# Tag Info

9

We assume that, under the risk-neutral measure, the stock process $\{S_t, t \ge 0\}$ satisfies an SDE of the form \begin{align*} dS_t = r S_t dt + \sigma dW_t, \end{align*} where $r$ is the constant interest rate, $\sigma$ is the constant volatility, and $\{W_t, t \ge 0\}$ is standard Brownian motion. For $0 \le t \le T$, \begin{align*} S_T = S_t e^{r(T-t)} ...

7

I'll expand on Mark's and SRKX's answers which are both correct but brief. To be clear the words long and short have been generalized in finance. They used to mean that you owned a stock or had sold a stock short. Now they are often used to say you make money when a value goes up (long) or make money when some value goes down (short). In this case ...

7

$$\begin{array}{rcl} (1) & \partial_KC_t(T,K) & \leq 0 \\ (2) & \partial^2_KKC_t(T,K) & > 0 \\ (3) & \partial_T C_t(T,K) & \geq 0 \\ \end{array}$$ If $(1)$ doesnot hold, it exists $K_1<K_2$ such that $C_t(T,K_1)<C_t(T,K_2)$. Then as barrycarter said in his comment, you sell $C_t(T,K_2)$ and you buy $C_t(T,K_1)$, so your ...

7

I think you've cottoned on to the main question - that is,to see why $\frac{1}{2} \sigma^2 S^2 \frac{\partial^2 C}{ \partial S^2 }$ is the same thing as $-\frac{1}{2} \sigma^2 S \frac{\partial C}{\partial S} + \frac{1}{2} \sigma^2 (S \frac{\partial}{\partial S})^2 C$. I think your confusion comes from dealing with the squared operator. Writing down what ...

6

the vega of a call is always positive. The holder of a call option is therefore long volatility whatever the spot price.

6

American put-call symmetry relies on the observation that trading $S$ for $K$ is optimal when $\frac1K$ is optimally traded for $\frac1S$. So long as the dynamics of the inverse process $\frac1S$ are sufficiently tractable, you can derive the symmetry formula. You don't have to have a "pure" GBM for this to work. For example, non-constant (and even price-...

6

It really, really, really depends on your parameters, i.e. $r$, $\sigma$, $K$, $T$, $S_0$. For example, here are some results from implementing the stopping criteria I explain in my answer here. These are the number of iterations requires in order for there to be an approximate 0.95 probability that the MC call price differs from the exact call price by ...

5

Let $f_0(S_T) =f(S_T|S_0)$ be the risk-neutral PDF for the underlying asset price at time $T$ (conditional on the price $S_0$ at present time $t=0$). The probability that the price is above a strike price $K$ at time $T$ is $$P(S_T \geqslant K) = \int_K^\infty f_0(x) \, dx.$$ This is just definitional regardless of the shape of the distribution (eg. ...

4

Mark Joshi's answer is absolutely right, but just to elaborate a bit: The Vega of an option is the sensitivity of its value with respect to volatility $\nu = \frac{\partial V}{\partial \sigma}$. For calls, it makes sense that the Vega is always positive, not matter the level of the underlying. If you take the Black-Sholes model, you can find the ...

4

Given one satisfies margin requirements anyone can short exchange traded options as long as local regulators permit (American retail investors at present are not permitted, for example, to trade futures options . As long as there is a market and one finds a willing counterpart nothing speaks against shorting options contracts. Some brokers might require a ...

4

I think you might confuse two things here. In the Black-Scholes formula, the term $$\Phi \left( d_2 \right) = \mathbb{Q} \left( \left. S_T > K \right| \mathfrak{F}_t \right)$$ is the conditionally probability of ending up in-the-money under the risk-neutral probability measure $\mathbb{Q}$. Similarly, \Phi \...

4

It's pretty simple to derive with basic knowledge of stochastic calculus. But since you are looking for the easy answer here it is: $$C_t=e^{-r(T-t)}\sigma\sqrt{T-t} (D \Phi(D)+\phi(D))$$ where $D=\frac{F_{t,T}-K}{\sigma \sqrt{T-t}}$ and $\Phi(\cdot)$ and $\phi(\cdot)$ are respectively the normal cdf and pdf. $F_{t,T}=S_te^{r(T-t)}$ is the forward price.

4

The stated equation claims that: Value of call spread <= present value of its maximum payoff. This is almost self evident. For a formal proof, suppose untrue. Then sell the call spread, invest proceeds at r, giving a certain profit at T.

4

First note that delta is the derivative w.r.t. to the spot and not the strike. The latter is often called "dual delta". Also, you don't need any knowledge of Black-Scholes as this is a model-independent result. The result follows from the general expression of the call price C_0 = e^{-r T} \mathbb{E}_\mathbb{Q} \left[ \left( S_T - K \right)...

4

"how to find the right risk neutral measure [...]?" Specifically addressing this question, let us work on a probability space $(\Omega,\mathcal{F},\mathbb{P})$ filtered with $(\mathcal{F}_t)_{t \geq 0}$. We assume that the asset process under the physical measure $\mathbb{P}$ is: $$\text{d}S_t=\alpha(t,S_t)\text{d}t+\sigma\text{d}W_t$$ where $\alpha(t,S_t)$...

