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这是南佛罗里达州大学统计推断课程的代写成功案例。
EEL6029 课程简介
Credit Hours: 3
This first-tier graduate course aims at establishing the background in mathematical statistics, and abstract modeling of complex cyber systems, data analytics, and Bayesian intelligence for graduate students in electrical and systems engineering.
Statistical inference is the process of using data analysis to infer properties of an underlying distribution of probability.Inferential statistical analysis infers properties of a population, for example by testing hypotheses and deriving estimates. It is assumed that the observed data set is sampled from a larger population.
Prerequisites
Statistical inference makes propositions about a population, using data drawn from the population with some form of sampling. Given a hypothesis about a population, for which we wish to draw inferences, statistical inference consists of (first) selecting a statistical model of the process that generates the data and (second) deducing propositions from the model.
Konishi & Kitagawa state, “The majority of the problems in statistical inference can be considered to be problems related to statistical modeling”. Relatedly, Sir David Cox has said, “How the translation from subject-matter problem to statistical model is done is often the most critical part of an analysis”
EEL6029 Statistical Inference HELP（EXAM HELP， ONLINE TUTOR）
Suppose $n$ balls are placed at random in $n$ cells; cells can contain more than one ball.
(a) Show that the probability that exactly one cell remains empty is
$$
\frac{n(n-1)\left(\begin{array}{l}
n \
2
\end{array}\right)(n-2) !}{n^n}=\frac{\left(\begin{array}{l}
n \
2
\end{array}\right) n !}{n^n}
$$
(b) The $\mathrm{R}$ function defined as
sim1 <- function $(n)$
$\quad$ length (unique (sample $(1: n, n$, replace $=$ TRUE $)))$
sim1 <- function $(n)$
length (unique (sample $(1: \mathrm{n}, \mathrm{n}$, replace $=$ TRUE)))
is one way to simulate an assignment of balls to cells and count the number of occupied cells. Use this function to compute the probability of exactly one empty cell for $n=4,6,8,10$ by simulation, and compare the simulated results to values computed with the formula of part (a). The function replicate may be useful.
(a) $n$ balls are assigned at random into $n$ cells. $S$ has $n^n$ elements since multiple balls per call are allowed.
Assume equally likely outcomes. Equivalently, assume balls are assigned independently.
Exactly one cell empty means:
- one empty cell
- one with two
- $\mathrm{n}-2$ with 1
Choices: - $n$ for the one empty cell
- $n-1$ for the cell with two balls
- $\left(\begin{array}{l}n \ 2\end{array}\right)$ for the balls to use for the two-ball cell.
- $(n-2)$ ! arrangements for the other balls in their cells.
So the number of ways to get one empty is
$$
n(n-1)\left(\begin{array}{l}
n \
2
\end{array}\right)(n-2) !=\left(\begin{array}{l}
n \
2
\end{array}\right) n !
$$
The probability of this arrangement is
$$
\frac{\left(\begin{array}{c}
n \
2
\end{array}\right) n !}{n^n}
$$
(b) We can define a function sim1empty to compute the probability of one empty cell by simulation using $\mathrm{N}$ simulation replicates:
sim1empty <- $\operatorname{function}(n, N)$
$\quad \operatorname{mean}(r \operatorname{replicate}(N, \operatorname{sim} 1(n)==n-1))$
sim1empty <- function $(n, N)$
$\operatorname{mean}(\operatorname{replicate}(N, \operatorname{sim} 1(n)==n-1))$
Probability estimates and standard errors for $n=4,6,8,10$ can then be computed and displayed with
$\mathrm{N}<-10000$
$\mathrm{n}<-\operatorname{seq}(4,10$, by $=2)$
phat <- sapply $(\mathrm{n}$, sim1empty, $\mathrm{N})$
se <- sqrt (phat $*(1-$ phat) $/ \mathrm{N})$
plot(n, phat)
segments(n, phat – $2 *$ se, $\mathrm{n}$, phat $+2 $ se) $\mathrm{N}<-10000$ $\mathrm{n}<-\operatorname{seq}(4,10$, by $=2)$ phat <- sapply $(n$, sim1empty, $N)$ se <- $\operatorname{sqrt}($ phat $(1$ – phat) / N)
plot (n, phat)
segments $(\mathrm{n}$, phat $-2 *$ se, $\mathrm{n}$, phat $+2 *$ se $)$
The formula derived in part (a) can be computed as p1empty <- function( $n)$ choose $(n, 2) *$ factorial $(n) / \mathrm{n} \sim \mathrm{n}$ and added to the plot with $\operatorname{points}(n$, p1empty $(n), \operatorname{col}=$ “red”)
An urn contains 11 balls numbered $0,1, \ldots, 10$. A ball is selected at random. Suppose the number on the selected ball is $k$. A second urn is filled with $k$ red balls and $10-k$ blue balls. Five balls are selected at random with replacement from the second urn.
