To perform simple calculations to compute certain quantities relating to Brownian motion, and to understand how these quantities can be important in pricing financial derivatives

这是一份Bath巴斯大学MA50089作业代写的成功案

In the univariate two-way model, we measure one dependent variable $y$ on each experimental unit. The balanced two-way fixed-effects model with factors $A$ and $B$ is
$$
\begin{aligned}
y_{i j k} &=\mu+\alpha_{i}+\beta_{j}+\gamma_{i j}+\varepsilon_{i j k} \
&=\mu_{i j}+\varepsilon_{i j k}, \
i &=1,2, \ldots, a, \quad j=1,2, \ldots, b, \quad k=1,2, \ldots, n
\end{aligned}
$$
where $\alpha_{i}$ is the effect (on $y_{i j k}$ ) of the $i$ th level of $A, \beta_{j}$ is the effect of the $j$ th level of $B, \gamma_{i j}$ is the corresponding interaction effect, and $\mu_{i j}$ is the population mean for the $i$ th level of $A$ and the $j$ th level of $B$. In order to obtain $F$-tests, we further assume that the $\varepsilon_{i j k}$ ‘s are independently distributed as $N\left(0, \sigma^{2}\right)$.

Let $\bar{\mu}{i .}=\sum{j} \mu_{i j} / b$ be the mean at the $i$ th level of $A$ and define $\bar{\mu}{. j}$ and $\bar{\mu}{. .}$ similarly. Then if we use side conditions $\sum_{i} \alpha_{i}=\sum_{j} \beta_{j}=\sum_{i} \gamma_{i j}=\sum_{j} \gamma_{j}=0$,

MA50089 COURSE NOTES ：

0 can be tested by $T^{2}$ or any of the four MANOVA test statistics. To test $H_{0}: \sum_{i} c_{i} \overline{\boldsymbol{\mu}}{i}=\mathbf{0}$, for example, we can use $$ T^{2}=\frac{n b}{\sum{i=1}^{a} c_{i}^{2}}\left(\sum_{i=1}^{a} c_{i} \overline{\mathbf{y}}{i . .}\right)^{\prime}\left(\frac{\mathbf{E}}{v{E}}\right)^{-1}\left(\sum_{i=1}^{a} c_{i} \overline{\mathbf{y}}{i . .}\right) $$ which is distributed as $T{p, v_{E}}^{2}$ when $H_{0}$ is true. Alternatively, the hypothesis matrix
$$
\mathbf{H}{1}=\frac{n b}{\sum{i=1}^{a} c_{i}^{2}}\left(\sum_{i=1}^{a} c_{i} \overline{\mathbf{y}}{i . .}\right)\left(\sum{i=1}^{a} c_{i} \overline{\mathbf{y}}{i . .}\right)^{\prime} $$ can be used in $$ \Lambda=\frac{|\mathbf{E}|}{\left|\mathbf{E}+\mathbf{H}{1}\right|}
$$

To perform simple calculations to compute certain quantities relating to Brownian motion, and to understand how these quantities can be important in pricing financial derivatives.

这是一份Bath巴斯大学MA30089作业代写的成功案

Let $(P(\cdot \mid X=x)){x \in E}$ be a regular version of the conditional probability with respect to $X$. Then, for any $Y \in \mathcal{L}^{1}(\Omega, \mathcal{F}, P)$ : $$ \int Y(\omega) d P(\omega \mid X=x)=E[Y \mid X=x], \quad P{X}-a . s .
$$
Proof: First, we observe that $Y$, being a random variable, is measurable. ${ }^{3}$ Now from (1.6) it follows that
$$
E\left[I_{F} \mid X=x\right]=P(F \mid X=x)=\int I_{F}(\omega) P(d \omega \mid X=x)
$$
for every $x \in E, P_{X}(x) \neq 0$. Now let $Y$ be an elementary function so that
$$
Y=\sum_{i=1}^{n} \lambda_{i} I_{F_{i}}
$$

MA30089 COURSE NOTES ：

If $Y:(\Omega, \mathcal{F}) \rightarrow\left(\mathbb{R}, \mathcal{B}{\mathbb{R}}\right)$ is a positive random variable, then $$ E[Y \mid X=x] \geq 0, \quad P{X}-a . s .
$$
Proof: Since
$$
\int_{[X \in B]} Y d P \geq 0 \quad \forall B \in \mathcal{B}
$$
it follows that
$$
\int_{B} E[Y \mid X=x] P_{X}(d x) \geq 0 \quad \forall B \in \mathcal{B}
$$
and therefore $E[Y \mid X=x] \geq 0$, almost surely with respect to $P_{X}$.