台大應數所(碩士班) 102年度一般考試 機率統計解答

題目 https://exam.lib.ntu.edu.tw/sites/default/files/exam/graduate/102/102061.pdf

1. Let X,Y be independent standard normal random variables. Find the distribution of 2XY / \sqrt{X^2+Y^2}.

(方法) 考慮極坐標 X=r\cos\theta,Y=r\sin\theta where r>0,0 \leq \theta < 2\pi，此時r, \theta為獨立的隨機變數。2XY / \sqrt{X^2+Y^2}=r\sin2\theta，因此我們只要證明\sin\theta與\sin 2\theta的分布相同即可。

(解答) Define X=r\cos\theta, Y=r\sin\theta where r\geq 0 and \theta \in [0,2\pi). Let g: \mathbb{R}^2 \mapsto \mathbb{R} be an arbitrary bounded continuous function. Recall that the distribution of the random variables is characterized by the expected value of g(r,\theta):

E[g(r,\theta)] = \int\int_{x,y \in \mathbb{R}} g(\cdot,\cdot) \frac{1}{2\pi} \exp(-\frac{x^2+y^2}{2}) dxdy,
where r = \sqrt{x^2+y^2} and \theta is expressed by the quation below:
\theta = \begin{cases} \arctan(y/x) &  (x\geq0,y\geq0) \\ \arctan(y/x) +\pi & (x<0) \\ \arctan(y/x) + 2\pi &  (x\geq0, y<0) \end{cases}
By a variable transformation, the right hand side can be further rewritten as:
= \int_{\theta \in [0,2\pi)} \int_{r>0} g(r,\theta) \frac{1}{2\pi} r \exp(-\frac{r^2}{2})drd\theta,
from which we can see that r, \theta are independently distributed and the marginal distributions of r, \theta are 1. a distribution with a probabilty density function f_r(r) = r \exp(-\frac{r^2}{2})\mathbb{1}(r>0) and 2. {\rm Unif}(0,2\pi).

Recall that r\sin\theta follows N(0,1) and here r, \theta are independent random variables. Therefore if \sin\theta and \sin2\theta follow the identical distribution, then it follows that r\sin2\theta = \frac{2XY}{\sqrt{X^2+Y^2}} also follows N(0,1).

Again let g: \mathbb{R}\mapsto\mathbb{R} be an arbitrary bounded continuous function and consider the expected value of E[g(U)] where U=\sin\theta:

We devide the integral into three parts: [0,\frac{\pi}{2}), [\frac{\pi}{2}, \frac{3\pi}{2}) and [\frac{3\pi}{2}, 2\pi), so that \theta and u are one-to-one. (The first interval and the third interval will be marged later.)

The above integral (the right hand side) can be splitted into three parts:
\int_{0}^{\pi/2} g(\sin\theta) \frac{1}{2\pi} d\theta + \int_{\pi/2}^{3\pi/2} g(\sin\theta) \frac{1}{2\pi} d\theta + \int_{3\pi/2}^{2\pi} g(\sin\theta) \frac{1}{2\pi} d\theta.

By a variable transformation the above integrals can be rewritten as:
\int_{0}^{1} g(u) \frac{1}{2\pi} \frac{1}{\sqrt{1-u^2}} du + \int_{-1}^{1} g(u) \frac{1}{2\pi} \frac{1}{\sqrt{1-u^2}} du + \int_{-1}^{0} g(u) \frac{1}{2\pi} \frac{1}{\sqrt{1-u^2}}du.

By merging the above three integrals, we have:
E[g(U)] = \int_{-1}^{1} g(u) \frac{1}{\pi} \frac{1}{\sqrt{1-u^2}} du,
which implies that the probability density function of U=\sin\theta is:
\frac{1}{\pi} \frac{1}{\sqrt{1-u^2}} \mathbb{1}_{[-1,1]}(u).

By repeating a similar argument on \sin2\theta, we also have the distribution of \sin2\theta is identical with that of \sin\theta.

Finally, we thus conclude that the distribution of the \frac{2XY}{\sqrt{X^2+Y^2}} is also a standard normal.

2. Let U_1, \ldots, U_n, \ldots be iid Uniform(0,1) random variables and X have the distribution:
P(X=x)=\frac{1}{(e-1)x!}\mathbb{1}_{\mathbb{N}}(x).
Find the distribution of Z=\min\{U_1, \ldots, U_X\}.

(方法) 考慮Z的cdf。計算Z的CDF時可以先取給定X的條件期望再對X取期望。

(解答) Note that
P(Z \leq Z) = 1-P(Z>z).
Consider 1-P(Z>z):
P(Z>z) = EE[\mathbb{1}(Z>z)\mid X] = E[P(Z>z\mid X)].
Consider the part P(Z>z\mid X).
P(Z>z\mid X) = \prod_{i=1}^X P(U_i > z) = \begin{cases} 1 & (z<0) \\ (1-z)^X & (0 \leq z \leq 1) \\ 0 & (z>1) \end{cases}
Note that for 0 \leq z \leq 1, we have
P(Z>z) = E[P(Z>z\mid X)] = E[(1-z)^X].
Furthermore,
E[(1-z)^X] = \frac{1}{e-1} \sum_{x=1}^\infty \frac{(1-z)^x}{x!} = \frac{e^{1-z}-1}{e-1}.
Finally, we conclude that the CDF of Z is as follows:
P(Z\leq z) = \begin{cases} 0 & (z<0) \\ 1-\frac{e^{1-z}-1}{e-1} & (0 \leq z \leq 1) \\ 1 & (z>1) \end{cases}.
We can also characterize the distribution of Z by its probability density function:
f_Z(z)= \frac{e}{e-1} e^{-z} \mathbb{1}_{[0,1]}(z).

