### 物理代写|电动力学代写electromagnetism代考|PHYC20014

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## 物理代写|电动力学代写electromagnetism代考|Sample Space RT with T Uncountable

The labelling set (or dimension set) $T$ in the domain $\mathbf{R}^{T}$ of (4.1) is a countable set of dimensions or labels. But the labels can be taken to be an uncountable set $\mathbf{T}$, such as the continuum $] 0,1]$, and then the domain is
$$\mathbf{R}^{\mathrm{T}}:=\prod_{s \in \mathrm{T}} \mathbf{R}{s}, \quad \mathbf{R}{s}=\mathbf{R} \text { for each } s \in \mathbf{T} \text {. }$$
With this change in meaning of $\mathbf{R}^{\mathrm{T}}$, the concepts and notation introduced for the definition (4.4) of the integral of a function $h$ in $\mathbf{R}^{\mathbf{T}}$ then carry over unchanged in the new context of uncountable $\mathbf{T}$.

The following example illustrates the use of $\int_{\mathbf{R}^{\mathbf{T}}} h$, with uncountable $\mathbf{T}$, by means of a calculation of broadly stochastic integral type.

Example 12 Suppose, at each instant $s$ of the time interval $] 0, \tau]$, a share takes random value $x(s)$ (or $\left.x_{s}\right)$. Suppose $x(0)=x_{0}=1$ (with probability 1), and suppose, at each time $s(0<s \leq \tau)$, the share price takes value $x(s)$, $0 \leq x(s) \leq 1$, with uniform probability on $[0,1]$. Suppose the value at time $s$ is independent of the value taken at any other time. Suppose an investor takes an initial shareholding $z(0)$, or $z_{0}$, of 1 share (so $\left.z_{0}=1\right)$, and suppose the shareholding or number of shares $z(s)$ varies randomly at $m$ fixed times $\tau_{k}$ between initial time 0 and terminal time $\tau$,
$$0=\tau_{0}<\tau_{1}<\tau_{2}<\cdots<\tau_{m}=\tau ; \quad M:=\left{\tau_{1}, \tau_{2}, \ldots, \tau_{m}\right} .$$
Thus $] 0, \tau]$ can be a period of $m$ days, with shareholding changing randomly at the end of each day. Suppose the random value $z_{\tau_{k}}$ (or simply $z_{k}$ ) of shares at time $\tau_{k}$ is independent of the value $z_{k^{\prime}}$ at any other time $\tau_{k^{\prime}}$, and independent of the value $x(s)$ of the share at any times. To keep things uncomplicated suppose that, at any time $\tau_{k}$, the shareholding $z_{k}$ is 1 with probability $0.5$, and $z_{k}=0$ with probability 0.5. What is the expected payout at terminal time $t$ from this shareholding?

The intention is to apply a stochastic integral calculation. But the probability distributions are deliberately chosen so that the expected payout is fairly obvious on intuitive grounds. Then it can be seen whether the stochastic integral calculation confirms what common sense indicates.

## 物理代写|电动力学代写electromagnetism代考|Stochastic Integrals for Example 12

Example 12 above has two independent stochastic processes,
$$Z_{\mathrm{T}} \simeq z_{\mathrm{T}}\left[\mathbf{R}^{\mathbf{T}}, F_{Z_{\mathrm{T}}}\right], \quad X_{\mathrm{T}} \simeq x_{\mathbf{T}}\left[\mathbf{R}^{\mathrm{T}}, F_{X_{\mathrm{T}}}\right]$$
with joint process $\left(Z_{\mathbf{T}}, X_{\mathrm{T}}\right)$ expressed by
$$\left(Z_{\mathbf{T}}, X_{\mathrm{T}}\right) \simeq\left(z_{\mathrm{T}}, x_{\mathrm{T}}\right)\left[(\mathbf{R} \times \mathbf{R})^{\mathbf{T}}, F_{\left(Z_{\mathrm{T}}, X_{\mathrm{T}}\right)}\right]$$

where each sample path $z_{\mathrm{T}}=(z(s): 0<s \leq \tau)$ is constant for $\tau_{j-1} \leq s<\tau_{j}$, $(1 \leq j \leq m)$.

