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5

The fourier transform is $$\hat{w}_c= \int_{-\infty}^\infty (\sqrt{y}-K)^+ e^{-i\phi y}dy = \int_{K^2}^\infty (\sqrt{y}-K) e^{-i\phi y}dy$$ Now do a change of variable with $t=\sqrt{i\phi y}$ and solve the resulting integral.

3

Formally, a long call payoff can be split as follows: $$(S_T-K)^+ = (S_T-K)\cdot 1_{\{S_T>K\}}$$ $$= S_T\cdot 1_{\{S_T>K\}} - K\cdot 1_{\{S_T>K\}},$$ that is, long an asset-or-nothing digital call payoff and short a cash-or-nothing digital call payoff. Here, $1_A$ is $1$ if event $A$ takes place, and it is $0$ otherwise.

3

We should be able to replicate the payoff exactly in each of the two regions $S_{T}\leq k_{1}$ and $S_{T}\geq k_{2}$. From the first, $$a_{0}+a_{1}S_{T}+a_{3}(k_{2}-S_{T}) =S_{T}$$ so, matching coefficients, $a_{0}+a_{3}k_{2}=0$ and $a_{1}-a_{3}=1$. From the second, $$a_{0}+a_{1}S_{T}+a_{2}(S_{T}-k_{1})=P_{0}$$ so, matching coefficients, $a_{0}-a_{2}k_{1}=P_{... 2 From @noob2's link, it looks like the product behaves like a basket of a long position in the underlying and short an out-of-the-money call option; thus the discount vs the price of the underlying is the implicit value of the call. As with any option, if you sell one and it stays out of the money, you get to keep the premium, so if the price of the ... 2 You can check my answer to this question for general details on how to solve this kind of problem. Let$C_X(S_T)$and$P_Y(S_T)$be a call and a put option with strikes$X$and$Yrespectively, then: \begin{align} (\text{i}) \quad V_T &= (S_T-A)1_{\{A\leq S_T\leq B\}}+(B-A)1_{\{S_T>B\}} \\ &=(S_T-A)1_{\{S_T\geq A\}}+(B-S_T)1_{\{S_T\geq B\}} \\... 1 Somewhere must be a little error, here I used r=0.02 and \sigma=0.25. In black you have the payoff and in red the current price of the portfolio. Note that as the time to maturity decreases, the red line converges towards the black line. In grey and and yellow (on the secondary axis), you can see the individual call option prices which form your ... 1 For question (i), you simply buy one EU call with strike A and sell one EU call with strike B - this is called a bull call spread. Try using the put-call-parity to construct the corresponding bull put spread yourself. 1 The payoff of an European call option is (S_T-K)^+. At maturity, if the spot price is greater than (or equal to) the strike price, then holding an asset-or-nothing call option has payoff S_T, writing a cash-or-noting call option K, which together give the payoff of the European call in this scenario. If the spot price is less than the strike, then all ... 1 Suppose you would like to compute \begin{align} Q_1(x_1,x_2;B) &= \Bbb{E}[X_1\max(B-X_2,0)]\\ Q_2(x_1,x_2;B) &= \Bbb{E}[X_2\max(X_1-B,0)] \end{align} where you know the marginal probability density functions p_{X_1}(u) and p_{X_2}(v). Let's start by focusing on Q_1. By definition, the expectation equivalently writes: Q_1(x_1,x_2;B) = \... 1h(\theta)=\nabla H(\theta)=\mathbb{E}\left[(\theta-Z)G^2(Z)e^{-\theta Z+\frac{1}{2}\theta^2}\right]\$ so just take a bunch of paths and evaluate $$(\theta-Z)G^2(Z)e^{-\theta Z+\frac{1}{2}\theta^2}$$on them and take the average.

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You can think about them as noise traders in the sense of Glostem and Milgrom (1985). It it is a fairly wide used assumption that there is someone out there that soaks up residual supply/demand. Usually one thinks about this guys as large mutual funds or pension funds.

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There is no advantage to spreading your entry into multiple tranches, unless you have private information (i.e. not priced-in to the market) that prices will fall first, before rising. If the security is following a random walk, with each incremental return being independent, it doesn't matter whether the price has risen or fallen since t=0. I think you ...

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