# Tag Info

7

A simple intuitive answer why the OTM Call is more expensive than the OTM Put is because of the skewness of the log-normal distribution. Think about it, what is the probability that the stock price is above 110 at expiration and what is the probability it is below 90? This should answer your question. Written in probability terms: The median of the ...

7

The put call parity is given as follows: $$c_t-p_t = S_t - \frac{X}{e^{r(T-t)}}$$ If you assume $r=0$, you get $$c_t-p_t = S_t - X$$ So, $c_t \neq p_t$. The rationale behind it is much more financial than mathematical. You have to look at the payoff on both side of the equation, and you see that both portfolio will give the same payoff at time $T$ (the ...

7

Let's carefully distinguish which exercise type we consider. European-style call option $$\max\{S_0-Ke^{-rT},0\}\leq C_E \leq S_0.$$ European-style put option $$\max\{Ke^{-rT}-S_0,0\}\leq P_E\leq Ke^{-rT}.$$ American-style call option $$\max\{S_0-K,C_E\}\leq C_A\leq S_0.$$ American-style put option $$\max\{K-S_0,P_E\}\leq P_A\leq K.$$ Because ...

6

On 10/24/17, Wells Fargo announced that they would pay a dividend of 0.39 to holders of record on 11/3/17. Thus, if you buy the stock after this date (through the exercise of the call) you do not get the dividend. This means that the potential arbitrage, instead of being 0.30, is -0.09. This is now sufficiently close to zero that there is likely no ...

6

Put-call parity says that a call and put (worth $C$ and $P$ respectively) with the same strike $K$ have the following relationship with the spot rate $S$, risk-free rate $r$, and time to maturity $T$ -- $$C - P = S - e^{-rT} K$$ Taking the first derivative with respect to $S$, $$\frac{\partial C}{\partial S} - \frac{\partial P}{\partial S} = 1$$ which ...

5

At the heart of the (relative) pricing theory is the concept of no arbitrage and replication. I'll focus on equities here because as stated in the comments it may be more complicated for commodities. Forwards deliver a payout linear in the future value of the underlying asset. Hence they can be replicated statically by a simple cash & carry replication ...

5

Elaborating on my comment: consider a 100 point in the money collar, one day before expiration. You are effectively claiming the price of this is 99.5. But if you call the pit and get a two way price of 99-100 there is nothing to do.

5

Once upon a time, all option contracts ceased trading on the third Friday of every month. There was no after hours trading for the underlying. When the exchanges closed, everything was done. This is no longer true. Contracts do not exclusively cease trading on the third Friday, although some still do. Likewise, the underlying can continue trading after ...

5

Vega is the partial derivative of the option price (as a function of parameters -- current stock price $S_t$, strike price $K$, implied volatility $\sigma$, etc.) with respect to $\sigma$ -- holding other parameters fixed: $$vega = \frac{\partial}{\partial \sigma} V(S_t,K,\tau,r,\sigma)$$ You are confusing the stochastic process with the parameter $S_t$ ...

5

This is a consequence of transforming a Put on $S_T$ with strike $K$ into a Call on $(K S_0)/S_T$ with strike $S_0$ under the stock measure. The new set of parameters $r_p$, $q_p$, $\kappa_p$, ... etc . are those that correspond to the Heston dynamics for the process $((K S_0)/S_t, v_t)$ under the stock measure. General results on that kind of symmetry can ...

5

Diviends or not, the put-call parity (with European options) always hold: $C(S,K) - P(S,K) = F - K*DF$ In the RHS, dividends will impact the forward $F$ (higher dividends imply lower forward). So the LHS should be lower as well: the Call costs less and the Put costs more. The proof is straightforward, you just notice that at maturity $T$ you have: $S_T - K ... 4 First, we have$P(t)+S(t)=C(t)+B(t,T)\cdot K$, Then,$\frac{\partial P(t)}{\partial S(t)} + \frac{\partial S(t)}{\partial S(t)} = \Delta^{\text{put}}_{t}+1$and$\frac{\partial C(t)}{\partial S(t)} + \frac{\partial [B(t,T)\cdot K]}{\partial S(t)} = \Delta^{\text{call}}_{t}+0$. Finaly,$\Delta^{\text{call}}_{t}-\Delta^{\text{put}}_{t}=1$. This relationship ... 4 Since American style options allow early exercise, put-call parity will not hold for American options (unless they are held to expiration). In practice, there is also a difference between calls and puts for European options as well. The full description is here: What causes the call and put volatility surface to differ? 4 You have forgotten the combinatorial factors for binomial probabilities on your terms. You need $${n\choose k} p^n(1-p)^{n-k},$$ not just $$p^n(1-p)^{n-k}.$$ The second term should have a factor of$6$and the third should have a factor of$15,$etc. 4 The market standard formula approximation for cash settled swaptions applies Black/shifted Black/Bachelier around the forward swap rate so that with this formula parity between payer and receiver swaptions occurs around the forward swap rate, and in particular the zero wide collar struck at the forward swap rate is worth zero (a zero wide collar is the ... 4 I don't mean to suggest such a large topic, but it would certainly be worth reading about delta-hedging with regards to your question. Since such a large percentage of options are delta-hedged, the net price change of shares in the underlying due to exercise on expiration would be ~0. As @Emma mentioned, deep in the money options have a high delta. This ... 4 You should see the risk free rate as the return on the strategy. That’s because you actually have to invest money , namely usd 19000 minus usd 38, for the one month period. Hence, there is no arbitrage in the market data you observe. 4 There are usually two ways to write proofs of equalities (like put-call parity) in quantitative finance. By replication, by constructing arbitrage. Both of these are actually the same, since the first one is done by making, say, two portfolios,$A$and$B$, and showing that they have the same outcome at time$t=T$. Then, by argument of LOOP (law of one ... 4 I am not sure why the question you link to does not provide an answer. I’ll try to answer it but it is really similar to what has already been said there. Bottom line is: if the value$K$is reachable by the underlying asset$S$, that is$K$belongs to the domain of process$S$, then the butterfly should be strictly positive. First note that the butterfly is ... 3 Put-call parity says that the difference between a call and a put is equivalent to the difference in the current stock price (adjusted down for dividends) and the strike price discounted at the risk-free rate. $$Call - Put = S_0*e^{-div} - K*e^{-rt}$$ So, if you want to have 120 dollars in the future, you would need to need to have$120 worth of "K" or 2....