The calculation in $(4.10)$ enabled us to disregard the random variation in $x(s)$ for $\tau_{j-1}<s<\tau_{j},(1 \leq j \leq m)$, so the joint processes can be expressed as
$$\left(Z_{M}, X_{M}\right) \simeq\left(z_{M}, x_{M}\right)\left[(\mathbf{R} \times \mathbf{R})^{M}, F_{\left(Z_{M}, X_{M}\right)}\right]$$
and the latter formulation enabled us to perform a calculation for the expected gain in portfolio value (or shareholding value).

Stochastic integrals $\int_{0}^{\tau} \cdots$ on domain $\left.\left.\mathbf{T}=\right] 0, \tau\right]$ can be formulated from version (4.16). The objective is to express the gains (or losses) in portfolio value $w(t)(0<t \leq \tau)$ in terms of joint sample paths $(z(s), x(s))(0<s \leq \tau)$ of the joint process $\left(Z_{\mathbf{T}}, X_{\mathrm{T}}\right)$.
\begin{aligned} &w(t)=\sum_{0 \leq s<s^{\prime} \leq t} z(s)\left(x\left(s^{\prime}\right)-x(s)\right), \quad \text { or } \ &w(t)=\int_{0}^{t} z(s) d x(s), \quad(0<t \leq \tau) . \end{aligned}
Thus $w(t)$ depends on the joint outcomes $((z(s), x(s)): 0<s \leq t)$, or $w(t)=$ $h\left(z_{\mathbf{T}}, x_{\mathbf{T}}\right)$ where $\left.\left.\mathbf{T}=\right] 0, t\right]$ and $h$ is the deterministic function given by the Stieltjes integral $\int_{0}^{t} z(s) d x(s)$ – if and when the latter integrals exist. These integrals are sample path versions of a stochastic integral $\int_{0}^{t} Z(s) d X(s)$, and can be examined further, in terms of particular sample paths, in order to try to understand whether or not they exist, and what other kinds of issues can arise with them.

Before undertaking this task, the random variability in the outcomes $(w(t)$ : $0<t \leq \tau)$ can be examined. Write $U_{\mathbf{T}}$ for the joint processes $\left(Z_{\mathbf{T}}, X_{\mathrm{T}}\right)$, so a sample path for $U_{\mathbf{T}}$ is $u_{\mathbf{T}}=\left(z_{\mathbf{T}}, x_{\mathbf{T}}\right)$. Then (4.16) gives
$$W_{\mathbf{T}}=h\left(U_{\mathbf{T}}\right) \simeq h\left(u_{\mathbf{T}}\right)\left[(\mathbf{R} \times \mathbf{R})^{\mathbf{T}}, F_{U_{\mathbf{T}}}\right],$$
where $F_{U_{\mathbf{T}}}$ is the joint distribution function $F_{\left(Z_{\mathbf{T}}, X_{\mathrm{T}}\right)}$ mentioned in (4.16), and $h\left(U_{\mathbf{T}}\right)$ is the stochastic integral $\int_{0}^{t} Z(s) d X(s)$. Thus $W_{\mathbf{T}}$ is a contingent process depending on the joint values of the processes $Z_{\mathrm{T}}, X_{\mathrm{T}}$.

The details of $F_{\left(Z_{M}, X_{M}\right)}$ were described above, but not those of $F_{\left(Z_{T}, X_{\mathrm{T}}\right)}$. These are provided in Section $4.4$ below.

## 物理代写|电动力学代写electromagnetism代考|stochastic integrals

Example 13 demonstrates that stochastic integrals can fail to exist even in the relatively simple case of Example 12. It has been pointed out that (4.15) delivers $\mathrm{E}[W(\tau)]$ without constructing stochastic integrals. The joint distribution function used in that calculation is $F_{U_{M}},=F_{\left(Z_{M}, X_{M}\right)}$ as in (4.17).

