# Tag Info

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 ...

3

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)\\ =&\ ...

2

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, + ...

2

Touch option is simply a barrier option in QuantLib. You could create one like a down-in barrier type. You can also set the payoff to a binary payoff. The payoff is represented by the StrikePayOff class. A comprehensible example is available on github here.

2

At 5pm you get in your car and drive down a highway that has multiple exits: exit 1, exit 2, exit 3 etc. The objective is to get home a quickly as possible, i.e to Maximize -T, where T is the time you arrive home.. Let's say if you take Exit 1 you can be home at 6:30, if you take exit 2 you can be home at 6:15, if you take exit 3 6pm and if you take exit 4 ...

2

I solved it the following way, just want make sure I'm not missing something obvious. Set up a portfolio $PF$ consisting of long $S$ and short $P$ at time $t = 0$. Choose arbitrary time $0 < t < T$. If $S_t > P_t$ then $PF_t = S_t - P_t$ which coincides with the value of the option. If $S_t$ hits $P_t$ from above, then dissolve the portfolio by ...

2

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 ...

1

The value of an option is the premium that is paid to own this option. For this paylater option, since nothing is paid upfront, the value of the option is zero. That is, \begin{align*} e^{-rT}E\big((S_{T}-K)^{+}-P1_{S_{T}>K}\big)=0, \end{align*} or \begin{align*} E\big((S_{T}-K)^{+}-P1_{S_{T}>K}\big)=0. \end{align*}

1

The option payoff at maturity $T$ is defined by \begin{align*} (S_T-P_T)1_{\left(\inf_{0 \le t <T}\frac{S_t}{P_t}\right) > 1}. \end{align*} Let $Q$ be the risk-neutral probability measure and $E$ be the corresponding expectation operator. Let $Q_p$ be a probability measure defined by \begin{align*} \frac{dQ_p}{dQ}\big|_t = \frac{P_t}{e^{rt} P_0}. ...

1

from a practitioner perspective, i can say there's no such thing as a 0 year swap (obviously). The shortest tenor that you could trade would be a contract on one month LIBOR or more likely 3 month LIBOR. Then the instrument you are asking about is a 5 year expiration caplet (payoff in 5 years = max (0, LIBOR- strike).)

1

You introduce a discretized auxiliary variable which represents $S_t$ to solve $N$ PDEs on $[t, t+\tau]$ using finite differences which will give you the present value of the option at time $t$ conditional on $S_t$. Then you solve one PDE using finite differences on $[0, t]$ to obtain the the present value at time $0$. This is the same methodology than that ...

1

I think it's ok $$S_T = e^{\ln S_T}$$

1

There is a problem in your last step. Note that \begin{align*} P_{t, T_2}E_{Q_{T_2}}\left(\frac{1}{P_{T_1, T_2}} \mid \mathcal{F}_t \right) &= P_{t, T_2}E_{Q_{T_2}}\left(\frac{P_{T_1, T_1}}{P_{T_1, T_2}} \mid \mathcal{F}_t \right)\\ &=P_{t, T_2} \times \frac{P_{t, T_1}}{P_{t, T_2}}\\ &=P_{t, T_1}. \end{align*}

1

The option payoff is equivalent to $Z_{\tau \wedge T}-1$ where $\tau=\inf\{t | Z_t = 1\}$ provided that $Z_t$ is assumed to be continuous. Since $Z_t=S_t/P_t$ is a martingale under $Q_P$, we have $E_P[Z_{\tau \wedge T}]=Z_0$ and the option value is $P_0 (Z_0 - 1)=S_0-P_0$ regardless of the model.

1

The dividend adjustment in the option formula represents compensation for dividend income that an ordinary stockholder will have entitlement to before expiry, but which you the option owner will not be entitled to. If at expiry you exercise into a stock that has not gone exdiv, then you are entitled to the dividend, and so should not include it in your ...

1

Certainly, you must agree that $$C_{T}-P_{T}=\left(S_{T}-K\right)^{+}-\left(K-S_{T}\right)^{+}=S_{T}-K.$$ Therefore, since $$C_{t}=e^{-r\left(T-t\right)}E_{Q}\left[C_{T}\right]\text{ and }P_{t}=e^{-r\left(T-t\right)}E_{Q}\left[P_{T}\right]$$ it follows by the linearity of $E$ that $$C_{t}-P_{t}=e^{-r\left(T-t\right)}E_{Q}\left[C_{T}-P_{T}\mid ... 1 You are mostly right, I don't really get what you don't understand. The answer in the book is quite clear, but let me put it that way : Selling a put and buying a call on the same underlying S with same maturity and same stike K is always equivalent to a long position in a forward contract on S with delivery price K. The easiest way to see that is ... 1 There's no best method. The question is : what is the behavior of the volatility structure (atm and skew) when the underlying moves? Each method assumes something different. In the real market, one method might work well for a period of time (in the sense that it minimizes residual p/l), but then another method might take over as best. Practitioners ... 1 I actually discuss this question at length in chapter 1 of More Mathematical Finance. The essential point is that if you can write$$ X=YZ  with $Y,Z$ independent $E(Z)=1$ and $Z>0$ then $X$ is more uncertain than $Y.$ It then follows from Jensen's inequality that the price of an option on $X$ that has a convex pay-off will be at least as high as the ...

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