# Tag Info

7

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 ...

6

The point is the following: Delta, $\Delta$, is defined as $\frac{\partial C}{\partial S}$, where $C$ is the value of the call option, and $S$ is the price of the underlying asset. So, given that the value of a call option for a non-dividend-paying underlying stock in terms of the Black–Scholes parameters is $$C = N(d_{1})S - N(d_{2})Ke^{-rT},$$ $$\Delta ... 6 Dividends do not matter for the determination of the upper bound. Indeed, the maximum profit which the holder of a put option can make (be it through a European or an American exercise feature) is exactly equal to the strike price X. This can be seen by simply looking at the payout function: the maximum profit is finite and located on the downside when the ... 6$$\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 ... 5 You only have one asset in your portfolio which means that you can only statically hedge. By the definition of self financing, V_0=\phi_0 S_0, V_1=V_0+\phi_1 (S_1-S_0), and V_1= \phi_1 S_1. Putting these last two together, V_0=\phi_1 S_0 . Hence \phi_1=\phi_0 and you have a static position. Intuitively, this is because you cannot trade in ... 5 By definition the fair value of an option is given by an expectation value of the payoff, \mathbf{E}\left[\textrm{payoff}(\textit{paths})\right]. The probability distribution of the paths is the risk neutral measure. This is just an integral expression of the form you wrote. This applies to all option prices. Many options are, of course, special in the ... 5 importance sampling is well known to be tricky. See the extensive discussion in Glasserman's book. I presume that you are simply meanshifting and multiply by the ratio of normal densities. For this sort of problem, I'd use a more stratified algorithm instead and force every path to end in the money. To do this I'd compute the uniform that goes to 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*} 5 You simply required 2 things: 1) Risk free rate, and 2) Standard Deviation. For the interest rate you can use LIBOR of nearest maturity. Convert your LIBOR rate into continuous compound rate by taking log. Additional: VIX also uses LIBOR as an proxy for risk free interest rate and they also select LIBOR of nearest maturity of option contract. Standard ... 5 I_{\{S_{T}-K>0\}} is NOT independent of \mathcal{F}_{t}, since \begin{align*} S_T=S_t \, e^{(r-\frac{1}{2}\sigma^2)(T-t) + \sigma (W_T^*-W_t^*)}, \end{align*} where S_t \in \mathcal{F}_t, though e^{(r-\frac{1}{2}\sigma^2)(T-t) + \sigma (W_T^*-W_t^*)} is independent of \mathcal{F}_t. However, since W_T^*-W_t^* is independent of ... 4 This is a bit of an old question, but I thought I'd contribute to add more weight to to what some people have been saying. A CSO (calendar spread option) is NOT a calendar spread of options. If you read it carefully, you can see the Hull quote Max Li posted is talking about a calendar spread, not a CSO. A CSO needs to be priced the same way as a spread ... 4 If \mu is large, then it is more likely for the call to finish in the money. Your and my intuitions suggest that this means that the option is more valuable. But this is wrong. A call option is an insurance policy. A call option is useful because it protects you in the case that the value of the stock goes down. That is why call options are valuable for ... 4 The price difference is so large -- that the only possible reason is that you have spot and strike confused between the two functions. And indeed: R> fOptions.BAW <- BAWAmericanApproxOption(TypeFlag, S, X, Time, + r, b, sigma, title = NULL, description = NULL) R> quantlib.BAW <- AmericanOption("call", X, S, b, r, Time, + ... 4 This drift comes from making the discounted stock a martingale in the risk-neutral measure \mathbb Q You start with a stock in \mathbb P having this form: dS_t = \mu S_t dt + \sigma S_t dW_t You also have a discount factor e^{rt}. The idea is to remove the drift of the discounted process in \mathbb Q so you get (after applying Girsanov's ... 4 I know two papers explaining how to calibrate this kind of models, and one of them explain the impact of the quality of the fit on a pricing model: Aït-Sahalia, Y. (2002, January). Maximum likelihood estimation of discretely sampled diffusions: A closed-form approximation approach. Econometrica 70 (1), 223-262. Azencott, R., Y. Gadhyan, and R. Glowinski ... 4 Let \{P_t \mid t \geq 0\} be a compound Poisson process, where \begin{align*} P_t = \sum_{i=1}^{N_t} (V_i -1), \end{align*} and N_t is a Poisson process with intensity \lambda and jump times \tau_i, i = 1, \ldots, \infty. Let Y_i=\ln V_i and f(x) be the density function. Then \begin{align*} P_t - \lambda t E(V_1) &= P_t - \lambda t ... 4 When a pay-off is piecewise linear plus jumps, it the same as the portfolio of calls and digital calls. Its price must agree with that of the portfolio by no arbitrage. Every time there is a jump we add in a digital call and every time there is a change in gradient we add in calls equal to the gradient change. Here we have a call struck at K. Just below ... 4 It is not the fact that volatility is time varying that creates the skew per se, but the fact that volatility is negatively correlated with the spot. That is to say, as the stock/index price declines volatility will tend on average to increase, and vice versa. Time varying volatility itself would create a more symmetric 'smile'. Edit: Suppose that you ... 4 Let t=1 and T=2. The value at time t is given by \begin{align*} &\ e^{-r(T-t)}\max\left(E\left((S_T-K)^+\mid \mathcal{F}_{t}\right), \, E\left((K-S_T)^+\mid \mathcal{F}_{t}\right)\right) \\ =&\ e^{-r(T-t)}E\left((K-S_T)^+\mid \mathcal{F}_{t}\right) +e^{-r(T-t)}\max\left(E\left((S_T-K)\mid \mathcal{F}_{t}\right), \, 0\right)\\ =&\ ... 4 Let's define t=0, T_1 = 1 and T_2 = 2. I believe the interviewer is looking for the price of the "global" option V_t for t \leq T_1 \leq T_2 . Let's define the payoff at time T_1: it is the maximum between the value of a call or a put on the same underlying with maturity at T_2.\text{Payoff}_{T_1} = \max( c_{T_1}, p_{T_1} )$$where ... 4 Fubini's theorem is only used to reverse the order of integration. We have: \int_{-\infty}^{\infty}{e^{i\nu k} \left( C \int_k^{\infty} \left( e^x - e^k \right) q(x) dx \right) dk} = \int_{-\infty}^{\infty}{\int_k^{\infty}{C e^{i\nu k} \left( e^x - e^k \right) q(x) dx} dk}  Now, let f(x, k) = C e^{i\nu k} \left( e^x - e^k \right) q(x), ... 4 Peter Jaeckel has written various papers on this. "by implication" and "Let's be rational" are the most recent ones. He also provides code on his website www.jaeckel.org. (Note: the question asked for literature.) 3 There is a good quick well-known approximation for at-the-money options:$$\textrm{Call,Put} = 0.4 S \sigma \sqrt{T}. See further discussion at What are some useful approximations to the Black-Scholes formula?.

