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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.). ...


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 ...


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

By definition, the payoff of a log-contract of maturity $T$ writes $$ \phi(S_T) = \ln\left(\frac{S_T}{S_0}\right) $$ Let $\Pi_t$ denote the $t$-value of such a contingent claim. We are interested in the price at $t=0$, best known as the option premium. Theory tells us that the latter premium can be computed as $$ \Pi_0 = e^{-rT} E^{\mathbb{Q}} \left[ \phi(...


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 $\mathcal{F}_t$,...


5

Time $T$ boundary condition is correct $u(T,x)=(x-K_1)^+-(x-K_2)^+$. Time $x\to 0$ boundary condition is known and is equal to $0$. Time $x\to\infty$ boundary condition is also known and is correct $\lim_{x\to\infty}u(t,x)=(K_2-K_1)e^{-r(T-t)}.$ You need to be precise if you want your boundary be "absorbing" or "reflecting".


5

stochastic vol and Levy process models are popular. Jump diffusion less so. FT techniques are definitely used. These days most of the focus is on valuation adjustments for vanilla products rather than how to price structured products. It tends to use both MC and lattice methods. If you want to be topical, I'd advise something related to valuation ...


5

Starting from the Black-Scholes model that $$ \dfrac{dS}{S} = \mu \:dt + \sigma\:dW_t $$ where $W_t$ is a standard Brownian motion, and $\sigma$ and $\mu$ are constant where $\sigma > 0$. Here $W_t$ is a Brownian motion under the physical measure $\mathbb{P}$. We can then use Girsanov's theorem to change the measure to risk neutral measure $\mathbb{Q}$ ...


5

Thanks to @Phun and @oliversm I solved the problem. So I'm posting here the solution in case someone will need it. Under Black-Scholes assets dynamics are determined by a Geometric Brownian Motion, and we can define the price of a security at time $t+\Delta t$ as: $$S_{t+\Delta t}=S_{t}\exp\left(\left(r-\frac{1}{2}\sigma^{2}\right)\Delta t+\sigma\sqrt{\...


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

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 \int_{\...


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)\\ =&\ e^{-r(T-t)}...


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 $c_{...


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 $f(x, k) = C e^{i\nu k} \left( e^x - e^k \right) q(x)$, $\int_{-\infty}^...


4

The first portfolio $\Pi^{(1)}_t$ is a self-financing hedging portfolio. It is typically what you get when you delta hedge an option position (here short hence the minus sign, but it could be long without loss of generality) with shares of the underlying asset. If the only source of risk comes from the randomness of the underlying asset price $S_t$, then one ...


4

My Answer You should set your limit order to: $s (v+1)^{-0.0314192 \sqrt{t}}$ where $s$ is the current price, $t$ is the time in years you're willing to wait, and $v$ is the annual volatility as a percentage. If you want to be $p$ percent sure (instead of 0.98), set your limit order to: $s (v+1)^{-\sqrt{\pi } \sqrt{t} \text{erf}^{-1}(1-p)}$ Of course, ...


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.)


4

1: Follow the calculations in The Complete Guide to Option Pricing Formulas. The book has many formulas, sample values and outputs. Highly recommended for validating your results. Apparently, this is one of most popular books used by real-world quants (simple and fast). 2: You can still use QuantLib to price with year fractions. I have an example: ...


3

In the Heston Model we have \begin{align} C(t\,,{{S}_{t}},{{v}_{t}},K,T)={{S}_{t}}{{P}_{1}}-K\,{{e}^{-r\tau }}{{P}_{2}} \end{align} where,for $j=1,2$ \begin{align} & {{P}_{j}}({{x}_{t}}\,,\,{{v}_{t}}\,;\,\,{{x}_{T}},\ln K)=\frac{1}{2}+\frac{1}{\pi }\int\limits_{0}^{\infty }{\operatorname{Re}\left( \frac{{{e}^{-i\phi \ln K}}{{f}_{j}}(\phi ;t,x,v)}{i\phi }...


3

Answering my own question as it could be useful for others. Actually package fOptions is vectorized. The only constraint (and that make sense) is that you can't compute at the same time 2 different greeks, or mix up calls and puts. So assuming that you want to compute the delta of a set of puts, the code will be the following: fOptions::GBSGreeks(...


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

This is not a valid statement. Most obvious reason is that this situation would provide an arbitrage opportunity for those that take the 30 seconds to check open interest :-) But more specifically for every call that is in the money there is a person (firm, hedge fund . . you pick) that is short the call and another person (etc) that is long it. Simply ...


3

The term of art in our industry for this type of option pricing formula is a series solution. As Farahvartish indicates in the comments, a series solution is not considered to be an "analytical solution" due to the reliance on a converging infinite sum for actual numeric output.(*) Series solutions have been employed at least since the 1990s, when they ...



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