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4

Write $$\frac{S_n - n\mu}{\sqrt{S_n}} = \frac{S_n - n\mu}{\sqrt{n}\sigma}\cdot \frac{\sqrt{n}\sigma}{\sqrt{S_n}}.$$ where $\sigma^2 = \text{Var}(X_1)$. By CLT, $\frac{S_n - n\mu}{\sqrt{n}\sigma} \Rightarrow N(0, 1)$. By WLLN, $\frac{S_n}{n} \to \mu$ in probability. The result follows from the Slutsky's theorem. Clearly, $b^2 = \frac{\sigma^2}{\mu}$.


3

Hints: Since $(W_t)_{t \geq 0}$ is a Brownian motion, it has independent normally distributed increments; in particular, $W_3-W_2, W_2-W_1,W_1$ are independent random variables which are Gaussian with mean $0$ and variance $1$. This implies that $(W_1, W_2-W_1,W_3-W_2)$ is (jointly) Gaussian with mean $\mu=0$ and covariance matrix $$C = \begin{pmatrix} 1 ...


3

$F_Y(y) := P(Y\le y)$ is the probability that the random variable $Y$ is less than or equal to a given real value $y$. This function is then known as the cumulative distribution function. For a continuous random variable it is the integral of the probability density function up to $y$, while for a discrete random variable it is the partial sum up to $y$ ...


3

We consider $m_n=\max\limits_{j=1}^n|a_j|$, $s_n^2=\sum\limits_{j=1}^na_j^2$, $W$ some random variable distributed like every $W_j$, and we simplify the quantities involved in Lindeberg's condition as follows; $$\begin{align} \sum_{j=1}^nE( X_{nj}^2 \cdot \mathbf 1\{ |X_{nj}|\gt \varepsilon \sigma_n \}) &= \sum_{j=1}^n \frac{a_j^2}{n} E\left( W^2 \cdot ...


3

Hints \begin{align} \operatorname{Var}(X) &= \mathbb{E}[X^2] - \mathbb{E}[X]^2 \\ \mathbb{E}[X-2(X-1)^2] &= \mathbb{E}[X-2(X^2-2X+1)] = \mathbb{E}[- 2X^2 + 5X - 2] \\ &= -2\mathbb{E}[X^2] + 5\mathbb{E}[X] - \mathbb{E}[2] \quad \textrm{(by linearity)}. \end{align}


2

Change of variables: $({x}/{\theta})^{\beta}=t$ with $$dx=d(t^{1/\beta}\theta)=\frac{\theta}{\beta}t^{1/\beta-1}dt$$ yields $$\frac{\beta}{\theta^{\beta}}\int_0^{\infty}x^{\beta+k-1}e^{-({x}/{\theta})^{\beta}}dx=\frac{\beta}{\theta^{\beta}}\int_0^{\infty}(t^{1/\beta}\theta)^{\beta+k-1}e^{-t}\frac{\theta}{\beta}t^{1/\beta-1}dt$$ ...


2

You are using a binomial model of the number of errors. For a binomial distribution for errors with $N$ cases and an error rate probability $p$ the mean number of errors is $\mu=Np$ and the standard deviation of the number of errors is $\sigma=\sqrt{Np(1-p)}$. Plugging in $N=80$ and $p=0.07$ (which is 7%) gives $\mu=5.6$ and $\sigma=2.2821$. For the second ...


2

Chebyshev's inequality says $$P(|X-m| \geq k\sigma) \leq \frac{1}{k^2}$$ where $m$ is the mean of $X$ and $\sigma$ is the standard deviation of $X$. This inequality is tight. This can be seen by considering random variables $X_x$ which are $0$ with probability $1-1/x^2$, $1$ with probability $1/2x^2$, and $-1$ with probability $1/2x^2$, where $x\geq 1$. ...


2

The upper bound on how many standard deviations a data point can be in terms of distance from the mean is given by Samuelson's inequality. Assume we define the sample standard deviation for a random sample $x_1, x_2, \ldots , x_n$ to be $$s = \sqrt{{{1} \over {n-1}} \sum_{i=1}^n \left( x_i - \bar x \right) ^2}$$ Then Samuelson's inequality, adjusted for the ...


2

This depends on how you define your standard deviation. If you have a random experiment with a given standard deviation and you draw values then it is entirely possible but increasingly unlikely that all the values fall within n standard deviations. If you calculate the standard deviation via the values you draw then it depends on the distribution of the ...


2

First I checked if it is indeed a density $$\int_0^\infty\int_0^ye^{-y}dxdy=1$$ and $X,Y$ are not independent.So I proceeded as you already did $(1)U=g_1(X,Y)=X+Y$ and $(2)V=g_2(X,Y)=X$ then $J=\begin{bmatrix}\frac{\partial u}{\partial x}&&\frac{\partial u}{\partial y}\\\frac{\partial v}{\partial x}&&\frac{\partial v}{\partial ...


2

See e.g. Cryptographically secure pseudorandom number generator. A high-quality pseudorandom generator will be extremely difficult to distinguish from random. However, in principle, if you are lucky enough to guess which algorithm your opponent is using (including the size of the seed), and have a lot of time on your hands, you could try all possible seeds, ...


2

For (b), we want $E(X^2-4XY+4Y^2)$, which is $E(X^2)-4E(XY)+4E(Y^2)$. However, we cannot say in general that $E(XY)=E(X)E(Y)$. To calculate $E(XY)$, recall perhaps that $\text{Cov}(X,Y)=E(XY)-E(X)E(Y)$. We know the covariance, and $E(X)$ and $E(Y)$, so we know $E(XY)$. For (c), as you know the $\pi$ makes no difference. We use the formula ...


