Probability and Statistics I
概率与统计 I

Final Exam 期末考试SDS, CUHK(SZ)

13:30-16:30, December 21, 2023
2023 年 12 月 21 日 13:30-16:30

Name: 名称：
Student ID: 学生证：

Answer the questions in the Answer Book.
回答答题卡上的问题。

I Regular Questions (100 points)
I 常规问题（100 分）

Let $X$ equal the number of flips of a fair coin that are required to observe the same face on consecutive flips.
让 $X$ 等于连续掷一枚公平的硬币观察到相同面所需的掷硬币次数。
(a) (4 points) Find the pmf of $X$ . Hint: Draw a tree diagram.
(a) （4 分）求 $X$ 的 pmf。提示：画树状图。
(b) (3 points) Find the moment generating function of $X$ .
(b) （3 分）求 $X$ 的矩生成函数。
© (3 points) Use the mgf to find the values of the mean of $X$ .
(3分) 使用mgf求出 $X$ 的平均值。

Solution: 解决方案

(a)

f (x) = {(\frac{1}{2})}^{x - 1}, \forall x = 2, 3, 4, \dots

(b)

M (t) = \frac{e^{2 t}}{2 - e^{t}}, t < \ln 2

M^{'} (t) = \frac{4 e^{2 t} - e^{3 t}}{{(2 - e^{t})}^{2}}

and

E [X] = M^{'} (0) = 3

M^{'} (t) = \frac{4 e^{2 t} - e^{3 t}}{{(2 - e^{t})}^{2}}

和

E [X] = M^{'} (0) = 3

。

Teaching Objective: Probability Mass Function, Moment Generating Function. Difficulty Level: Moderate.
教学目标概率质函数、矩产生函数。难度：中等中等
2. A batch of 4 products is known to contain 2 that are defective. The products are to be tested, one at a time, until the defective ones are identified. Denote by

X

the number of tests made until the first defective is identified and by

Y

the number of additional tests until the second defective is identified.
2.已知一批 4 件产品中有 2 件有缺陷。这些产品将逐一进行测试，直到找出次品为止。用

X

表示直到找出第一个次品为止的测试次数，用

Y

表示直到找出第二个次品为止的额外测试次数。
(a) (4 points) Find the joint probability mass function of

X

and

Y

.
(a) (4 分) 求

X

和

Y

的联合概率质量函数。
(b) (3 points) Find

P (X \leq Y)

.
(b) （3 分）找出

P (X \leq Y)

.
(b) (3 points) Find

E [X + Y]

.
(b) （3 分）找出

E [X + Y]

Solution: 解决方案

(a)

Note that

X

and

Y

take positive values such that

X + Y \leq 4

, i.e.,

X = 1, \dots, 3

and

Y = 1, \dots, 4 - X

. Each pair of values for

X

and

Y

are equally likely, and there are 6 such pairs. So
请注意，

X

和

Y

取正值，使得

X + Y \leq 4

，即

X = 1, \dots, 3

和

Y = 1, \dots, 4 - X

。

X

和

Y

的每一对值的可能性相同，这样的值有 6 对。所以

f (x, y) = \frac{1}{6}, x = 1, 2, 3 y = 1, \dots, 4 - x

(b)

There are 4 pairs of

X

and

Y

such that

X \leq Y

, so

P (X \leq Y) = \frac{4}{6}

.
有 4 对

X

和

Y

，使得

X \leq Y

，所以

P (X \leq Y) = \frac{4}{6}

。
©

E [X + Y] = \frac{1}{6} (2 + 3 + 4 + 3 + 4 + 4) = \frac{20}{6}

Teaching Objective: Joint Probability Mass Function, Bivariate Random Variable. Difficulty Level: Fundamental.
教学目标联合概率质量函数、双变量随机变量。难易程度：基础：基础。
3. Let

X

be a continuous random variable with probability density function (pdf) defined as follows
3.假设

X

是一个连续随机变量，其概率密度函数（pdf）定义如下

f (x) = \frac{1}{\sqrt{2 π}} e^{- \frac{1}{2} x^{2}}, x \in (- \infty, \infty)