4

This is an approximation (to first order) based on the idea that the option gives you access to the underlying, but with leverage. Let the duration of the underlying be $D_B$. The expression $\lambda=\Delta_c\frac{B}{C}$ is called the elasticity of the option (link), defined as "the percentage change in option value per percentage change in the underlying ...

3

The argument that the American and European call are worth the same is model independent. So it holds for the binomial model. So there is no need to check to see if the early exercise occurs because it won't. Of course, if you have written general purpose code, it is much easier to test for early exercise and always have the test fail than to try and deal ...

3

There is a good quick well-known approximation for at-the-money options: $$\textrm{Call,Put} = 0.4 S \sigma \sqrt{T}.$$ See further discussion at What are some useful approximations to the Black-Scholes formula?.

3

You are right for delta but wrong for other greeks. Delta of stock is 1 so Covered call Delta = 1 - "long call delta" is correct However a stock doesn't have gamma , theta or vega. So these greeks of your covered call positions will be just that of short call. i.e Covered Call Gamma = - "Long Call gamma" Covered Call Theta = - "Long Call theta" ...

3

Formally, a long call payoff can be split as follows: $$(S_T-K)^+ = (S_T-K)\cdot 1_{\{S_T>K\}}$$ $$= S_T\cdot 1_{\{S_T>K\}} - K\cdot 1_{\{S_T>K\}},$$ that is, long an asset-or-nothing digital call payoff and short a cash-or-nothing digital call payoff. Here, $1_A$ is $1$ if event $A$ takes place, and it is $0$ otherwise.

3

this is how i would explain your approximation. First start with notation: Define $K_{atm}$ to be the atm strike. Define $\Delta K := K2 - K1$ where $K2 > K_{atm} > K1$. This corresponds to \$\Delta K = StrD in your notation. Now assume a black scholes world, within this world we can approximate the Call and Put price of an atm option with: C_{atm}... 3 A few years ago I asked a similar question on MO: https://mathoverflow.net/questions/22828/big-picture-concerning-ito-integral-stratonovich-integral-and-standard-results My take today is that you really don't need this heavy mathematical machinery for standard BS but as soon as you move on to more sophisticated (and realistic) models you surely do, so it ... 3 A butterfly in general has a payoff of the form \begin{align*} (X_T-K_c)^+ + (K_p-X_T)^+-(X_T-K_{atm})^+-(K_{atm}-X_T)^+, \end{align*} where X_T is the asset value at maturity T, while K_c, K_p, and K_{atm} are strike levels. 3 Let \sigma \in \mathbb{R}_n^n and let W be an n-dimensional standard Brownian motion. Define a new one-dimensional process \tilde{W} by $$\tilde{W}_t = \frac{1}{\vert \vert \sigma \vert \vert} \sum_{i = 1}^n \sigma_i W_t^{(i)}.$$ Then it is easy to show that \tilde{W} is a continuous martingale, starting at \tilde{W}_0 =... 3 The intrinsic value of a call is the price of the underlying minus the strike (S0-K), so if you find a european call whose value is less that that you would: Sell (or short) the underlying at S0 Use the proceeds to buy the call at C and wait. At maturity, the price of the underlying is Sm, and you will make a profit in either case: If Sm < K, the call ... 3 how to construct the portfolio (St−K)+ or how to make this arbitrage If you have this scenario on your hands then you construct the portfolio by putting as much capital as you can into the trade. It's an all reward and no risk scenario. Max it out! You "make" the arb by buying the call, shorting the equivalent amount the underlying at the current price ... 3 Look at the values of d_1 and d_2 when t \rightarrow T:d_1 = \frac{\ln(S/K) + \left(r - D + \dfrac{1}{2}\sigma^2\right)\tau}{\sigma \sqrt{\tau}}$$and$$d_2 = d_1 - \sigma \sqrt{\tau}$$with \tau = T -t. Therefore, t \rightarrow T is equivalent to \tau \rightarrow 0 Case 1: S > K d_1 \sim \frac{\ln(S/K) }{\sigma \sqrt{\tau}} \... 3 If W is a standard Brownian motion then \frac{1}{T}\int_0^T W_t dt has standard deviation \sqrt{\frac{T}{3}}. For this reason if S_t=S_0e^{(\alpha -\frac{1}{2}\sigma^2)t + \sigma W_t} is the GBM spot price then the average spot price \frac{1}{T}\int_0^T S_t dt has approximate volatility \frac{\sigma}{\sqrt{3}}. 3 This situation can arise with some non-vanilla options. For example, a digital put option, which pays 1 if the underlying price S is below a strike K at expiry, can exhibit "negative theta". Assuming zero interest rate and dividend yield to keep it simple, the price is$$\\P = N(-d_2), \quad d_2 = \frac{\log \frac{S}{K}}{\sigma \sqrt{T}} - \frac{\...

3

I am assuming you are short EUR and long USD based on your description of your hedges. I am also assuming the size of your hedges and your fx position are the same. In the first example of a hedge of this position, you cannot be delta neutral. A long at-the-money (ATM) EUR call will have a higher delta than the short out-of-the-money (OTM) call--you will ...

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