3

If the riskless security cost $100$ at time $t=0$ and $120$ at time $T$ then the risk free rate, $r$, is $20\%$. So that, $r=0.2$. Denote the initial stock price as $S_0$ and price of the call option as $c$. Suppose that at time $t=0$ you buy one stock and sell $\Delta$ options. Your portfolio value at time $t=0$ is $$P_0 = -\Delta\times c + S_0$$. At time $... 3 Let's talk about your first equation: If you exercised your option early, you got this payoff. But if you are a rational investor you'd realize that this is less than what you would get if you would just sell your option itself. i.e. the payoff at time t will be more than S(t)-K because the option is worth more than that as it also has some time value. so ... 3 It costs 0.03 dollars for the option to (sell 1 pound/buy 1.5 dollars. Now divide everything by 1.5: It costs 0.02 dollars for the option to (sell 2/3 pound / buy 1 dollar). Now convert to pounds at spot rate: It costs 0.0133 pounds for the option to (sell 2/3 pound / buy 1 dollar). Done 3 In developed equity markets you have at least those three entities: Stocks, futures and options. As far as my experience goes you indeed often hedge options with futures. What is more interesting is the relationship concerning stocks and futures: In theory stocks are the underlying and futures are the derivatives. In practice it is, interestingly enough, ... 3 Financing Risk e.g. You sell the stock short and buy the Call and sell the Put. Lets say you can only get a 1w repo borrow on the stock yet your options are 3m options. You have the risk after that week is up, that the stock goes special, so your financing costs erode the arbitrage. Similarly the Fed could cut causing your financing assumptions to change, ... 3 This situation can arise with some non-vanilla options. For example, a digital put option, which pays$1$if the underlying price$S$is below a strike$K$at expiry, can exhibit "negative theta". Assuming zero interest rate and dividend yield to keep it simple, the price is$\\P = N(-d_2), \quad d_2 = \frac{\log \frac{S}{K}}{\sigma \sqrt{T}} - \frac{\... 3 There are two ways to look at it, a mathematical way or an alternative, intuitive way. The alternative way can be to look at F as an alternative S with 0 interest rate discounting because we still have the cash (minus a small posted margin, and ignoring this) which earns the interest rate. So for the F’s value itself every day’s time value of money effect ... 3 Whether arithmetic or geometric averaging, you always get \begin{align*} \mathrm{AsianCall} - \mathrm{AsianPut} = e^{-rT} (\mathbb{E}[\bar{S}]-K). \end{align*} So, let’s compute the expectation. You know that\bar{S}=\exp\left( \frac{1}{T} \int_0^T \ln(S_t)\mathrm{d}t \right)$where$\ln(S_t) =\ln(S_0)+\left(r-q-\frac{1}{2}\sigma^2\right)t+ \sigma W_t$. ... 3 Aren’t both statements incorrect? In a puttable bond, the investor is LONG the put. In a callable bond , the investor is SHORT the call. It may be true that put price = call price for ATM options, but being long a put is always better than being short a call. 3 Into the first equation we can substitute$C$and$P$as given by the other two equations, we get:$(S-K)^+ -(K-S)^+ +TV_C - TV_P = S-PV(K)S-K+TV_C-TV_P=S-PV(K)TV_C-TV_P=K-PV(K)$If interest rates are zero then$PV(K)=K$and then we indeed have$TV_C=TV_P\$ Note: as suggested in the comments above a slightly different definition of Intrinsic Value ...

Only top voted, non community-wiki answers of a minimum length are eligible