# Tag Info

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You have a multidimensional problem - there isn't an answer of "this is what the greeks look like" for all cases, because it depends on the various levels of the different parameters. For example, if we limit ourselves purely to KO Call options, where the spot is 100, and there is no drift, with a time to maturity of 1 year (changing this is equivalent to ...

6

Let \begin{align*} \tau = \inf\{t: t \ge 0, S_t \le L \}. \end{align*} Then the down-out-call option has payoff \begin{align*} (S_T-K, 0)^+\pmb{1}_{\tau >T}, \end{align*} and the down-out version zero-coupon $T$-maturity bond has payoff \begin{align*} \pmb{1}_{\tau >T}. \end{align*} Moreover, for the down-in payoff $X$, since $L=K$, \begin{align*} X &...

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You're right that the "real" greeks of a digital option are unwieldy, e.g. delta is zero everywhere except at the barrier where it is an impulse. So sell-side trading desks model/price digital options as tightly struck call/put spreads that will sit and play nicely with the rest of the book. Here's a simple example: let's say a bank sells a digital call on ...

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Here are at least three mistakes in your code: p += s0 * exp(...) should be p *= exp(...). Your volatility and rates are per annum, so divide the days by 365 (or 255) in your function asset_price. In asset_price you multiply by days inside the loop. However, the loop is already iterating over the days - so you don't take two steps of one day but two steps ...

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Presumably the option can be exercised for intrinsic at any point. Note the interviewer asked for a static hedge using the stock, not a dynamic hedge. Hence you must find a buy and hold portfolio that will always give you at least the value of the option (if you’re short it which I suppose is the question) until it is exercised. Note that the maximum ...

