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概率论 Probability ST118-15/ST119-10 

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这是一份warwick华威大学ST118-15/ST119-10 的成功案例

概率论 Probability ST118-15/ST119-10 


and it follows that when $X$ and $Y$ are independent
$$
w\left(u_{1}, u_{2}\right)= \begin{cases}\frac{1}{2 \pi \sigma^{2}} \frac{2 u_{1}}{1+u_{2}^{2}} e^{-u_{1}^{2} / 2 \sigma^{2}}, & u_{1}>0,-\infty0 \ 0, & u_{1} \leq 0\end{cases}
$$
and
$$
w_{2}\left(u_{2}\right)=\frac{1}{\pi\left(1+u_{2}^{2}\right)}, \quad-\infty<u_{2}<\infty
$$
respectively.

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ST118-15/ST119-10  COURSE NOTES ：

Let $F$ and $G$ be two absolutely continuous DFs; then
$$
H(x)=\int_{-\infty}^{\infty} F(x-y) G^{\prime}(y) d y=\int_{-\infty}^{\infty} G(x-y) F^{\prime}(y) d y
$$
is also an absolutely continuous DF with PDF
$$
H^{\prime}(x)=\int_{-\infty}^{\infty} F^{\prime}(x-y) G^{\prime}(y) d y=\int_{-\infty}^{\infty} G^{\prime}(x-y) F^{\prime}(y) d y
$$
If
$$
F(x)=\sum_{k} p_{k} \varepsilon\left(x-x_{k}\right) \quad \text { and } \quad G(x)=\sum_{j} q_{j} \varepsilon\left(x-y_{j}\right)
$$
are two DFs, then
$$
H(x)=\sum_{k} \sum_{j} p_{k} q_{j} \varepsilon\left(x-x_{k}-y_{j}\right)
$$










统计模型介绍 Introduction to Statistical Modelling ST117-15

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这是一份warwick华威大学ST117-15的成功案例

统计模型介绍 Introduction to Statistical Modelling ST117-15


If a quantile function $S(p)$, has a standard distribution in the sense that some measure of position, e.g., the median, is zero and some linear measure of variability, scale, e.g., the $I Q R$, is one, then the quantile function
$$
Q(p)=\lambda+\eta S(p)
$$
has a corresponding position parameter of $\lambda$ and a scale parameter of $\eta$.
Thus the two parameters $\lambda$ and $\eta$ control the position and spread of the quantile function. Obviously if we know the parameters of a quantile function, we can use this result in reverse to create a standard distribution:
$$
S(p)=[Q(p)-\lambda] / \eta
$$

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ST117-15 COURSE NOTES ：

$$
E\left(x^{2}\right)=\int_{0}^{1}\left[1 /(1-p)^{2 \beta}\right] d p
$$
By analogy with the previous calculation, this is seen to be
$$
\mu_{2}^{\prime}=1 /(1-2 \beta), \text { for } 0<\beta<0.5 \text {. }
$$
Notice that we now have an even tighter constraint to give a finite variance. On substituting in the expression for $\mu_{2}$ and simplifying, we finally obtain
$$
\mu_{2}=\beta^{2} /\left[(1-2 \beta)(1-\beta)^{2}\right] \text {, for } 0<\beta<0.5
$$










集合与数字 Sets and Numbers MA138-10

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这是一份warwick华威大学MA138-10的成功案例

集合与数字 Sets and Numbers MA138-10


Show that if $K_{n}$ or $K_{p, \eta}$ has an even number of edges, then it has an optimal biplanar drawing, which is self-complementary.

Concerning upper bounds for $\operatorname{cr}{2}(G)$, in terms of $m$, we proved in a joint paper with Shahrokhi [21] a general upper bound for the $k$-page book crossing numbers of graphs: $$ \mu{k}(G) \leq \frac{1}{3 k^{2}}\left(1-\frac{1}{2 k}\right) m^{2}+O\left(\frac{m^{2}}{k n}\right),
$$
which together with $\mathrm{cr}{2}(G) \leq \mu{4}(G)$ gives a general upper bound on $\mathrm{cr}{2}(G)$ $$ \mathrm{cr}{2}(G) \leq \frac{7}{384} m^{2}+O\left(\frac{m^{2}}{n}\right) .
$$

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MA138-10 COURSE NOTES ：

The girth formula yields
$$
\mathrm{cr}{2}\left(K{p, q}\right) \geq p q-4(p+q-2) .
$$
One can use the counting argument with $H=K_{10,10}, G=K_{p, q}$, and the fact that $\mathrm{cr}{2}\left(K{10,10}\right) \geq 28$ from to obtain:
Theorem 1. For $10 \leq p \leq q$, we have
$$
\mathrm{cr}{2}\left(K{p, q}\right) \geq \frac{p(p-1) q(q-1)}{290} .
$$
For $p \leq 9$ we make a finer analysis of $\operatorname{cr}{2}\left(K{p, q}\right)$.