# Tag Info

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In general, quantitative finance requires mathematics, finance, and numerical programming. The mix of the three and the areas of focus within the three will depend on the particular area you intend to work in. For example, option pricing, risk, and asset management are all related but derivative modeling would draw more on stochastic processes and ...

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Question 1. Actually, the assumption of trade data format is that you have timestamp, size and price (not bid/ask) of trade. Sometimes, trades(ticks) are included to Level 1 data (also called BBO) which assumes bid and ask information. However, bars are constructed on trades, not quotes. Question 2. Yes, T value is derived from equation 3. The process is ...

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It is correct that $$\mathbf{P}(t^{-1/2}W(t) \in[a,b])=Φ(b)−Φ(a), \forall t\in(0,\infty)$$ due to the stationary increments property of the Wiener process and the fact that you normalized the random variable by dividing by its standard deviation. $\mathbf{P}$ is a probability measure on an abstract space, not a random variable. Hence, you probably mean ...

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You seem to use the term "volatility" to describe two very different quantities: (1) the diffusion coefficient of your SDE and (2) the standard deviation of the log-returns under your modelling assumptions. While the first may be negative, the second may not. [Interpretation 1] Consider a probability space $(\Omega,\mathcal{F},\mathbb{P})$ and a standard ...

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In the black scholes model, today's stock price, risk free rate and stock volatility are considered independent variables. They are inputs to the model. Hence the cross partial derivatives are zero. You could invent a model where you tried to explain the current stock price in terms of risk free rate and volatility. Then indeed the partial derivatives ...