4

As specified I will assume your option is perpetual; I will also assume that it is written on an asset whose price $(S_t)_{t \geq 0}$ follows a Geometric Brownian Motion (GBM) with drift coefficient $rS_t$ and diffusion coefficient $\sigma S_t$ under the risk-neutral measure $\mathbb{Q}$ $-$ we assume a constant risk-free rate: $$dS_t = rS_tdt + \sigma S_t ... 4 there are a number of ways to do this. You do have to make some modelling assumptions, however. eg continuity, BS model holds, or log stock price process is independent of level. The most common way is to take the pay-off and geometrically reflect in the barrier. (i.e. pass to log coordinates and reflect). i.e. write the function as f(x) where x= \log ... 4 The difference is that the barrier option is weakly path dependent while the lookback option is strongly path dependent. In case of a knock-out barrier option, conditional on the option being alive at the pricing time you don't need to carry any additional state variables except for the current asset price. The payoff doesn't directly depend on the level of ... 4 I nearly agree with @phlsmk's answer, but with some small differences. First off, the delta of a digital is not "zero everywhere except at the barrier where it is an impulse". This is what it is at t=T. before this, it is smoothed out, exactly like a regular option is. The problem is on what the delta may become. This is not the only place where it ... 3 No, the pricing of a double barrier knock-out option cannot be decomposed into single barrier options. Here are a few references that apply the method of images to the valuation of double barrier options: A very clear and easy to follow exposition can be found in Chapter 3.5 of the Ph.D. thesis by Konstandatos (2003). If you don't have access to that, then ... 3 Their formula looks correct. As is usually the case, there are multi ways to derive this result. I will outline two of them here. Reflection Principle & Measure Change The solution to the risk-neutral dynamics of S is $$S_t = S_0 \exp \left\{ \left( r - \frac{1}{2} \sigma^2 \right) t + \sigma W_t^* \right\},$$ where W^*... 3 The error is, you are not storing the random numbers for the same path at the end: xbefore = x + c*tau + sigma*sqrt(tau)*randn() A = muA + sigmaA*randn(); xafter = xbefore + A; But then at end you set a different path here by creating a new random number: xT = log(S0)+(c+muA*lambda)*T+sqrt((sigma^2+(muA^2+sigmaA^2)*lambda)*T)*randn(); randn() generates ... 3 Assuming \theta>0 (take \tilde{X}=\mu-X if it is not the case) Let us denote \text{erfi}(x) the imaginary error function Let us denote \tau_L,resp.\tau_U the hitting time of Lresp.U where L<U 1) Using Ito's lemma, prove that :$$Y_t = \text{erfi}\left(\sqrt{\frac{\theta}{\sigma^2}}\left(X_t-\mu\right)\right) \text{ is a martingale}... 3 Since there is a closed form in the BS case for continuous barrier options, you probably won't find a huge amount of work on this since it's not needed. In the discrete case, I did a paper with Tang: http://ssrn.com/abstract=1441142 Pricing and Deltas of Discretely-Monitored Barrier Options Using Stratified Sampling on the Hitting-Times to the Barrier 3 As Daneel mentioned in his comment, you can't simply split your expectation of product into a product of two expecations as the two quantities are far from being independent... Now, to answer your question w.r.t. how you could compute the expectation of the joint event of being in the money while having hit the barrier, you were right in using the reflexion ... 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 As I mentioned above, I am not sure what the variable r is. If we ignore that, or assume the questioner wanted to say its the risk free interest rate, then it has no effect on the number of paths. Then it is clear that after 50 steps going from \1024 to \2500 requires a net of 4 up movements with the given x=y^{-1}=1.25. Thus the number of steps ... 2 I'd recommend M. Joshi and T. Leung "Using Monte Carlo simulation and importance sampling to rapidly obtain jump-diffusion prices of continuous barrier options". Though it assumes jump-diffusion process for the returns it is straightforward to obtain the scheme for a diffusion process. Also Paul Glasserman's [book][2] [2]: http://www.amazon.com/Financial-... 2 I do not have any reference, but I think H is for hitting. 2 May be I have overlooked something, but I believe that \begin{align*} Q(t, S) = \mathbb{P}\left(\tau_{B} \le T \mid \mathcal{F}_t\right). \end{align*} Then \{Q(t, S), \, 0<t < T\} is a martingale, and the PDE follows immediately, by noting that \begin{align*} dQ &= Q_t dt + Q_S dS + \frac{1}{2}Q_{SS} d\langle S, S\rangle_t\\ &=\Big(\... 2 Formally, let \begin{align*} \tau = \inf\{t: t \ge 0, S_t \ge 50 \}. \end{align*} Then \begin{align*} \text{Payoff} &= \left(S(31)-33 \right)^+\pmb{1}_{\tau >31} + 50\times \pmb{1}_{\tau \le 31}. \end{align*} That is, an up-out barrier call plus 50 digital up-in barrier options. 2 The goal of this exercise is to replicate the payoff of the Secured Barrier Call by a linear combination of the known products: European up-out call (cost 12), digital strike 33 (cost 0.73) and digital strike 50 (cost 0.7). Looks to me it is sufficient to buy: 1x up-out call 50 x digital strike 50 The payout at expiry of this linear combination would be: ... 2 There are a few issues that need to be separated here. Issue "zero" is whether your MC is able to correctly represent the dynamics you've chosen for your assets. If you implement your MC properly, by construction it should converge in distribution to the postulated dynamics. No bias there. Variance yes potentially, because of discretisation, but no ... 2 For the first question, you can just plug in t for T and S for K: \sigma^2 \left(t, S \right)=\left. \sigma^2 \left(T,K\right) \right|_{T=t,K=S} For the Monte Carlo part, the barrier would apply to the history of the stock price over some window (which could be from today to the option maturity, but other variations are possible) instead of just the ... 2 Standard call options are trivially more expensive than up/down and out call options. However, for high strikes, down and out options will very likely never be knocked out, therefore their prices should be close to standard call options. For low strikes, down and out call options are almost worthless, therefore , the down and out call options curve price ... 2 for an intuitive answer, if we start with a vanilla call as our base, then with an up & out call, we would like the underlying to go up in price yes. But as the price increases, we also increase the probability of kicking out and losing our payout - so we don't want it to go up too much. If the barrier is so far away that the probability of reaching it ... 2 When you simulate a sample path of a standard Brownian motion, you are generating a sequence (B_t)_{t \in \mathbb{\Pi}} where \mathbb{\Pi} := \{t_0, ..., t_n\} is your time partition. You can view that sequence as n draws of the same random variable, although no one could say that this isn't also 1 draw each of n independent normal random variables. ... 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 ... 1 Assume you are long an up-and-in put and short and up-and-in call of the same maturity, strike and barrier. When S_t = B for t \in [0, T], then both barrier options knock-in and turn into vanilla options. You are now long a put and short a call with the same maturity and strike. From put/call parity, you know that S_t + C_t = K e^{-r (T ... 1 Idea Let B be a standard brownian motion starting from x_0=0, m_T = \inf_{u\leq T}B_u and M_T =\sup_{u\leq T}B_u. Let's define if it exists for A\in\sigma(B_u,u\leq T), \mathbb{P}(A | B_T=x_T)\stackrel{\rm def}{=}\lim_{\varepsilon\to 0}\mathbb{P}(A|B_T\in(x_T-\varepsilon,x_T+\varepsilon))\begin{split} \mathbb{P}(\tau_U\leq T \cap \tau_U\leq ...

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