(a) (4 points) Use the moment generating function for

X

to show that

E [X] = 0

Var (X) = 1

.
(a) (4 分) 利用

X

的矩生成函数证明

E [X] = 0

Var (X) = 1

.
(b) (3 points) Define a random variable

Y = a X + b

. What is the distribution of

Y

? Show the proof in details.
(b) （3 分）定义随机变量

Y = a X + b

。

Y

的分布是什么？请详细证明。
© (3 points) What is

E [(Y - b)^{177}]

?
(3分) 什么是

E [(Y - b)^{177}]

？

Solution: 解决方案

(a)

The MGF of

X

X

的 MGF 为

M (t) = \exp [\frac{t^{2}}{2}]

We can obtain that

M^{'} (t) = t \exp [\frac{t^{2}}{2}]

and

E [X] = M^{'} (0) = 0; M^{''} (t) =

(t^{2} + 1) \exp [\frac{t^{2}}{2}]

and

Var (X) = M^{''} (0) - (E [X])^{2} = 1

.
我们可以得到

M^{'} (t) = t \exp [\frac{t^{2}}{2}]

和

E [X] = M^{'} (0) = 0; M^{''} (t) =

(t^{2} + 1) \exp [\frac{t^{2}}{2}]

和

Var (X) = M^{''} (0) - (E [X])^{2} = 1

。
(b)

E [\exp (t Y)] = \exp [t b + \frac{1}{2} a^{2} t^{2}]

Therefore,

Y \sim N (b, a^{2})

. 因此，

Y \sim N (b, a^{2})

。
©

Since

Y - b = a X, E [(Y - b)^{177}] = a^{177} E [X^{177}]

. 因为

Y - b = a X, E [(Y - b)^{177}] = a^{177} E [X^{177}]

E [X^{177}] = \int_{- \infty}^{\infty} x^{177} \frac{1}{\sqrt{2 π}} e^{- \frac{1}{2} x^{2}} d x = 0 (the integrand is an odd function).

Therefore,

E [(Y - b)^{177}] = 0

. 因此，

E [(Y - b)^{177}] = 0

。

Teaching Objective: Moment Generating Function, Normal Distribution. Difficulty Level: Moderate.
教学目标矩生函数、正态分布。难度：中等中等。
4. Two envelopes have i.i.d rewards

R_{1}, R_{2}

that are drawn from a uniform

[0, 1]

distribution. Because the rewards are i.i.d, selecting any of the envelopes gives you a

50 %

chance of selecting the one with the largest reward. Suppose that you select envelope one, but are allowed to switch to envelope two upon seeing the realization of the first envelope. A friend of yours proposes the following switching policy: pick a number

t \in (0, 1)

. If

R_{1} > t

, then keep

R_{1}

and otherwise switch to

R_{2}

.
4.两个信封的奖励

R_{1}, R_{2}

都来自均匀分布

[0, 1]

。因为奖励是 i.i.d，所以选择任何一个信封都有

50 %

的机会选择奖励最大的那个。假设你选择了信封一，但允许你在看到第一个信封实现后切换到信封二。你的一个朋友提出了如下的转换策略：选择一个数字

t \in (0, 1)

。如果

R_{1} > t

，则保留

R_{1}

，否则切换到

R_{2}

。
(a) (4 points) What is the probability of selecting the envelope with higher reward under this strategy for arbitrary

t

?
(a) (4 分) 在任意

t

的情况下，该策略下选择奖励更高的信封的概率是多少？
(b) (3 points) What would be the optimal choice of

t

?
(b) (3 分)

t

的最佳选择是什么？
© (3 points) What would be the expected reward under the optimal

t

?
(3分) 最佳

t

下的预期收益是多少？

Solution: 解决方案

(a)

Let

τ (t)

be the probability of selecting the envelope with higher reward with this strategy. Then
假设

τ (t)

是使用该策略时选择奖励更高的信封的概率。那么

\begin{aligned} τ (t) & = P (R_{2} > R_{1}, R_{1} < t) + P (R_{1} > R_{2}, R_{1} > t) \\ = \int_{0}^{t} \int_{u}^{1} d v d u + \int_{t}^{1} \int_{0}^{u} d v d u \\ = \int_{0}^{t} (1 - u) d u + \int_{t}^{1} u d u \\ = t - 0.5 t^{2} + 0.5 (1 - t^{2}) \\ = 0.5 + t - t^{2} \end{aligned}