Compare this with (4.8), where the sample space for the elementary form of the random variable $W(\tau)$ is $\mathbf{R}$, with distribution function $F_{W_{\tau}}$. The difference between the elementary and joint basic representations of joint random variability was described in Section 3.5. In the elementary format (4.8), the expected value of $W(\tau)$ is obtained by integration on $\mathbf{R}$ with respect to $F_{W_{\tau}}$. Likewise for any contingent observable $f(W(\tau))$ that might arise.

In contrast, $(4.20)$ and $(4.16)$ employ $(\mathbf{R} \times \mathbf{R})^{\mathbf{T}}$ as sample space for $W_{\tau}$; and the joint distribution function is $F_{U_{T}},=F_{\left(Z_{T}, X_{T}\right)}$ (or $F_{Z_{T} X_{T}}$ ); but this distribution function was left unspecified. At this point it is useful to pursue this approach to $\mathrm{E}[W(\tau)]$ a bit further. Here is a summary of what is involved:

• With $W(\tau) \simeq w(\tau)\left[\mathbf{R}, F_{W(\tau)}\right]$, expected gain in portfolio value at time $\tau$ can be calculated as
$$\mathrm{E}[W(\tau)]=\int_{\mathbf{R}} w(\tau) F_{W(\tau)}\left(I_{\tau}\right)$$
• With $W(\tau)=f\left(X_{M}, Z_{M}\right) \simeq f\left(x_{M}, z_{M}\right)\left[(\mathbf{R} \times \mathbf{R})^{M}, F_{X_{M} Z_{M}}\right]$, expected gain in portfolio value at time $\tau$ can be calculated as
\begin{aligned} \mathrm{E}[W(\tau)] &=\mathrm{E}\left[f\left(X_{M}, Z_{M}\right)\right] \ &=\int_{(\mathbf{R} \times \mathbf{R})^{M}} f\left(x_{M}, z_{M}\right) F_{X_{M} Z_{M}}\left(I_{X_{M}}(M) \times I_{Z_{M}}(M)\right) \end{aligned}
Interval notation $I_{X_{M}}, I_{Z_{M}}$, refers to events (or sets of potential occurrences) of random variables $X_{M}, Z_{M}$, respectively.

## 物理代写|电动力学代写electromagnetism代考|Sample Space RT with T Uncountable

$$\mathbf{R}^{\mathrm{T}}:=\prod_{s \in \mathrm{T}} \mathbf{R} {s}, \quad \mathbf{R } {s}=\mathbf{R} \text { 对于每个 } s \in \mathbf{T} \text {。}$$

0=\tau_{0}<\tau_{1}<\tau_{2}<\cdots<\tau_{m}=\tau ; \quad M:=\left{\tau_{1}, \tau_{2}, \ldots, \tau_{m}\right} 。0=\tau_{0}<\tau_{1}<\tau_{2}<\cdots<\tau_{m}=\tau ; \quad M:=\left{\tau_{1}, \tau_{2}, \ldots, \tau_{m}\right} 。

## 物理代写|电动力学代写electromagnetism代考|Stochastic Integrals for Example 12

(从吨,X吨)≃(和吨,X吨)[(R×R)吨,F(从吨,X吨)]

(从米,X米)≃(和米,X米)[(R×R)米,F(从米,X米)]

## 物理代写|电动力学代写electromagnetism代考|stochastic integrals

• 和在(τ)≃在(τ)[R,F在(τ)], 投资组合价值的预期收益τ可以计算为
和[在(τ)]=∫R在(τ)F在(τ)(我τ)
• 和在(τ)=F(X米,从米)≃F(X米,和米)[(R×R)米,FX米从米], 投资组合价值的预期收益τ可以计算为
和[在(τ)]=和[F(X米,从米)] =∫(R×R)米F(X米,和米)FX米从米(我X米(米)×我从米(米))
间隔符号我X米,我从米, 指随机变量的事件（或潜在发生的集合）X米,从米， 分别。

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