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3

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 ...


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

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 ...


2

Call-put parity writes (to see this, notice that $(S_T-K)^+ - (K-S_T)^+ = S_T - K $ and take the discounted risk-neutral expectation $E^{\mathbb {Q}} [. \vert \mathcal {F}_0 ]$ on both sides): $$ C(K,T) - P(K,T) = DF ( F(0,T) - K ) $$ with $DF = e^{-rT} $ the discount factor, and $F(0,T)$ the fair forward price given by $$ F(0,T) = (S_0 - D^*)e^{rT} $$ ...


2

The intuitive explanation is given in @Alex C's comment. You should stick to that if you understand it. Yet, if you are more comfortable with a mathematical approach: Payoff of being long a forward contrat with maturity $T$: $(S_T - X)$. Interpretation: at time $T$, you pay a certain price $X$ in exchange for which you receive the underlying $S_T$ Payoff ...


2

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 ...


2

Let's call R the riskless security (100 today, 120 at time T). And call S the stock = 50, and either 70 or 30 at time T. One way to look at it is: A] Consider: buy 2 call options (C), short the stock (S), invest 50 (proceeds from S) in R. At time T: S=70: 2C=40, buy back S=-70, proceeds from R=60. net: 30 S=30: 2C=0, buy back S=-30, proceeds from ...


2

If $PV_{t, T}(\text{Divs}) \ge K\big(1-e^{-r(T-t)}\big)$, since $P_{Eur}(S_t, K, T-t) >0$, the identity \begin{align*} C_{Eur}(S_t, K, T-t) = P_{Eur}(S_t, K, T-t) + (S_t-K) -PV_{t, T}(\text{Divs}) +K\big(1-e^{-r(T-t)}\big), \end{align*} implies that \begin{align*} C_{Eur}(S_t, K, T-t) > (S_t-K). \end{align*} That is, it is not rationale to exercise the ...


1

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


1

Let $\{X_t \mid t \ge 0\}$ be the foreign exchange rate rate from $£$ to $\$$. Moreover, let $C(X_0, K, T)$ and $P(X_0, K, T)$ be the prices of the respective call and put options with strike $K$ and maturity $T$. Then \begin{align*} \frac{1}{X_0}P(X_0,\, K,\, T) = K C\left(\frac{1}{X_0},\, \frac{1}{K},\, T \right). \end{align*} Based on the given condition, ...


1

A call lets to purchase one unit of underlying for some strike price x. So a call on GBP in USD lets us buy 1 unit of GBP for price x. However, since this is FX, lets clarify this to be USD x and USD 1 gets us GBP 1/x. A put lets you sell one unit of underlying for some strike price y (= 1/x). So a put on USD in GBP lets us sell 1 unit of USD for price 1/x. ...


1

What a difficult problem. The first line gave $165 e^{-rt} -3 S e^{-dt} = 15$ [since 50+55+60 = 165]. In the second line we want to evaluate $110 e^{-rt} -2 S e^{-dt} $. We notice that this is exactly two thirds of the left side of the above, because 110 is two thirds of 165 and 2 is two thirds of 3. So we take two thirds of the right hand side of the first ...


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In this case, call option is deep in the money while put option is deep out of money. As maturity is very near, any change in stock price would have equivalent impact on the call option price. Decline in one dollar in stock price lead to almost one dollar decline in call option price. Whereas for put, its worth would increase but put is still deep out of ...


1

Futures payoff is indeed $S_t-F_0$, but the $t$ in question is the maturity date of futures. In this derivation $t$ denotes maturity date of the option, which is always before the futures maturity. Therefore, on the day of option maturities, the futures did not expire yet, but the value of the futures position is $F_t-F_0$ (in mark-to-market sense, you can ...


1

the call version pays $$ I_{S_T > K } S_T $$ the put version pays $$ -I_{S_T < K } S_T $$ Subtract to get a pay-off $$ S_T. $$ (ignoring the probability zero event of $S_T=K.$) So the prices subtract to give $S_0.$



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