(b)

τ (t)

is a concave function that is maximized at

t^{*} = 0.5

, resulting in

τ (0.5) = 3 / 4

. ©

τ (t)

是一个凹函数，在

t^{*} = 0.5

处达到最大，从而得到

τ (0.5) = 3 / 4

。©

The expected reward of this policy is
这项政策的预期回报是

\begin{aligned} E [R_{1} ∣ R_{1} > t] P (R_{1} > t) + E [R_{2} ∣ R_{1} \leq t] P (R_{1} \leq t) \\ = & \int_{t}^{1} u d u + E [R_{2}] t \\ = & 0.5 (1 - t^{2}) + 0.5 t \\ = & 0.5 (1 + t - t^{2}) . \end{aligned}

So at

t^{*} = 1 / 2

the expected reward is

5 / 8

.
因此，在

t^{*} = 1 / 2

时，预期奖励为

5 / 8

。

Teaching Objective: Bivariate Probability Density Function. Difficulty Level: Moderate.
教学目标：二元概率密度函数。难度：中等中等。
5. Let

X

be a continuous random variable with probability density function
5.假设

X

是一个连续的随机变量，其概率密度函数为

f_{X} (x) = {\begin{cases} \frac{3}{8} x^{2} & 0 \leq x \leq 2 \\ 0 & otherwise \end{cases}

Let

Y = U (X)

, where

U (X) = \sqrt{X}

.
让

Y = U (X)

，其中

U (X) = \sqrt{X}

.
(a) (10 points) Find the probability density function

f_{Y} (y)

Y

.
(a) （10 分）求

Y

的概率密度函数

f_{Y} (y)

。

Solution: 解决方案

(a)

f_{Y} (y) = \frac{3}{4} y^{5}

over

0 \leq y \leq \sqrt{2}

f_{Y} (y) = \frac{3}{4} y^{5}

胜过

0 \leq y \leq \sqrt{2}

。

Teaching Objective: Transformation of Random Variables. Difficulty Level: Fundamental.
教学目标随机变量的变换。难度：基础基础。
6. Let

X

and

Y

be continuous random variables with a joint probability density function given by:
6.假设

X

和

Y

是连续的随机变量，它们的联合概率密度函数由以下公式给出：

f_{X, Y} (x, y) = {\begin{cases} c (2 x + y) & 0 \leq x \leq 1, 0 \leq y \leq 1 \\ 0 & otherwise \end{cases}

(a) (3 points) Find the value of the constant

c

that makes

f_{X, Y} (x, y)

a valid probability density function.
(a) (3 分) 找出使

f_{X, Y} (x, y)

成为有效概率密度函数的常数

c

的值。
(b) (3 points) Compute the marginal probability density functions

f_{X} (x)

and

f_{Y} (y)

.
(b) （3 分）计算边际概率密度函数

f_{X} (x)

和

f_{Y} (y)

。
© (4 points) Are

X

and

Y

independent? Justify your answer.
(4分)

X

和

Y

是独立的吗？请说明理由。

Solution: 解决方案

(a)

c = \frac{2}{3}

.
(b)

The marginal probability density functions are

f_{X} (x) = \frac{4}{3} x + \frac{1}{3}

and

f_{Y} (y) = \frac{2}{3} + \frac{2}{3} y

.
边际概率密度函数为

f_{X} (x) = \frac{4}{3} x + \frac{1}{3}

和

f_{Y} (y) = \frac{2}{3} + \frac{2}{3} y

。
©

X

and

Y

are not independent.