(a) Find the probability that the sample from the second urn consists of three red and two blue balls.
(b) Given that the sample from the second urn consists of three red and two blue balls, find the conditional probability that the ball selected from the first urn had the number $k=6$.
(a) Given that ball $k$ is chosen from the first urn, the probability of choosing three red and two blue balls from the second when sampling with replacement is the binomial probability
$$
P(\text { three red } \mid \text { ball } k)=\left(\begin{array}{l}
5 \
3
\end{array}\right)\left(\frac{k}{10}\right)^3\left(1-\frac{k}{10}\right)^2
$$
The probability of choosing ball $k$ from the first urn and three red balls from the second is therefore
$$
\begin{aligned}
P(\text { three red and ball } k) & =P(\text { three red } \mid \text { ball } k) P(\text { ball } k) \
& =\left(\begin{array}{l}
5 \
3
\end{array}\right)\left(\frac{k}{10}\right)^3\left(1-\frac{k}{10}\right)^2 \times \frac{1}{11},
\end{aligned}
$$
and the unconditional probability of choosing three red balls from the second urn is
$$
\begin{aligned}
P(\text { three red }) & =\sum_{k=0}^{10} P(\text { three red and ball } k) \
& =\sum_{k=0}^{10}\left(\begin{array}{l}
5 \
3
\end{array}\right)\left(\frac{k}{10}\right)^3\left(1-\frac{k}{10}\right)^2 \times \frac{1}{11} \
& \approx 0.1515
\end{aligned}
$$
This can be computed in $\mathrm{R}$ as
$$
\begin{aligned}
& >\operatorname{sum}(\operatorname{dbinom}(3,5,(0: 10) / 10) / 11) \
& {[1] 0.1515}
\end{aligned}
$$
is
$$
\begin{aligned}
P(\text { ball } k=6 \mid \text { three red }) & =\frac{P(\text { three red and ball } k=6)}{P(\text { three red })} \
& =\frac{\left(\begin{array}{l}
5 \
3
\end{array}\right)\left(\frac{k}{10}\right)^3\left(1-\frac{k}{10}\right)^2 \times \frac{1}{11}}{P(\text { three red })} \
& \approx \frac{0.03142}{0.1515} \approx 0.2074
\end{aligned}
$$
This can be computed as
$$
(\operatorname{dbinom}(3,5,6 / 10) / 11) /(\operatorname{sum}(\operatorname{dbinom}(3,5,(0: 10) / 10)) / 11)
$$
[1] 0.2073807
The probabilities for all $k$ can be computed and graphed as
$\mathrm{k}<-0: 10$
$\mathrm{f}<-d b i n o m(3,5, \mathrm{k} / 10) / 11$
$\mathrm{p}<-\mathrm{f} / \operatorname{sum}(\mathrm{f})$
plot $(\mathrm{k}, \mathrm{p}$, type $=$ “h”, ylim $=c(0,0.3), l w d=3)$
$\mathrm{k}<-0: 10$
f <- dbinom(3,5, k / 10) / 11
$p<-f / \operatorname{sum}(f)$
plot $(\mathrm{k}, \mathrm{p}$, type $=” h “, y \lim =c(0,0.3), \operatorname{lwd}=3)$
This plot shows the distribution of probability on values of $k$ given that three red balls were obtained in the sample from the second urn.
Sampling without replacement is also possible; this would use hypergeometric probabilities:
fwo <- dhyper $(3, k, 10-k, 5)$
pwo <- fwo / sum(fwo)
$\operatorname{lines}(\mathrm{k}+0.05$, pwo, type $=” \mathrm{~h} “, \operatorname{col}=$ “red”, $\operatorname{lwd}=3)$
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