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6

At the first glance, what you are asking for is a model admitting arbitrage, so there is a zero chance of losing money and positive chance of yielding profits. Well, many equilibrium models start with assuming arbitrage is not possible (otherwise it would be trivial wouldn't it). But, in my opinion, what you actually seek is the Efficient Markets ...


6

Let $\tau = T-t$. Then \begin{align*} S_T = S_t e^{(\mu - \frac{1}{2}\sigma^2) \tau + \sigma \sqrt{\tau}\, Z}, \end{align*} where $Z$ is a standard normal random variable, independent of $\mathcal{F}_t$. Moreover, \begin{align*} E\left(S_T 1_{\{S_T >K\}}\mid \mathcal{F}_t \right) &= E\left(S_t e^{(\mu - \frac{1}{2}\sigma^2) \tau + \sigma \sqrt{\tau}\, ...


5

Swap Just to be clear, (3.4c) leads to (3.5a) when we assume lognormal $R(\tau)$. Lognormal $R(\tau)$ means we can write $$R(\tau) = R_0 e^{-\frac{1}{2}\sigma^2 \tau + \sigma \sqrt{\tau} Z}$$ with $Z$ normal, and I'm assuming a zero mean -- which I think is required. Then for (3.4c) we have for the expectation value: $$ E\left[(R(\tau) - R_0)^2 \right] = ...


4

Trinomial trees give incomplete markets so there is a range of possible risk neutral prices. So you have to find the possible probabilities that make the tree risk-neutral and see what prices you get. You have the correct expressions. Now just have to parametrize the set of solutions. It is one-dimensional and all the probabilities are positive so you need ...


4

Quick answer The payoff you mention is that of a call spread, i.e. long a call $C_1$ struck at $K_1$ and short a call $C_2$ struck at $K_2$, with $K_2>K_1$. The price of the instrument is therefore: $V = C_1 - C_2$. [First way] If you are stuck because this payout seems 'unsual' to you, an easy way to reach your goal (assuming you know how to use ...


4

Sorry, but despite being used as a popular example in machine learning, no one has ever achieved a stock market prediction. It does not work for several reasons (check random walk by Fama and quite a bit of others, rational decision making fallacy, wrong assumptions ...), but the most compelling one is that if it would work, someone would be able to become ...


3

Use Dynkin's formula to write the expectation: $\mathbb{E}[e^{-r\tau} \phi(S_\tau)]= g(S_0)+\mathbb{E}[\int_ 0 ^ \tau (A g -rg) dt]$ where $\phi$ is the payoff. Use the infinitismal generator $A$ to derive an ODE which describes the solution Use the fact that American options must be equal to or greater than their intrinsic value to derive boundary ...


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 ...


2

For a martingale $\{M_t \mid t\geq 0\}$ and the stochastic integral \begin{align*} I_t = \int_0^tZ_s dM_s, \end{align*} we have that \begin{align*} E((I_t)^2) = E\bigg( \int_0^tZ_s^2 d\langle M\rangle_s\bigg), \end{align*} where $\langle M\rangle$ is the quadratic variation. That is, the ito's isometry holds for a martingale integrator only. However, in ...


2

The claim payoff you describe, $g(M)$, looks to me like a tight butterfly spread that pays off only in one state of the world. Can't you just replicate that by short two calls with strike $K_0$ and long two calls, with strikes one either side at $K_0\pm 1$? Then the price of your option would be $C(K_0+1)+C(K_0-1)-2\cdot C(K_0)$. This is effectively the ...


2

This will depend on the nature of your tree. For a re-combining binomial tree, the number of nodes, including the initial one, will be \begin{align*} \sum_{i=1}^n i = \frac{n(n+1)}{2}. \end{align*} For the paths, as at each time $j$, there are two possibilities from each node, the total path number is $2^n$.


2

For question a). From the assumptions, in particular, that $R=0$, \begin{align*} \pi_l + \pi_m + \pi_u &=1\\ \frac{1}{2}\pi_l + \pi_m + 2\pi_u&=1. \end{align*} Set $\pi_m=x$, and solve for $\pi_l$ and $\pi_u$, \begin{align*} \pi_l &= \frac{2}{3}(1-x)\\ \pi_m &= x\\ \pi_u &= \frac{1}{3}(1-x), \end{align*} where $0<x<1$. The option ...