X

和

Y

不是独立的。

Teaching Objective: Bivariate Probability Density Functions. Difficulty Level: Fundamental.
教学目标：二元概率密度函数。难易程度：基础：基础。
7. Consider infinitely many i.i.d experiments of throwing a pair of six-sided dice. The outcome of each experiment is the sum of the two numbers. Let

N \geq 1

be the number of experiments needed for the number 5 or 7 to first appear as the outcome. Let further

X

be that number at the

N

-th experiment, i.e.,

X

is either a 5 or 7 .
7.考虑无穷多个投掷一对六面骰子的 i.i.d 实验。每次实验的结果都是两个数字之和。设

N \geq 1

是数字 5 或 7 第一次作为结果出现所需的实验次数。再设

X

为第

N

次实验中的数字，即

X

不是 5 就是 7 。
(a) (5 points) Find

P (N = n)

for

n > 1

.
(a) （5 分）找出

n > 1

的

P (N = n)

。
(b) (5 points) Find

E [X]

.
(b) （5 分）找出

E [X]

。

Solution: 解决方案

(a)

For each experiment, the number 5 appears with probability

\frac{4}{36}

and the number 7 appears with probability

\frac{6}{36}

. Hence, the outcome is either 5 or 7 with probability

\frac{10}{36}

.
在每次实验中，数字 5 出现的概率为

\frac{4}{36}

，数字 7 出现的概率为

\frac{6}{36}

。因此，结果要么是 5，要么是 7，概率为

\frac{10}{36}

。

\begin{aligned} P (N = n) & = P ({first n - 1 outcomes \neq 5, 7} \cap {n-th outcome = 5 or 7}) \\ = {(\frac{26}{36})}^{n - 1} (\frac{10}{36}) . \end{aligned}

(b)

\begin{aligned} E [X] & = E [E [X ∣ N]] \\ = \sum_{n = 1}^{\infty} E [X ∣ N = n] P (N = n) \\ = (\frac{4}{10} \cdot 5 + \frac{6}{10} \cdot 7) \sum_{n = 1}^{\infty} P (N = n) \\ = \frac{62}{10}, \end{aligned}

where the third equality follows from
其中第三个等式源于

\begin{aligned} E [X ∣ N = n] & = \frac{5 \cdot P (X = 5, N = n) + 7 \cdot P (X = 7, N = n)}{P (N = n)} \\ = \frac{4}{10} \cdot 5 + \frac{6}{10} \cdot 7 . \end{aligned}

Teaching Objective: Conditional Expectation, Tower Property. Difficulty Level: Fundamental.
教学目标条件期望、塔属性。难度等级：基础：基础。
8. Consider the following signal

(X) +

noise

(Y) =

observation

(Z)

model:
8.考虑以下信号

(X) +

噪声

(Y) =

观察

(Z)

模型：

X \sim N (μ_{1}, σ_{1}^{2}), Y \sim N (0, σ_{2}^{2}), Z = X + Y .

Assume that

X

and

Y

are independent. The goal is to say something about

X

upon observing

Z

.
假设

X

和

Y

是独立的。我们的目标是在观察

Z

时对

X

做出一些判断。
(a) (2 points) Find the joint pdf of

X

and

Z

. Hint:

(X, Z) = (x, z)

is equivalent to

(X, Y) = (x, z - x)

.
(a) (2 分) 求

X

和

Z

的联合 pdf。提示：

(X, Z) = (x, z)

等价于

(X, Y) = (x, z - x)

。
(b) (2 points) Compute the correlation coefficient between

X

and

Z

.
(b) （2 分）计算

X

和

Z

之间的相关系数。
© (4 points) Find the conditional pdf of

X

given

Z

, i.e.,

f (x ∣ z)

. Hint: Compare the joint pdf found in part (a) with that of a bivariate normal distribution.
(4分) 给定

Z

，即

f (x ∣ z)

，求

X

的条件pdf。提示：将第（a）部分求得的联合 pdf 与二元正态分布的 pdf 进行比较。
(d) (2 points) Find the conditional expectation and variance of

X

, given

Z = z

.
(d) (2 分) 在给定

Z = z

的条件下，求

X

的条件期望和方差。

Solution: 解决方案

(a)

\begin{aligned} f_{(X, Z)} (x, z) & = f_{(X, Y)} (x, z - x) \\ = \frac{1}{\sqrt{2 π σ_{1}^{2}} \sqrt{2 π σ_{2}^{2}}} \exp [- \frac{{(x - μ_{1})}^{2}}{2 σ_{1}^{2}} - \frac{(z - x)^{2}}{2 σ_{2}^{2}}] . \end{aligned}