3. Let X_1, \ldots, X_n be a random sample from a population with a probability denity funtion:
f(x\mid \theta_0) = \theta_0 x^{\theta_0-1} \mathbb{1}_{(0,1)}(x),
where \theta_0>0. Find the UMVUE of \theta_0.

(方法) Beta分配屬於指數族。在找到\theta的完備充分統計量T(\bm{x})後，根據T(\bm{x})造一個期望值為\theta的統計量。

(解答) Note that the joint probability density function of x_1, \cdots, x_n is written as:
\frac{\theta^n}{x_1\cdots x_n} \exp((-\theta)\sum_{i=1}^n (-\ln x_i)) \mathbb{1}(0<x_{(1)}, x_{(n)}<1).
Therefore we can see that the distribution belongs to an exponential family. Furthermore, note that a subset in \mathbb{R}^1 , \{-\theta \mid \theta > 0\} contains an interior point, (i.e., it can contain an interval), thus \sum_{i=1}^n (-\ln x_i) is a complete sufficient statistic for \theta.

Next let us find a distribution of Y =-\ln X where X \sim {\rm Beta}(\theta,1). The distribution of a random variable is characterized by the expectation of g(Y) where g(\cdot) is an arbitrary bounded continuous function.

Consider E[g(Y)]:
E[g(Y)] = E[g(-\ln X)] = \int_0^1 g(-\ln x) \theta x^{\theta-1} dx = \int_0^\infty g(y) \theta \exp(-\theta y) dy,
where the third equality is obtained by a variable transformation. From the last equation, we can see that the probability density function of Y is \theta \exp(-\theta y) \mathbb{1}_{(0,\infty)}(y). So Y \sim \exp(\theta) = {\rm Gamma}(1,\theta) with E[Y] = \frac{1}{\theta}.

By the reproductive property of a Gamma distribution, we have \sum_{i=1}^n (-\ln X_i) \sim {\rm Gamma}(n,\theta).

Since E[\frac{1}{T}] = \frac{\theta}{n-1}, we have E[\frac{n-1}{T}] = \theta. Since T is a complete sufficient statistic, this implies that E[\frac{n-1}{T}] is the UMVUE for \theta.

4. Let X_1, \ldots, X_n \stackrel{\rm iid}{\sim} N(\mu, \sigma^2). Find a size \alpha unbiased test for hypotheses H_0: \theta \in [\theta_1, \theta_2] vs H_1: \theta \notin [\theta_1, \theta_2].

(方法) 此題並無指定如何建構不偏檢定，最簡單的方法是考慮隨機化檢定(randomized test)。

(解答) This problem does not require any specific form for the unbiased test that we have to find, thus let \varphi(\bm{x}) \stackrel{\rm def}{=} \alpha. And this is a size \alpha unbaised test for the given hypotheses.

Note that
\alpha = \sup_{\theta \in [\theta_1, \theta_2]} E_\theta[\varphi(\bm{x})] \leq \inf_{\theta \notin [\theta_1, \theta_2]} E_\theta[\varphi(\bm{x})] = \alpha.

So we conclude that the above test is an unbiased test.

(補充) 我們也可以建造一個非randomized的不偏檢定。(日後再補充) (或見Casella的習題)

5. Let X_1, \cdots, X_n \stackrel{\rm iid}{\sim} {\rm Po}(\lambda) and suppose that \lambda follows a Gamma distribution with parameters \alpha, \beta, which are known constant. Find a Bayes estimator for \lambda under the squared error loss function.

(方法) 平方誤差之下的貝氏估計量為後驗期望。出題者可能想知道考生是否理解貝氏估計量的定義，因此我們應該從定義出發。

(解答) Let \delta(\bm{x}) be the desired statistic. We are required to find \delta(\bm{x}) s.t the average risk

is minimized. By swapping the order of the integral, the above equation can be written as follows:

In the above equation, let us consider the minimization of the conditional expectation given \bm{x}:

This can be rewritten as:
E[(\delta(\bm{x})-E[\lambda\mid\bm{x}]+E[\lambda\mid\bm{x}]-\lambda)^2\mid \bm{x}],

which can further be rearranged as:

Note that E[(\lambda-E[\lambda\mid\bm{x}])^2\mid \bm{x}] = V[\lambda\mid\bm{x}].

So we have we rewrite it as

In the above inequality, the equality holds when \delta(\bm{x}) = E[\lambda\mid\bm{x}] (a.s.).

Therefore, the Bayes estimator under the squared error loss is the posterior mean. Now we consider the posterior distribution of \lambda. Note that

where \approx means that the equation is equal if we only focus on the part \lambda.

The right hand side is:
\approx e^{-n\lambda} \lambda^{x_1+\ldots+x_n} \lambda^{\alpha-1} \exp(-\beta \lambda) = \lambda^{\sum_{i=1}^n x_i + \alpha-1}  \exp(-(n+\beta)\lambda),

which implies that the posterior distribution is {\rm Gamma}(\sum_{i=1}^n x_i + \alpha, n+\beta).

So its posterior mean is \frac{\sum_{i=1}^n x_i + \alpha}{n+\beta}. And this is the desired Bayes estimator.

6. Let X_i \sim {\rm Bin}(n_i, p_i) ~ (i=1,\ldots,m) be indepenent random variables. Derive the likelihood ratio test for the hypotheses H_0: p_1 = \ldots = p_m vs H_1: \exists i \neq j ~s.t.~ p_i \neq p_j.

(方法) 依照Likelihood Ratio Test的定義建立檢定即可。最後我們也可以討論其檢定的漸進分布，以便給出

(解答)