9

Most of the literature on "Strategic experimentation" (or Bandits) uses Poisson processes. Here players can use either a risky or safe arm and one of them generates a fixed stream of payoffs (usually the safe arm) the other one generates lump-sum payments whose arrival times are described by a Poisson process. Some examples are: Klein, Nicolas, and Sven ...


5

Suppose $B$ is an $n\times n$ matrix of Poisson transition rates, where $B_{ij}\geq 0$ for $i\neq j$ denotes the rate at which state $i$ transitions to state $j$, and $B_{ii}\leq 0$ gives the rate at which state $i$ transitions to all other states. Each row of $B$ sums to 0. Then if $p(t)$ denotes the probability distribution at time $t$, by definition of $...


5

To show that transformation is measure preserving you need to show that full preimage $S^{-1}(A)$ of any set A in the Borel $\sigma$-field on [0,1) is again in the same $\sigma$-field, i.e. it is measurable, and that $Pr\{S^{-1}(A)\} = Pr\{A\}.$ As you have correctly pointed out, it will suffice to show it for any base of usual topology, namely, all open ...


4

This construction you describe is not fully general. In fact it characterizes strictly stationary time series. You see that it's shift-invariant. This operator $S$ is essentially a shift operator. For comparison, here's the usual definition of, let's say discrete-time, processes: Definition A stochastic process is a sequence $\{ X_t \}$ of Borel measurable ...


3

First, you are in fact given $p$. You can think of return as a security that costs 1 in current period and pays off $R$ in the next period. The price vector and the payoff matrix are thus $$ p = \begin{bmatrix} 1 \\ 1 \end{bmatrix}, \ X = \begin{bmatrix} 1.1 & 1.0 & 0.9 \\ 1.0 & 1.0 & 1.0 \end{bmatrix} $$ and we need to find positive vector ...


3

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 ...


3

Klette and Kortum (2004) develop a parsimonious model of innovating firms rich enough to confront firm-level evidence. It captures the dynamics of individual heterogenous firms, describes the behavior of an industry with simultaneous entry and exit, and delivers ageneral equilibrium model of technological change. At the root of the model is a Poisson process ...


2

It is just an opaque way to denote that the non-definite integration limit will take the value of the variable itself. Namely, $$\int_a g(x) dx \equiv \int_a^x g(s)ds$$ ... where $s$ is just the dummy variable of integration. Since sometimes we use the notation $\int _D f(x) dx$ to denote fully definite integration "over the domain $D$ of $x$", I would ...


2

Let $G$ be the cumulative distribution function for the buyers' willingness to pay. If there is no cost, expected profit is simply expected revenue. The revenue is quantity times price. If the unit of the good is actually sold, this is just the price. Now the probability that the good is sold at price $p$ is the probability that the willingness to pay is not ...


2

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

Can't comment, or I'd ask for more specifics first. If you're trying to convert an AR(1) process fitted against a discrete time series to a continuous time process, I found a relevant resource here on page 4. The computations are provided for estimating the coefficients of a CAR(2) process from an AR(2) process, but of course you can substitute a 0 for the ...


2

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 ...


2

Using the given formulas (inserting the expression for the $X$'s in the sum), we arrive at $$Y_t = (1-\beta_0)\alpha_0\cdot t + \beta_0\sum_{j=0}^{t-1}X_j +\sum_{j=1}^{t}W_j $$ We do not need a theorem to obtain that, given the description of the problem. Manipulating, $$Y_t= (1-\beta_0)\alpha_0\cdot t - \beta_0(X_t-X_0) + \beta_0Y_{t} +\sum_{j=1}^{t}...


1

Usually additive functionals are defined for (strong) Markov processes with continuous sample paths (diffusions) but I suppose you do have a Markov---AR(1)---time series and $\{ Y_t \}$ is indeed additive. So, in your case, $$ X_t = a_0 + a_1 X_{t-1} + W_t, $$ and you would like \begin{align*} Y_t - Y_{t-1} &= r_1 + (M_t - M_{t-1}) + r_2 (X_t - X_{t-1}...


1

He's just thinking of $\mathbb{S}$ as being a deterministic and $\omega$ as being unobservable. Then we observe $X(\omega)$ as a form of incomplete information about $\omega$. $\mathbb{S}$ and $X$ then help us deduce a joint probability distribution over $\{X_t\}_{t=0}^\infty$.


1

It is possible to consider cases of $\omega$ being a point in infite dimensional space, e.g. sequence of shock, but such interpretation would be unproductive, as then you will get no simplifications when compared to direct specification of the process on filtered probability space and only produced unwanted additional entities to complicate matters. This ...


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