这是一份leeds利兹大学MATH5325M01作业代写的成功案例

Continuous random variables. In this case $\Omega$ is an interval of the real line. Probabilities are specified using a probability density function $f(y)$ with $f(y) \geq 0$ for $y \in \Omega$ and 0 otherwise. Areas under $f(y)$ correspond to probabilities:
$$
P(a \leq y \leq b)=\int_{a}^{b} f(y) d y .
$$
Hence the probability of $y$ taking on any given value is zero and
$$
\int_{-\infty}^{\infty} f(y) d y=1 .
$$
Analogous to the discrete case, the mean and variance of a continuous random variable $y$ are defined as
$$
\mu=\mathrm{E}(y) \equiv \int_{-\infty}^{\infty} y f(y) d y, \quad \operatorname{Var}(y) \equiv \int_{-\infty}^{\infty}(y-\mu)^{2} f(y) d y
$$

MATH5325M01 COURSE NOTES ：

The above formulation supposes $r$ is a positive integer. However, the negative binomial distribution can be defined for any positive values of $r$, by using the gamma function in place of factorials:
$$
f(y)=\frac{\Gamma(y+r)}{y ! \Gamma(r)} \pi^{r}(1-\pi)^{y}, \quad y=0,1,2, \ldots .
$$
In generalized linear modeling the following parametrization is convenient:
$$
\mu=\frac{r(1-\pi)}{\pi}, \quad \kappa=\frac{1}{r} .
$$
Using this notation, the probability function of $y$ is
$$
f(y)=\frac{\Gamma\left(y+\frac{1}{\kappa}\right)}{y ! \Gamma\left(\frac{1}{\kappa}\right)}\left(\frac{1}{1+\kappa \mu}\right)^{\frac{1}{\kappa}}\left(\frac{\kappa \mu}{1+\kappa \mu}\right)^{y}, \quad y=0,1,2, \ldots
$$
with
$$
\mathrm{E}(y)=\mu, \quad \operatorname{Var}(y)=\mu(1+\kappa \mu)
$$