这是一份 Imperial帝国理工大学 MATH97114作业代写的成功案例
Integrating both sides and taking exponentials of both sides we get the following expression for the unit zero coupon bond:
$$
P_{t}(x)=e^{-\int_{0}^{x} f_{t}(s) d s} .
$$
If we rewrite using the notation $x=T-t$, we get:
$$
P_{t}(x)=e^{-x r_{t}(x)}
$$
for the relationship between the discount unit bond and the spot rate. In terms of the forward rates it reads:
$$
r_{t}(x)=\frac{1}{x} \int_{0}^{x} f_{t}(s) d s .
$$
This relation can be inverted to express the forward rates as function of the spot rate:
$$
f_{t}(x)=r_{t}(x)+x \frac{\partial}{\partial r} r_{t}(x) .
$$
MATH97114 COURSE NOTES ：
If $X^{(1)}, X^{(2)} \ldots$ form a sequence of random vectors, each with the same law in $\mathbb{R}^{m}$, then the strong law of large numbers tells us that the estimators
$$
\mu_{N}=\frac{1}{N} \sum_{i=1}^{N} X^{(i)}
$$
and
$$
Q_{N}=\frac{1}{N-1} \sum_{i=1}^{N}\left(X^{(i)}-\mu_{N}\right) \otimes\left(X^{(i)}-\mu_{N}\right)
$$