(b)

\begin{aligned} Cov (X, Z) & = Cov (X, X + Y) \\ = Var (X) = σ_{1}^{2} . \\ Var (Z) & = Var (X + Y) \\ = σ_{1}^{2} + σ_{2}^{2} . \end{aligned}

Hence,

ρ = \frac{σ_{1}}{\sqrt{σ_{1}^{2} + σ_{2}^{2}}}

. 因此，

ρ = \frac{σ_{1}}{\sqrt{σ_{1}^{2} + σ_{2}^{2}}}

。
©

\begin{aligned} f (x ∣ z) & = \frac{f_{(X, Z)} (x, z)}{f_{Z} (z)} \\ = \frac{\frac{1}{\sqrt{2 π σ_{1}^{2}} \sqrt{2 π σ_{2}^{2}}} \exp [- \frac{{(x - μ_{1})}^{2}}{2 σ_{1}^{2}} - \frac{(z - x)^{2}}{2 σ_{2}^{2}}]}{\frac{1}{\sqrt{2 π (σ_{1}^{2} + σ_{2}^{2})}} \exp [- \frac{{(z - μ_{1})}^{2}}{2 (σ_{1}^{2} + σ_{2}^{2})}]} \\ = C_{1} (z) \exp [- \frac{{(x - μ_{1})}^{2}}{2 σ_{1}^{2}} - \frac{(z - x)^{2}}{2 σ_{2}^{2}}] \\ = C_{2} (z) \exp [- \frac{1}{2} (\frac{σ_{1}^{2} + σ_{2}^{2}}{σ_{1}^{2} σ_{2}^{2}}) {(x - \frac{μ_{1} σ_{2}^{2} + z σ_{1}^{2}}{σ_{1}^{2} + σ_{2}^{2}})}^{2}] \\ = C_{2} (z) \exp [- \frac{{(x - \frac{μ_{1} σ_{2}^{2} + z σ_{1}^{2}}{σ_{1}^{2} + σ_{2}^{2}})}^{2}}{2 (\frac{σ_{1}^{2} σ_{2}^{2}}{σ_{1}^{2} + σ_{2}^{2}})}] \end{aligned}

where

C_{1} (z)

and

C_{2} (z)

are two functions of

z

. We must have

C_{2} (z) = \frac{1}{\sqrt{2 π \frac{σ_{2}^{2} σ_{2}^{2}}{σ_{1}^{2} + σ_{2}^{2}}}}

, and the conditional distribution is a normal distribution with parameters
其中

C_{1} (z)

和

C_{2} (z)

是

z

的两个函数。我们必须有

C_{2} (z) = \frac{1}{\sqrt{2 π \frac{σ_{2}^{2} σ_{2}^{2}}{σ_{1}^{2} + σ_{2}^{2}}}}

，条件分布是正态分布，参数为

(\frac{μ_{1} σ_{2}^{2} + z σ_{1}^{2}}{σ_{1}^{2} + σ_{2}^{2}}, (\frac{σ_{1}^{2} σ_{2}^{2}}{σ_{1}^{2} + σ_{2}^{2}}))

Another way to find

f (x ∣ z)

is by noticing that the joint pdf of

X

and

Z

has the same form as that of a bivariate normal distribution and hence the conditional distribution of

X

given

Z = z

is still normal.
找到

f (x ∣ z)

的另一种方法是注意到

X

和

Z

的联合 pdf 与二元正态分布的形式相同，因此给定

Z = z

的

X

的条件分布仍然是正态分布。
(d)

E [X ∣ Z = z] = \frac{μ_{1} σ_{2}^{2} + z σ_{1}^{2}}{σ_{1}^{2} + σ_{2}^{2}}, Var (X ∣ Z = z) = (\frac{σ_{1}^{2} σ_{2}^{2}}{σ_{1}^{2} + σ_{2}^{2}})

Teaching Objective: Normal Distribution, Independence, Conditional Expectation. Difficulty Level: Hard.
教学目标正态分布、独立性、条件期望。难度等级：难。
9. Let

