the smile butterfly volatility. Then you have the volatility quote. Your confusion is caused by the misuse of notations. Note that, other treatments are also available. See for example, FX Volatility Smile Construction by UWe Wystup ...

1

If you compute the cumulative sum of the 'Shares Purchased' column you will find that in Week 9 the company owns a total of 78,700 shares. Each share is worth 53.00 (see 'Stock Price' column), so the value of the shares held in Week 9 is 78700*53 = 4,171,100. The increase in share value is 4,171,100-2,557,800 = 1,613,300 The loss in the option position is -...

1

Let \begin{align*} C(S, K, t) = SN(d_1) - e^{-rt}KN(d_2) \end{align*} denote the Black-Scholes call option price with initial asset value $S$, strike $K$, and maturity $t$. Note that \begin{align*} \frac{\partial C}{\partial S} = N(d_1). \end{align*} For the above barrier option, note that \begin{align*} E_0 &= V_0 N(d_1)-e^{-rt}KN(d_2) -\bigg[V_0 \Big(...

1

We assume that \begin{align*} dX_t &= X_t(rdt + \sigma_x dW^x_t)\\ dY_t &= Y_t\Big[rdt + \sigma_y\Big(\rho dW^x_t + \sqrt{1-\rho^2} dW^y_t \Big)\Big], \end{align*} where $\rho$ is the correlation, which we assume to be less than 1, and $W^x$ and $W^y$ are two independent standard Brownian motions. Then \begin{align*} d(X_t/Y_t) &= \frac{1}{Y_t}...

1

It seems fairly simple to demonstrate an example in discrete space where delta>1. Consider a 2 step binomial tree on a dividend free stock, with interest rates at zero. Let the initial stock price be 100, and each step on the tree have risk neutral p(up)=p(down)=0.5. Let the tree be as follows: 100-(101,99)-(200,2,99,99) meaning that on the first step it ...

1

This is not the Taylor expansion with respect to $t$, instead, it is the Taylor expansion with respect to $S$. Moreover, the prices at time $t+\delta t$ is used for approximation. That is, \begin{align*} \frac{\partial V}{\partial S}\big|_t &\approx \frac{V(uS, t) - V(vS, t)}{uS - vS}\\ &\approx \frac{V(uS, t+\delta t) - V(vS, t+\delta t)}{S(u-v)}\\ &...

1

The "right" thing to do is to treat the options as derivative contracts. Let's say for simplicity that you are using Monte Carlo to compute VaR. Then you would simulate the equity prices on each iteration, and then apply an option-pricing formula to get the corresponding option prices on that iteration. This lets you obtain an accurate simulated portfolio ...

1

The Black-Scholes delta: $$\partial_SC=N\left(\dfrac{\ln\left(\frac{S_0}{K}\right) +(r - q + \frac{1}{2}\sigma^2)(T - t)}{\sigma\sqrt{T - t}}\right)$$ As you can see this delta would go to$1$ if $\sigma\to\infty$ (and $t<T$).

1

It's a combination of too few sample paths and/or too small an increment. Your estimation error on the price is magnified by the $dS^2$. Try using a larger sample or a larger increment. Alternatively, you can use a multiplier instead of a fixed increment; in my experience, it usually yields better results.

1

First, I think you made a mistake in your computations above. Where you wrote $(30-20)$, I think you really meant $(30-(-20))$ i.e. $30+20$, yielding a gamma P&L of $1000$ instead of $200$. Your total P&L over $[90,170]$ would then be $110$ instead of $-690$. It doesn't matter for my answer either way, just thought I'd point it out for confused ...

1

http://www.theoptionsguide.com/in-the-money-covered-calls.aspx Look at the chart on this site. Do you see how the line is upward sloping for a covered call until the stock reaches a price of 45? To simplify things, you have a delta of 1 until 45 and then a delta of 0 after 45 onwards. Delta is the slope of the line. Delta of a covered call changes depending ...

1

When pricing FX options, the underlying is the spot or forward exchange rate. The foreign currency is analogous to a stock where the owner of the foreign currency receives a "dividend yield" equal to the risk-free rate in the foreign currency.

1

I am assuming you are talking about probability of becoming at least $\Delta = 30$, otherwise probability is zero. Hard to give a complete answer as quite some information is missing. As you are seeking for the probability, the outcome definitely depends on which model of underlying you are using. Moreover, even if you are using the BS model, some parameter ...

1

I've started thinking about this, too. My gedanken conclusion turned out to be too simple once I found what I was after: http://www.investment-and-finance.net/derivatives/o/option-beta.html, which I've confirmed in Black & Scholes (1973) p10 (eq 15). In short: $$\beta_{\text{option}} = \frac{S\cdot\Delta}{O}{\beta_S}$$ where $S$ is the underlying ...

1

For Black-Scholes, $\Delta_C=\partial_{S} C=N(d_1)$, $d_1= \frac{\ln\left(\frac{S_t}{K}\right) + \left(r + \frac{\sigma^2}{2}\right)(T - t)}{\sigma\sqrt{T - t}}$ You may fit the volatility $\sigma$ to this term by $$\Delta_C({\hat{\sigma}})=0.25$$Note that $\Delta_P=1-\Delta_C$ by Put-Call-Parity.

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