# Tag Info

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 ...

1

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

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$T. Then \begin{align*} \frac{1}{X_0}P(X_0,\, K,\, T) = K C\left(\frac{1}{X_0},\, \frac{1}{K},\, T \right). \end{align*} Based on the given condition, ... 0 A call lets to purchase one unit of underlying for some strike price x. So a call on GBP in USD lets us buy 1 unit of GBP for price x. However, since this is FX, lets clarify this to be USD x and USD 1 gets us GBP 1/x. A put lets you sell one unit of underlying for some strike price y (= 1/x). So a put on USD in GBP lets us sell 1 unit of USD for price 1/x. ... 0 In foreign exchange a contract can equally be seen as a put or a call, depending on the point of view: a call on dollars or a put on sterling. This is not Put-call-parity, which is not needed for this problem, it is just two names for the same thing. All you need to do is to invert the strike and convert the price to the other currency: 0.03 usd is 0.02 gbp. ... 1 What a difficult problem. The first line gave165 e^{-rt} -3 S e^{-dt} = 15$[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 ...

Top 50 recent answers are included