Tag Archives: Asset Pricing in Continuous Time代写

计算和模拟Asset Pricing in Continuous Time MATH0086

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这是一份UCL伦敦大学 MATH0086作业代写的成功案例

计算和模拟Asset Pricing in Continuous Time MATH0086
问题 1.

Solving for $x_{4}$ in the last equation yields
$$
x_{4}=\frac{6}{3}=2
$$
Using $x_{1}=2$ in the third equation, we obtain
$$
x_{3}=\frac{4-5(2)}{6}=-1
$$


证明 .

Now $x_{3}=-1$ and $x_{4}=2$ are used to find $x_{2}$ in the second equation:
$$
x_{2}=\frac{-7-7(-1)+4(2)}{-2}=-4 .
$$
Finally, $x_{1}$ is obtained using the first equation:
$$
x_{1}=\frac{20+1(-4)-2(-1)-3(2)}{4}=3 .
$$

英国论文代写Viking Essay为您提供作业代写代考服务

MATH0086 COURSE NOTES ：



Modify Program so that it will compute $A^{-1}$ by repeatedly solving $N$ linear systems
$$
A C_{J}=E_{J} \quad \text { for } J=1,2, \ldots, N
$$
Then and
$$
\begin{gathered}
A\left[\begin{array}{llll}
C_{1} & C_{2} & \ldots & C_{N}
\end{array}\right]=\left[\begin{array}{llll}
E_{1} & \boldsymbol{E}{2} & \ldots & \boldsymbol{E}{N}
\end{array}\right] \
A^{-l}=\left[\begin{array}{llll}
C_{1} & C_{2} & \ldots & C_{N}
\end{array}\right] .
\end{gathered}
$$
Make sure that you compute the $\boldsymbol{L} \boldsymbol{I}$ factorization only once!






连续时间内的资产定价 Asset Pricing in Continuous Time MATH0085

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Posted on by

这是一份UCL伦敦大学 MATH0085作业代写的成功案例

连续时间内的资产定价 Asset Pricing in Continuous Time MATH0085
问题 1.

The cost of the new portfolio $h(t)$, which has to be bought at today’s prices, is given by
$$
\sum_{i=1}^{N} h_{i}(t) S_{i}(t)=h(t) S(t)
$$
whereas the cost for the consumption rate $c(t)$ is given by $c(t) \Delta t$. The budget equation for period $t$ thus reads
$$
h(t-\Delta t) S(t)=h(t) S(t)+c(t) \Delta t .
$$


证明 .

If we introduce the notation
$$
\Delta X(t)=X(t)-X(t-A t),
$$
for an arbitrary process $\mathrm{X}$, we see that the budget equation reads
$$
S(t) \Delta h(t)+c(t) \Delta t=0
$$

英国论文代写Viking Essay为您提供作业代写代考服务

MATH0085 COURSE NOTES ：



  • The value of our old portfolio, as usual given by
    $$
    \mathrm{h}(\mathrm{t}-\Delta t) S(t)
    $$
  • The dividends eamed during the interval $(t-A t, t]$. These are given by
    $$
    \sum_{i=1}^{N} h_{i}(t-\mathrm{At})\left[D_{i}(t)-D_{i}(\mathrm{t}-\mathrm{At})\right]=\mathrm{h}(\mathrm{t}-\mathrm{At}) \Delta D(t)
    $$
    The relevant budget equation is thus given by
    $$
    \mathrm{h}(\mathrm{t}-\Delta t) S(t)+\mathrm{h}(\mathrm{t}-\Delta t) \Delta D(t)=h(t) S(t)+c(t) \Delta t .
    $$