2

We assume the following Black-Scholes equation: \begin{align} \frac{\partial V}{\partial t} = -\frac{\sigma^2 S_t^2}{2}\frac{\partial^2 V}{\partial S_t^2} -r S_t \frac{\partial V}{\partial S_t} +r V.\tag{1} \end{align} From the assumption, \begin{align} V(t,\, S_t) = e^{rt}\tilde{V}(t,\, \tilde{S}_t).\tag{2} \end{align} Then \begin{align*} \frac{\partial ...


2

Note that $(S_T-K)^+ -S_T \le 0$, By the dominance principle, \begin{align*} 0 &\ge E\left(\frac{S_T-K)^+ -S_T}{e^{rT}}\right)\\ &= E\left(\frac{S_T-K)^+}{e^{rT}}\right) - E\left(\frac{S_T}{e^{rT}}\right)\\ &=C(T, K, S)-S. \end{align*} That is, \begin{align*} C(T, K, S) \le S. \tag{1} \end{align*} On the other hand, since \begin{align*} ...


2

To express such payoff in mathematical form, it is better to use indicator functions. I assume that the bottom of graphs (i.e., the vertex for the left one and the bottom segment for the right side one) represents zero. For the left-hand one, the payoff is given by \begin{align*} (K-S_T)\pmb{1}_{S_T \le K} + (S_T-K)\pmb{1}_{S_T \ge K} = (K-S_T)^+ + ...


2

For this type of question, you basically need only to write the payoff with certain indicator functions. In particular, for the above payoff, we have that \begin{align*} \textrm{Payoff} &= K\, 1_{S_T \le K} + (2K-S_T)\,1_{K < S_T \le 2K}\\ &=K\, 1_{S_T \le K} + (2K-S_T)\big(1_{S_T \le 2K} - 1_{S_T \le K} \big)\\ &=(2K-S_T)\,1_{S_T \le 2K} - ...


2

Another take on the question which uses stochastic calculus [Digression] Assume deterministic and constant rates without loss of generality. Also assume the absence of arbitrage opportunities and market completeness Let $B_t$ denote the time-$t$ value of a risk-free money market account in which 1 unit of currency $C$ has been invested at $t=0$: ...


1

For a standard European option (i.e. non path - dependent payoff): $$V_0 = \frac {1}{1+R} E [ V (S_T) ] $$ Because, in a 2 period binomial tree, the terminal stock price $S_T$ can take 3 distinct values: $S_{uu}=S_0u^2$, $S_{ul}=S_0ul$ and $S_{ll}=S_0l^2$, you can write the expectation: $$V_0 = \frac {1}{1+R} (q_{u}^2 V (S_{uu}) + {\color {red}{2}} q_u ...


1

I think you mix up marginal law, and law of the process. Your $Z_k$ must have three values, you just have to write the value of $\ln(H_k)$ for each possible value Let $X$ taking values $(x_1,x_2,...,x_n)$ and $p_i=P(X=X_i)$, then $P(f(X)=f(x_i))=\sum_{j=1}^n p_j\mathbf{1}_{f(x_j)=f(x_i)}$ if $f$ is a one-to-one mapping, you get $f(x_j)=f(x_i)\Rightarrow ...


1

your statement is quite imprecise. See https://en.wikipedia.org/wiki/Central_limit_theorem With : $(Z_k)_{k=1\dots n}$ i.i.d with $\mathbb{E}\left[Z_1\right] = \mu$ and $\text{Var}(Z_1)=\mathbb{E}\left[Z_1^2\right] -\mu^2=\sigma^2$ and by denoting $\mathcal{N}(m,v)$ a normal variance with mean $m$ and variance $v$ we have : $$ ...


1

In your answer, you don't include dividend. I am sorry to say it is wrong. Payoff function is $$ g(S_T) = (S_T - K_1)_+ - 2(S_T - \frac{K_1+K_2}{2})_+ + (S_T - K_2)_+ $$ BS pricing formula with dividend gives $$ V(t=0,S) = e^{-r}E(g(\tilde{d}S_T)) = \tilde{d} \left(BS_{call}\left(\frac{K_1}{\tilde{d}}\right) - 2BS_{call}\left(\frac{K_1+K_2}{2 ...