X_{1}, \dots, X_{n}, \dots

be i.i.d Bernoulli random variables with

p = \frac{1}{2}

.
9.设

X_{1}, \dots, X_{n}, \dots

为 i.i.d 伯努利随机变量，

p = \frac{1}{2}

为 i.i.d 伯努利随机变量。
(a) (2 points) Prove the law of large numbers for their sum using Chebyshev’s inequality, i.e.,
(a) （2 分）利用切比雪夫不等式证明它们的和的大数定律，即：

lim_{n \to \infty} P (| \frac{X_{1} + \dots + X_{n}}{n} - \frac{1}{2} | \geq ϵ) = 0 for any ϵ > 0

(b) (2 points) Compute

lim_{n \to \infty} P (X_{1} + \dots + X_{n} > \frac{3 n}{4})

.
(b) （2 分）计算

lim_{n \to \infty} P (X_{1} + \dots + X_{n} > \frac{3 n}{4})

t > 0

,
(2分) 证明对于任意

t > 0

、

P (X_{1} > r) \leq e^{- t r} M_{X_{1}} (t),

where

M_{X_{1}} (\cdot)

is the moment generating function of

X_{1}

.
其中

M_{X_{1}} (\cdot)

是

X_{1}

的矩生成函数。
(d) (2 points) For

r = \frac{3}{4}

, find the value of

t

r = \frac{3}{4}

，求

t

P (X_{1} + \dots + X_{n} > \frac{3 n}{4})

and compare it with the upper bound obtained by the Chebyshev’s inequality. Which one is tighter (i.e., smaller) when

n

P (X_{1} + \dots + X_{n} > \frac{3 n}{4})

的上界，并将其与切比雪夫不等式求出的上界进行比较。当

n

较大时，哪一个更严格（即更小）？

Solution: 解决方案

(a)

By Chebyshev’s inequality, we have
根据切比雪夫不等式，我们得出

P (| \frac{X_{1} + \dots + X_{n}}{n} - \frac{1}{2} | \geq ϵ) \leq \frac{1}{ϵ^{2}} \frac{1 / 4}{n}

Taking limits on both sides yield
两边都有限制

0 \leq lim_{n \to \infty} P (| \frac{X_{1} + \dots + X_{n}}{n} - \frac{1}{2} | \geq ϵ) \leq lim_{n \to \infty} \frac{1}{ϵ^{2}} \frac{1 / 4}{n} = 0

which implies that

lim_{n \to \infty} P (| \frac{X_{1} + \dots + X_{n}}{n} - \frac{1}{2} | \geq ϵ) = 0

.
这意味着

lim_{n \to \infty} P (| \frac{X_{1} + \dots + X_{n}}{n} - \frac{1}{2} | \geq ϵ) = 0

。
(b)

P (X_{1} + \dots + X_{n} > \frac{3 n}{4}) \leq P (| \frac{X_{1} + \dots + X_{n}}{n} - \frac{1}{2} | \geq \frac{1}{4}) .

By the Law of Large Number,

lim_{n \to \infty} P (| \frac{X_{1} + \dots + X_{n}}{n} - \frac{1}{2} | \geq \frac{1}{4}) = 0

, which implies that

lim_{n \to \infty} P (X_{1} + \dots + X_{n} > \frac{3 n}{4}) = 0

.
根据大数定律，

lim_{n \to \infty} P (| \frac{X_{1} + \dots + X_{n}}{n} - \frac{1}{2} | \geq \frac{1}{4}) = 0

，这意味着

lim_{n \to \infty} P (X_{1} + \dots + X_{n} > \frac{3 n}{4}) = 0

。
©

For

t > 0

, we have 对于

t > 0

，我们有

P (X_{1} > r) = P (e^{t X_{1}} > e^{t r})

Because

e^{t X_{1}} \geq 0

, we can apply the Markov’s Inequality and obtain that
因为

e^{t X_{1}} \geq 0

，我们可以应用马尔可夫不等式，得出

\begin{aligned} P (e^{t X_{1}} > e^{t r}) & \leq \frac{E [e^{t X_{1}}]}{e^{t r}} \\ = e^{- t r} M_{X_{1}} (t) \end{aligned}

(d)

When

r = \frac{3}{4}

, we have 当

r = \frac{3}{4}

时，我们有

e^{- t r} M_{X_{1}} (t) = e^{- \frac{3}{4} t} (\frac{1}{2} + \frac{1}{2} e^{t}) = \frac{1}{2} e^{- \frac{3}{4} t} + \frac{1}{2} e^{\frac{1}{4} t},

which is a convex function and attains the minimum when its derivative is zero, i.e.,
是一个凸函数，当其导数为零时达到最小值，即

- \frac{3}{8} e^{- \frac{3}{4} t} + \frac{1}{8} e^{\frac{1}{4} t} = 0 .