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Assuming standard BS dynamics for $S_t$, you have $\frac{Z_t}{Z_0}\equiv(\frac{S_t}{S_0})^p = \exp{((rt-\frac{\sigma^2}{2}t+\sigma W_t)(1-\frac{2r}{\sigma^2}))}$ Now, this is a lognormally distributed r.v. and the Gaussian inside the exponential has mean and variance $m=2rt - \frac{\sigma^2}{2}t -\frac{2r^2}{\sigma^2}t$ $v = (\sigma-\frac{2r}{\sigma})^2 ... 4 90.422798 is not the correct value for price for Benchmark 18A. If you were the original purchaser of the book in 1993 you would have received an errata correcting that benchmark. If you have a second printing from 1996 it contains the correct value for price of 99.422450. I hope that helps. 3 It is my understanding that a replicating portfolio for a put involves short selling stock and lending money. You cannot statically replicate an option. So this is not true in general, you'll need to re-balance your replicating portfolio (underlying + cash) dynamically if you want to replicate the option. This will imply sometimes buying stock and borrowing ... 3 It is a very badly worded question in my humble opinion. There are three "prices" to contend with. (1) If you want to buy a stock and pay for it now, you pay the current stock price S. (2) If you want to buy a stock and not have to pay for it until a future delivery date T, then you enter into a "forward" or (in the United States) a "futures contract" ... 3 The standard starting point with modelling a stock price process is to use the Black-Scholes model for the stock price. This simply asserts that the changes in the stock price are described by the following stochastic differential equation (SDE) $$\dfrac{\textrm{d}S}{S} = \mu\:\textrm{d}t + \sigma\:\textrm{d}W_t$$ where$W_t$is a standard Brownian motion (... 3 Your formula for$p$is $$p = \frac{e^{{(\alpha - \delta})h} - d}{u - d},$$ where$\alpha$is not expected return on stock but continuous risk free rate, i.e. 1%. If you use$\alpha$as 1%, you will get$p=0.009125828 $which is within$[0,1]$EDIT: With the information given in the question, it must satisfy following equality: $$S_0e^{\alpha - \delta}=... 3 It costs 0.03 dollars for the option to (sell 1 pound/buy 1.5 dollars. Now divide everything by 1.5: It costs 0.02 dollars for the option to (sell 2/3 pound / buy 1 dollar). Now convert to pounds at spot rate: It costs 0.0133 pounds for the option to (sell 2/3 pound / buy 1 dollar). Done 3 The 1 you are referring to is a vector of ones The expression (1 + R)^T \backslash 1 appears to be shorthand for a MATLAB equation such as: s = (ones(n,k) + R)' \ ones(n, 1) where R is an n by k matrix (i.e. specifying returns for n periods of k assets). 1 + R adds 1 to each element of the matrix R (eg. makes it a return like 1.02 instead of .02).... 3 I think you mixed several things up. I will try to help you out. Everything started with your claim that \Bbb E \bigl[W(T) \mid \mathcal F_t \bigr] = 0 which is wrong! if W is a Brownian notion, then$$ \Bbb E \bigl[W(T) \mid \mathcal F_t \bigr] = W(t), \quad t\leq T. $$This follows from the fact that Brownian motions are martingales. Here and in ... 3 As a starting point: for price dynamics dS(t) = rS(t)dt + \sigma S(t) dW^\mathbb{Q}(t), to show that Z(t)/Z(0) is a positive mean 1 Q-martingale, use Itô's formula to get: dZ(t)=p \sigma Z(t)dW^\mathbb{Q}(t) 3 The risk neutral measure is used to price assets (e.g. derivatives) and not to base your investment decisions on. In the first part of you question your simulation gives you the Risk-Neutral expectation of the stock at time T. If you want the expectation at time t, then why don't you just simulate from time 0 up to time t? (I might have misunderstood ... 3 Imho that's more of a probability question than finance really. If you take N attempts, then the probability of at least one (or more) failures is the complementary probability of never failing on those N attempts: p=1-p_{pass}^N Solving for p=50\%, you need to evaluate p_{pass}^{N}=0.5, where p_{pass}=0.99, getting that N\geq 69. 3 There is an open source hedge fund project which is implementing the ideas contained in the book and which has a github where you can see their code implementation of tick bars. Personally I always find it extremely enlightening to see code rather than mathematical symbolism, and maybe this will be the same for you. On the linked pages there are also links ... 3 Let P(t,T) denote the time t price of a zero-coupon bond maturing at time T and \mathbb{Q}_T be the associated equivalent martingale measure which uses P(t,T) as numeraire. Then, for any \mathcal{F}_T-measurable payoff \xi, the time t value of \xi is given by$$V_t=P(t,T)\cdot\mathbb{E}^{\mathbb{Q}_T} [\xi\mid\mathcal{F}_t].$$The ... 2 If you're lucky enough that the payment schedules (start/end dates, frequency, day count, business day adjustment etc.) are the same between the fixed leg of the interest rate swap and the "spread" leg of the basis swap, then you can simply use: OIS rate (%) = IRS Rate (%) - 0.01 * (basis spread (bps)) Otherwise, to do it accurately, you'll need to do a ... 2 What a difficult problem. The first line gave 165 e^{-rt} -3 S e^{-dt} = 15 [since 50+55+60 = 165]. In the second line we want to evaluate 110 e^{-rt} -2 S e^{-dt} . We notice that this is exactly two thirds of the left side of the above, because 110 is two thirds of 165 and 2 is two thirds of 3. So we take two thirds of the right hand side of the first ... 2 First of all I’ll work with column vectors because I find it easier than with row vectors as you did. I guess it’s a little bit easier if we modify your first equation a little bit. Notice that is really the first order condition of the following lagrangian:$$L(w, \lambda, \delta)= \frac{1}{2}{\bf w^TCw} - \lambda({\bf w^Tm} - \mu_v) - \delta({\bf w^Tu} - 1)... 2 If$PV_{t, T}(\text{Divs}) \ge K\big(1-e^{-r(T-t)}\big)$, since$P_{Eur}(S_t, K, T-t) >0, the identity \begin{align*} C_{Eur}(S_t, K, T-t) = P_{Eur}(S_t, K, T-t) + (S_t-K) -PV_{t, T}(\text{Divs}) +K\big(1-e^{-r(T-t)}\big), \end{align*} implies that \begin{align*} C_{Eur}(S_t, K, T-t) > (S_t-K). \end{align*} That is, it is not rationale to exercise the ... 2 The best explanation I came across so far is the one in Gravelle and Rees (2003) chapter 17. I could exactly write here what they state, but that would be copying. 2 Let\{X_t \mid t \ge 0\}$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, ... 2 I think they have expanded the third term on the LHS and simplified. 2 For general mathematical finance, you may start with the book Stochastic Calculus for Finance, and then the books Martingale Methods in Financial Modelling and Mathematical Methods for Financial Markets. After those preparations, you can start with some books in specialized areas such as the books Interest Rate Models by Brigo and Mercurio, Credit Risk by ... 2 Yes, your steps are valid This is a wrong use of the term "quantile". Here you need to compute a probability (through the normal cdf) and not a quantile (i.e. the value of a random variable corresponding to a given level of the cdf, e.g. the quantile 0.5 (or percentile 50%) is the median) 2 No, you can have \frac{1}{2n}\sum_{i=1}^{2n} C(S^i_T,K,T) = 0 $$First off, there's the obvious case where n=1 and u_1 = 0.5 More generally, for options way out of the money it is common to have$$ \frac{1}{n}\sum_{i=1}^{n} C(S^i_T,K,T) = 0 $$even for very large n. Antithetic sampling does not change that. 2 Here's one algebraic way to derive it:$$ \frac{(1 - e^{-\sigma\sqrt{h}})(1 + e^{\sigma\sqrt{h}})}{(e^{\sigma\sqrt{h}} - e^{-\sigma\sqrt{h}})(1 + e^{\sigma\sqrt{h}})} = \frac{1 - e^{-\sigma\sqrt{h}} + e^{\sigma\sqrt{h}} - 1}{(e^{\sigma\sqrt{h}} - e^{-\sigma\sqrt{h}})(1 + e^{\sigma\sqrt{h}})} = \frac{e^{\sigma\sqrt{h}} - e^{-\sigma\sqrt{h}}}{(e^{\sigma\sqrt{h}...

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