1

Use Taylor expansion of $T_{LR}$ around $p_0$: $$T_{LR}=-2n\left[\hat p \ln\left(\frac{p_0}{\hat p}\right) + (1-\hat p) \ln\left(\frac{1-p_0}{1-\hat p}\right)\right]$$ $$=\frac{n}{p_0(1-p_0)}(\hat p-p_0)^2+O_p(|\hat p-p_0|^3)=T_S+o_p(1)=T_W+o_p(1)$$


1

(a) Indeed, the probability density function of $X$ can be obtained by differentiating the cumulative distribution of $X$. So your answer $f(x)=F'(x)=3x^2e^{-x^3}$ is correct. (b) Here you simply use the definition of cumulative distribution function, $P(X<c)=F(c)$, so \begin{eqnarray} 0.5 &=& F(c) = 1-e^{-c^3}\\ e^{-c^3}&=&1/2 \\ c ...


1

Not the 'entire' course, but a few ideas that may be helpful, when you put them together. First, always pay attention to the support of a random variable. For example, $X$ has support $(0, \infty),$ which implies $P(X > 0) = 1.$ This is the reason that $P(-\sqrt{y} < X < \sqrt{y})$ becomes $P(0 < X < \sqrt{y}).$ Then when you move on to ...


1

Notice that: $\{M_n=k\}\equiv \{k \,\, \text{be} \,\, \text{in the sample and the others be greater than}\,\, k\}$ and $\# \{\text{greater than}\,\, k\}=N-k$ therefore $P(M_n=k)=\dfrac{\binom{1}{1}\binom{N-k}{n-1}}{\binom{N}{n}}=\dfrac{\binom{N-k}{n-1}}{\binom{N}{n}}$.


1

If you have $\lambda\rightarrow \infty$ then $$\varphi_{Y/\sqrt{\lambda}}(t)=\varphi_{Y}\left(\frac{t}{\sqrt{\lambda}}\right)=e^{-\lambda(1-\exp\{-\frac{t^2}{2\lambda}\})}\rightarrow e^{-\frac{t^2}{2}}$$ because $$\left(1-\frac{\lambda(1-\exp\{-\frac{t^2}{2\lambda}\})}{\lambda}\right)^\lambda=e^{-\frac{t^2}{2}}$$


1

Let $X \sim N(0,\sigma^2)$. Then we want to evaluate the expectation of $Y = X^n$. So lets first evaluate its distribution. $$P(X^n \le x) = P(X \le {x}^{1/n}) = \int_{-\infty}^{x^{1/n}}\frac{1}{\sqrt{2\pi\sigma^2}}e^{\frac{-y^2}{2\sigma^2}}dy$$ Then, if we differentiate with respect to $x$: $$f_Y(x) = f_{X^n}(x) = ...


1

The paper states directly before equation $2$ that $x_1, x_2, \cdots, x_n$ are uniformly drawn from the set $S$. The set $S$ was previously defined to be $B_{p, r}$, the $L^p$ ball of radius $r$ you mentioned above. The density of such as uniform random variable can be stated as: $$P(x_1, \cdots, x_n | p, r) = \frac{1}{\lambda(B_{p, r})}1_{(x_i \in ...


1

$$ \operatorname{E}\left( \frac 1 {2n} \sum_{i=1}^n X_i^2 \right) = \frac 1 {2n} \sum_{i=1}^n \operatorname{E}(X_i^2) = \frac 1 {2n} n\operatorname{E}(X_1^2), \tag 1 $$ so showing unbiasedness just requires finding that last expected value. \begin{align} \operatorname{E}(X_1^2) & = \int_0^\infty x^2 f(x)\,dx = \int_0^\infty x^2 ...


1

Part 2: I think the inequality is slightly incorrect, assuming "6m flood only happens once every thousand years" means "6m or greater floods happen no more frequently than once in 1000 years": instead of $F(6) < 0.001$ it should be $F(6) \leq 0.999$ and, thus, $a \leq 6/(\ln 10) \approx 2.6$. Part 3: An estimate is not biased if its expected value is ...


1

You are right, for this, you cannot use a CI based on proportions. Actually, the method of constructing CI based on proportions is a consequence of the Central Limit Theorem for Binomial Parameter Estimation. It isn't something arbitrary. We use a result from the Central Limit Theorem. Suppose $X_1,X_2,...,X_n$ form an i.i.d. sample from a relatively ...


1

I have been following this Question for several days, and it seems to me that some clarification is in order. MLE for Poisson mean. Suppose $X_i,$ for $i = 1,\dots,n$ are iid $Pois(\lambda).$ You have correctly shown that the MLE of $\lambda$ is $$\hat \lambda = \bar X = \frac{1}{n}\sum_{i=1}^n = T/n,$$ where the total $T \sim Pois(\Lambda)$ and $\Lambda = ...


1

A reasonable interpretation of "sum" or "multiply" for two probability distributions is the distribution of the sum or product of two independent random variables with those distributions. However, there's no natural way to have an "inverse".


1

This is not trivial. For example density function of sum of two continuous variables is convolution of densities of these variables.


1

Hint: $$ Var(X)=E[(X-E[X])^2]=E[X^2-2XE[X]+E[X]^2]=E[X^2]-E[X]^2. $$ Therefore, $E[X^2]=Var(X)+E[X]^2$. Then, $$ E[X-2(X-1)^2]=E[X-2(X^2-2X+1)]=E[-2X^2+5X-2]=-2E[X^2]+5E[X]-2. $$ Now, substitute for variance and expectation.



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