1

Let \begin{align*} V(t, S_t) = E\Big(e^{-r(T-t)} g(S_T)\mid \mathcal{F}_t \Big) \end{align*} be the risk-neutral value at time $t$ of the option payoff $g(S_T)$. Then $\{e^{-rt}V(t, S_t), 0 \le t \le T\}$ is a martingale. Consequently, \begin{align} -rV + \frac{\partial V}{\partial t} + (r-q)S\frac{\partial V}{\partial S_t}+\frac{1}{2}\sigma^2 S_t^2 ...


1

All these strategies can indeed be synthesised using simple instruments such as zero coupon bonds and European vanilla options (such as call and puts). Some info can be found here http://www.investopedia.com/slide-show/options-strategies/... although many other sites can provide relevant information, e.g. this one http://optioncreator.com/long-butterfly, ...


1

Here, we assume that the bottom is zero and the top is $K_2-K_1$. Then, in mathematical form, the ${\color{blue} {blue}}$ option payoff is given by \begin{align*} & \ (K_2-K_1)\pmb{1}_{S_T \le K_1} + (K_2-S_T)\pmb{1}_{K_1 < S_T \le K_2} \\ =& \ (K_2-K_1)\pmb{1}_{S_T \le K_1} + (K_2-S_T)\left(\pmb{1}_{S_T \le K_2} - \pmb{1}_{S_T \le K_1}\right)\\ ...


1

Form a smooth convex function, the second derivative is always non-negative. In particular, for any $\varepsilon >0$, \begin{align*} V(S) &= V\Big(\frac{1}{2}(S+\varepsilon ) + \frac{1}{2}(S-\varepsilon )\Big)\\ &\le \frac{1}{2}\Big(V(S+\varepsilon )+ V(S-\varepsilon) \Big). \end{align*} That is, $$V(S+\varepsilon )+ V(S-\varepsilon) - 2 V(S) ...


1

We show that \begin{align*} Y_t^i = \frac{1}{1+R}E\big( Y_{t+1} \mid S_t = S_t(i)\big).\tag{1} \end{align*} Note that \begin{align*} Y_{t+1} &= S_{t+1} \Delta_{t+1}(S_{t+1})\\ &=\frac{V_{t+2}(S_{t+1}u) - V_{t+2}(S_{t+1}l)}{u-l}. \end{align*} Then \begin{align*} \frac{1}{1+R}E\big( Y_{t+1} \mid S_t = S_t(i)\big) &= ...


1

For part a). As you posted, \begin{align*} (\pi_1+\pi_2)D_1 + (\pi_3+\pi_4)D_2 = \frac{D_1+D_2}{2}.\tag{1} \end{align*} Moreover, \begin{align*} \pi_3+\pi_4 = 1 - (\pi_1+\pi_2).\tag{2} \end{align*} Then \begin{align*} (\pi_1+\pi_2)(D_1-D_2)=\frac{D_1-D_2}{2}. \end{align*} That is, $\pi_1+\pi_2=1/2$. Similarly, $\pi_1+\pi_3=1/2$.


1

I believe a few things need to be said here. First, returns are usually calculated (END_VALUE-BEGIN_VALUE)/BEGIN_VALE. There are other ways, but this is what is usually used, and much arguments can be had on what "value" actual is. Second, data frequency should be aligned so daily standard deviation should be aligned to daily expected returns. Third, the ...


1

Let $B_t$ be the value of the risk-free asset at time $t$. Then $B_0=1$ and $B_{t+1} = (1+R) B_t$. Moreover, let $\beta_t$ be units invested in the risk-free asset at time $t$. It is clear that $\beta_0 = w_0 - \Delta_0 S_0$. Since the strategy is self-financing, \begin{align*} \Delta_{t-1} S_{t-1} + \beta_{t-1} B_{t-1} = \Delta_t S_{t-1} + \beta_t ...


1

This doesn't really suffice as an existence proof, but you can start with a series of mathematical results collectively known as no free lunch theorems. The linked paper proves the average performance of any optimization algorithm over arbitrary problem domains is independent of the algorithm. That is, no single algorithm can ever be better than others on ...



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