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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 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

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[ ...


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 ...


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".


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

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

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.)


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, ...


3

Its a stylized fact in academia that put options are overpriced. E.g., the monthly average return on S&P500 put options is around -40% for ATM options. The most often quoted reason for this phenomenon are hedging costs: A put is more difficult to hedge from a market maker's perspective, hence the prices artificially go up. An important paper on this ...


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 ...


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

if you let $\delta t$ be small enough, this won't happen. So the solution is to take more steps. The CRR tree is very out dated in any case.



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