本次新西兰代考主要为统计相关的final assessment
1.(a) 假设 X1;X2;X3;X4 是从 Pois-
子分布,均值 = 1
4
.让 X 表示样本均值:
X = X1 + X2 + X3 + X4
.展示你的工作。 [3 分]
(b) 假设 X 具有均值为 t 的泊松分布。让 Y Gamma(k; ),
其中 k 是一个正整数。
通过证明下面等式的左边和右边都可以是
被视为在泊松过程中相同事件的概率,其中预期
每单位时间出现的次数是 ,解释为什么下面的恒等式是正确的:
P(X k) = P(Y t)
[3 分]
【总分:6分】
2. 设 X 指数(1)。考虑与 X 无关的随机变量 S,
并具有以下属性:
S =
有的
1 概率 1=2
1 概率 1=2
让 Y = SX。
回答以下的问题:
(a) 设 y 是一个实数。用 :: 表示的空白处填写如下:
P(Y y) = P(SX yj : : :)P(: : 🙂 + P(SX yj : : :)P(: : :):
命名作为上述恒等式基础的定理。 [2 分]
(b) 找出 Y 的 cdf。展示你的工作。 [6 分]
(c) 求 Y 的 pdf。展示你的工作。 [2 分]
(d) 令 V =
磷
X. 找到 V 的 pdf,记住包括 v 的值范围
你的答案是有效的。 [5 分]
[总分:15 分]
1.(a) Suppose that X1;X2;X3;X4 are independent random observations drawn from a Pois-
son distribution with mean = 1
4
. Let X denote the sample mean:
X = X1 + X2 + X3 + X4
. Show your working. [3 marks]
(b) Suppose that X has a Poisson distribution with mean t. Let Y Gamma(k; ),
where k is a positive integer.
By showing that both the left side and the right side of the equation below can be
regarded as the probability of the same event in a Poisson process in which the expected
number of occurrences per unit of time is , explain why the following identity is true:
P(X k) = P(Y t)
[3 marks]
[Total: 6 marks]
2. Let X Exponential(1). Consider the random variable S, which is independent of X,
and has the following property:
S =
1 with probability 1=2
1 with probability 1=2
Let Y = SX.
Answer the following questions:
(a) Let y be a real number. Fill in the gaps indicated with : : : in the following identity:
P(Y y) = P(SX yj : : :)P(: : 🙂 + P(SX yj : : :)P(: : :):
Name the theorem that is the basis for the identity above. [2 marks]
(b) Find the cdf of Y . Show your working. [6 marks]
(c) Find the pdf of Y . Show your working. [2 marks]
(d) Let V =
p
X. Find the pdf of V , remembering to include the range of values of v for
which your answer is valid. [5 marks]
[Total: 15 marks]