Solving the equation yields

t = \ln (3)

.
解方程可得

t = \ln (3)

.
(e)

Because

X_{1}, \dots, X_{n}, \dots

X_{1}, \dots, X_{n}, \dots

\begin{aligned} P (X_{1} + \dots + X_{n} > \frac{3 n}{4}) & \leq {(\frac{1}{2} 3^{- 3 / 4} + \frac{1}{2} 3^{1 / 4})}^{n} \\ = {(\frac{16}{27})}^{\frac{n}{4}} . \end{aligned}

In pat (b), we obtained that by Chebyshev’s inequality,
在拍子 (b) 中，我们通过切比雪夫不等式得到了这一点、

P (X_{1} + \dots + X_{n} > \frac{3 n}{4}) \leq \frac{4}{n}

Hence, when

n

is large, the upper bound obtained in part (e) is tighter than that obtained by Chebyshev’s inequality.
因此，当

n

较大时，(e) 部分得到的上界比切比雪夫不等式得到的上界更紧密。

Teaching Objective: Markov’s Inequality, Chebyshev’s Inequality, Law of Large Num-
教学目标马尔可夫不等式、切比雪夫不等式、大数定律
ber. Difficulty Level: Moderate.
ber.难度级别：中等。
10. Let

X_{i}

be a negative binomial random variables with success probability

p

and number of successes

r_{i}

for

i \in {1, \dots, n}

. Suppose that

X_{1}, \dots, X_{n}

are independent. Define

Z = X_{1} + X_{2} + \dots + X_{n}

.
10.设

X_{i}

为负二项随机变量，成功概率为

p

，

r_{i}

为

i \in {1, \dots, n}

的成功次数。假设

X_{1}, \dots, X_{n}

是独立的。定义

Z = X_{1} + X_{2} + \dots + X_{n}

。
(a) (6 points) Determine the moment generating function (MGF) of

Z

(a) （6 分）确定

Z

的矩生成函数 (MGF)
(b) (4 points) Based on part (a), determine the distribution of

Z

.
(b) （4 分）根据 (a) 部分，确定

Z

的分布。

Solution: 解决方案

(a)

To find the moment generating function (MGF) of

Z

, we can use the property that the MGF of the sum of independent random variables is the product of their individual MGFs. Let

M_{X_{i}} (t)

be the moment generating function of

X_{i}

. Since

X_{1}, X_{2}, \dots, X_{n}

are independent we have
要找到

Z

的矩生成函数 (MGF)，我们可以利用独立随机变量之和的 MGF 是它们各自 MGF 的乘积这一特性。假设

M_{X_{i}} (t)

是

X_{i}

的矩生成函数。由于

X_{1}, X_{2}, \dots, X_{n}

是独立的，我们有

M_{Z} (t) = \prod_{i = 1}^{n} M_{X_{i}} (t)

(b)

Then using the properties of the negative binomial we see that

Z

has a negative binomial distribution with parameters

p

and

r = \sum_{i = 1}^{n} r_{i}

.
然后利用负二项分布的性质，我们可以看到

Z

具有参数

p

和

r = \sum_{i = 1}^{n} r_{i}

的负二项分布。

Teaching Objective: Moment Generataing Function Technique. Difficulty Level: Fundamental.
教学目标时刻产生函数的技巧。难度：基础基础。

Probability and Statistics I 概率与统计 I

I Regular Questions (100 points)I 常规问题（100 分）

Solution: 解决方案

Solution: 解决方案

Solution: 解决方案

Solution: 解决方案

Solution: 解决方案

(a)

Solution: 解决方案

Solution: 解决方案

Solution: 解决方案

Solution: 解决方案

Solution: 解决方案

Probability and Statistics I
概率与统计 I

I Regular Questions (100 points)
I 常规问题（100 分）