# Tag Info

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$\newcommand{\dd}{\, \mathrm{d}}$ If we apply Ito's lemma, then \begin{align*} \dd \xi_t &= -\xi_t \dd X_t + \frac 12 \xi_t (\dd X_t)^2\\ &= -\xi_t \left(\frac 12 \lambda_t^2 \dd t + \lambda_t \dd z_t\right) + \frac 12 \xi_t \lambda_t^2 \dd t \\ &= -\xi_t \lambda \dd z_t. \end{align*}

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I have never used calculus in my non-academic jobs. I didn't do sports, speak German (except when socializing), program Excel VBAs or directly apply what I have learned about product line management either, yet all these were taught to me at university. A lot of things you learn are indirectly useful, in that they let you see a more complete picture or that ...

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Yes, whether you do research and you need to study a new model and characterize optimal behavior, or if you are studying data and you are estimating any model that is not linear you use calculus on a regular basis. Even though computers are very powerful and can save the time and energy of doing calculations (including finding derivatives), often times ...

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Generally: $P(f\in[10,20]) = P(120 \leq S_t \leq 130) = P(S_t \leq 130) - P(S_t \leq 120)$ That is, the probability that the option is between 10 and twenty is the same that the stock is between 120 and 130. The probability that the stock is between 120 and 130 is the probability the stock is less than 130 minus the probability that it is less than 120. If ...

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(If urns are vacancies and balls are unemployed, what distinction between unemployed workers does the Red/Green dichotomy reflects?) Each ball has in front of it an identical box, each with the exact same lottery tickets, its ticket has a number on it, and each number corresponds to an urn. We say "Go!" and each ball draws "randomly" (i.e. with equal ...

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Maybe I'm familiar with a completely different background of mathematical finance, but I'll drop my two cents anyways. First, note that $$\int d Wt$$ is a stochastic integral and not easily integrable with standard methods. I hope you are preset with a tool of stochastic calculus. The following method based on no-arbitrage was first introduced by ...

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In economics, the Radon-Nikodym density $\frac{d \mathbb Q}{d \mathbb P}$ of the risk-neutral measure $\mathbb Q$ with respect to the physical measure $\mathbb P$ is the price of Arrow-Debreu securities. It is a price, not a claim. In the binomial setting, there are two AD securities, $1_u$ and $1_d$. The former entitles the holder to 1 unit of numeraire ...

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As you say (how did $\sin t$ in the initial problem become $\cos t$?), $\mathbb{Q}$ is a measure under which $W_t$ becomes $\tilde{W}_t - \sin t$, where $\tilde{W}_t$ is a $\mathbb{Q}$-Brownian motion. So Girsanov's theorem would say $$L_t = E[\frac{ d \mathbb{Q} }{ d \mathbb{P} }|\mathcal{F}_t] = e^{ - \int_0^t \sin s dW_s - \frac{1}{2} \int_0^t \sin^2 s ... 2$$dx_t = \mu dt + \sigma dz_t \\ y_t = f(t, x_t) A key idea here is that $\left( dx_t \right)^2=\left( \ldots \right)dt^2 + \left(\ldots\right) dzdt + \sigma^2 dz_t^2 = \sigma^2 dt$. The loose reasoning is that $\left( dz_t\right)^2 = dt$ and all the other terms (i.e. $dt^2$ and $dz\, dt$) are infinitely smaller than $dt$. Horribly loose intuition for $... 1 I will try to answer to your last question. I did not read the paper but in models with higher dimensions, it is always difficult to find an analytical solution. If there exists an analytical solution (for a very basic model with a one-state variable), it is possible to derive the initial conditions for your control and state variable from your differential ... 1 This is a complement/comment to Alecos' answer, who said that Note that the requirement that the probabilities are Uniform, impose the condition that, if we want to have proper distributions, the number of urns must be finite Denote the total size of the world as$N$. Denote urns as$U_N$, balls as$B_N$, and say that we randomly toss every ball ... 1 Red ball solo For any red ball in some urn, the probability that another red ball misses that urn is$1-\frac{1}{U}$. The probability that all other red balls miss that urn is the product of the chance they all miss that urn, or$\left(1-\frac{1}{U}\right)^{R-1}$. Red ball super ball The balls are symmetric, so$\Pr[\text{a red ball is a superball}] = \...

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