这是一份uwa西澳大学STAT3062的成功案例

随机过程及其应用|STAT3061 Random Processes and their Applications代写 UWA代写

Now, $\hat{\mu}{m}$ is the value minimizing Taking the derivative of this equation, with $\xi$ given by and setting the result equal to zero, $\hat{\mu}{m}$ is determined by
$$
2 \sum_{i=1}^{n} W\left(X_{i}-\mu_{m}\right)=0
$$
where
$$
\Psi(x)=\max [-K, \min (K, x)]
$$
is Huber’s $\Psi$. (For a graph of Huber’s $\Psi$, see Chapter 2.) Of course, the constant 2 in is not relevant to solving for $\hat{\mu}{m}$, and typically is simplified to $$ \sum{i=1}^{n} \Psi\left(X_{i}-\hat{\mu}_{m}\right)=0
$$

STAT3062 COURSE NOTES ：

$$
A(x)=\operatorname{sign}(|x-\theta|-\omega),
$$
where $\theta$ is the population median and $\operatorname{sign}(x)$ equals $-1,0$, or 1 according to whether $x$ is less than, equal to, or greater than 0 . Let
$$
B(x)=\operatorname{sign}(x-\theta),
$$
and
$$
C(x)=A(x)-\frac{B(x)}{f(\theta)}{f(\theta+\omega)-f(\theta-\omega)} .
$$
The influence function of $\omega_{N}$ is
$$
\mathrm{IF}{\omega{N}}(x)=\frac{C(x)}{2(.6745){f(\theta+\omega)+f(\theta-\omega)}} .
$$