Statistics 2 Problem Sheet 1

What to hand in: answers to Q1–Q3

Please write the day and time of your tutorial class on your homework.

There is no need to hand in an answer to Q4, but it is very important to do it, as R plays a large part in the course.

For reference:

• X ～ Exponential(λ) has PDF

$$f(x)= \begin{cases} λ \space exp(-λx) & x \geq 0, \\ 0& \text{otherwise.} \end{cases}$$

• X ～ Poisson(λ) has PMF

$$f(x)= \begin{cases} \frac{λ^x \space exp(-λ)}{x!} & x \in \{0, 1, 2, ...\}, \\ 0 & \text{otherwise.} \end{cases}$$

• X ～ Binomial(n, p) has PMF

$$f(x)= \begin{cases} (^n_x)p^x(1-p)^{n-x} & x \in \{0, 1,...,n\}, \\ 0 & \text{otherwise.} \end{cases}$$

• X ~ Normal(μ, σ2) has PDF

$$f(x)= \frac{1}{\sqrt{2\pi\sigma^2}}exp{-\frac{1}{2\sigma^2}(x-\mu)^2}, x \in R$$

• X ～ Uniform(a, b) has PDF

$$f_x(x)= \begin{cases} \frac{1}{b-a} & x \in [a, b], \\ 0 & \text{otherwise.} \end{cases}$$

2. Suppose that $X_1$, $X_2$, $X_3$ $\stackrel{\text{iid}}{\sim}$ Exponential(λ) represent the lifetimes of three lightbulbs in my living room. I installed them 2.5 years ago, and one has just blown (the others are still working). Show that the likelihood function corresponding to this is

$$L(λ; 2.5) ∝ λ exp(−7.5λ).$$

Hint: consider the random variable $Y = min \{X_1, X_2, X_3\}$ and start by trying to deduce its CDF.

3. Zero-inflated Poisson (ZIP). Suppose I construct a random variable X in the following way. First, I toss a biased coin with probability of heads p. If it comes up heads, then X = 0, otherwise X ～ Poisson(λ). We call this the zero-inflated Poisson distribution, and write X ～ ZIP(p, λ). Derive the probability mass function for X and the likelihood function for a realization x of $X \stackrel{\text{iid}}{\sim} ZIP(p, λ)$.

Hint: think about expressing X in terms of two other random quantities.

首先，我们可以把随机变量 (X) 的构造描述成两个不同的随机事件。一个是掷硬币得到正面，其概率是 (p)，如果这个事件发生，那么 (X = 0)。另一个是掷硬币得到反面，其概率是 (1-p)，然后 (X) 遵循 Poisson(λ) 分布。

所以，(X) 的概率质量函数（PMF）可以写成如下：

1. 当 (x = 0) 时:

($P(X = 0)$) 是掷硬币得到正面的概率与掷硬币得到反面后 (X) 遵循 Poisson 分布取值为 0 的概率的和，即:

$P(X = 0) = p + (1-p) \cdot \frac{\lambda^0 \space exp(-λ)}{0!} = p + (1-p) \cdot exp(-λ)$

2. 当 (x > 0) 时:

($P(X = x)$) 只能是掷硬币得到反面后 (X) 遵循 Poisson 分布取到 (x) 的概率，即:

$P(X = x) = (1-p) \cdot \frac{\lambda^x \space exp(-λ)}{x!}$

将这些结合起来，我们得到 (X) 的概率质量函数为:

$f(x) = \begin{cases} p + (1-p) \cdot exp(-λ) & x = 0 \\ (1-p) \cdot \frac{\lambda^x \space exp(-λ)}{x!} & x \in \{1, 2, ...\} \\ 0 & \text{otherwise} \end{cases}$

对于似然函数，对于给定的数据集中的观察值 ($x_1, x_2, ..., x_n$)（其中 (n) 是观察数），似然函数 (L) 为:

$L(p, \lambda | x) = \prod_{i=1}^{n} f(x_i)$

其中 ($f(x_i)$) 是在 ($x_i$) 处的 PMF 值。

这就是零膨胀泊松分布（ZIP）的概率质量函数和似然函数。

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