3

The above equation is the price of a call option. It has nothing stochastic inside it. It only depends on the current price and the time. So no Ito is needed. You should just compute the derivatives of your solution v (like you do for any deterministic multivariable function), plug them into the PDE and verify that it's satisfied.

3

"Intuitively, everything else being equal, if a stock has higher drift, shouldn't it have higher probability of finishing in-the-money (and higher probability of having higher payoff), and the call option should be worth more?" All these other answers are focusing on the wrong aspect of the question - it is true that the maths makes the drift drop out from ...

3

I don't know the BS formula you are trying to use. The price is the expected value of the discounted payoff under the risk neutral probability measure (I.e. Under which S is a martingale) So the you need to compute the risk neutral probabilities for S to go up or down. The probabilities given in the problem have no impact. They are just there to trick the ...

3

this is probably the most asked question in quantitative finance... There are many answers. One nice example to consider is what if the calls were struck at zero. The call then pays the stock price at time $T$ and so it's value today must the stock price today since we can replicate by holding one unit of stock. This will be true regardless of the drift of ...

3

if put call parity seems to be violated there could be things you are ignoring like dividends or hard to borrow fees. Hard to borrow will make puts more expensive

3

The typical investor is long. To protect the portfolio, he buys puts, thus driving up the price. To generate income against his long position, he sells covered calls, thus driving down the price. This is the most basic explanation for the difference in put call prices that are equidistant from the money. Obviously other factors are there as pointed out by ...

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