New answers tagged self-study
2
Perhaps it would be helpful to give this some clearer structure, via explicit assumptions. Suppose we are willing to assume a priori that each person is equally likely to be male or female, and we assume that the sexes are mutually independent. Then the "female-indicator" variables for the people in the group are:
$$X_1,...,X_{12} \sim \text{IID ...
1
You did everything correctly, you are just missing the last step. First of all, you can write $o((uh)^m) = u^m o(h^m)$, which will lead to
$$ \int_\mathbb{R} k(u) o((uh)^m)du = o(h^m)\int_\mathbb{R} k(u) u^mdu = o(h^m)$$
by the properties of the kernel function given in the assignment. Hence,
$$\mathbb{E}[f_n(t)] = f(t) + \frac{f^{(m)}(t)(-h)^m}{m!} \int_{\...
0
It seems to be a self-study question so let me just give you some hints to figure it out by yourself.
Mean is sensitive to outliers.
MAD is an acronym generally used for median of deviations from median. MAD is a robust measure of variability, so it is insensitive to outliers (see also other questions tagged as mad). You may want to use it if you have clear ...
2
The justification of both AIC and BIC is asymptotic. AIC is asymptotically efficient while BIC is consistent (which also an asymptotic notion) in model selection, where each property holds under a respective set of assumptions. It is not correct that AIC is better suited for small samples while BIC is suited for large samples.
Lag order selection by AIC or ...
1
I have figured it out. It is not the case that
$$(E(Y_{(n)}))^2=\theta^2.$$
$\hat\theta_2$ is the unbiased estimator, not $Y_{(n)}.$ If I carry through this correction, I get the book's answer.
3
By definition:
\begin{equation}
Corr(X,Z)=\frac{Cov(X,Z)}{\sqrt{Var(X)}\sqrt{Var(Z)}}
\end{equation}
You know that $Var(X)=3$ and $Var(Y)=4$. Since Z is a linear affin transformation of $X$ and $X$ and $Y$are independent, you get:
\begin{align}
Var(Z)&=Var(2X+Y)=4Var(X)+Var(Y)=12+4=16
\end{align}
You can calculate the covariance between X and Z as:
\...
0
Assuming $A$ has finite variance, the multivariate Central Limit Theorem says that $(\omega,\tau)$, suitably scaled, are asymptotically bivariate Normal. Since they are uncorrelated, and uncorrelated bivariate Normals are independent, they are asymptotically independent.
1
Your scanned answer correctly derives the probability that a single test is positive. Now, you primarily need to consider how the probability of having two positive tests differs, and then you apply the same formula just with $P(\text{two positive tests}|F_1)$ and $P(\text{two positive tests}|F_2)$ instead of $P(A|F_1)$ and $P(A|F_2)$.
Without additional ...
0
Your first step was unhelpful, I think. I would keep the residuals inside the square. First add and subtract the region mean
$$\sum_{i=1}^n (y_i-\bar y)^2=\sum_{i=1}^{n_1} (y_i-\bar y+\bar y_1-\bar y_1)^2+\sum_{i=n_1+1}^{n} (y_i-\bar y+\bar y_2-\bar y_2)^2$$
Now, rearrange
$$\sum_{i=1}^{n_1} (y_i-\bar y+\bar y_1-\bar y_1)^2=\sum_{i=1}^{n_1} (y_i-\bar y_1)^2+...
4
By the Law of Large Numbers
$$\frac{1}{n}\log\frac{L_n(\theta)}{L_n(\theta_0)} = -K_n(\theta_0,\theta)+o_p(1)$$
so
$$\log\frac{L_n(\theta)}{L_n(\theta_0)} = -nK_n(\theta_0,\theta)+o_p(n)$$
And that's all we need: the term that's proportional to $n$ dominates the $o_p(n)$ term.
If you want the epsilontics: write $r_n$ for the $o_p(n)$ term. For every finite $...
0
For a general case with $a,b$ we can derive the result of following integral,
\begin{eqnarray}
\int_{-\infty}^{\infty}\Phi(a+bx)^{2}\phi(x)dx\\&=&P\left(z_{1}\leq a+bx,z_{2}\leq a+bx\right)\\&=&P\left(z_{1}-bx\leq -a,z_{2}-bx\leq -a\right)\\&=&
\mathcal{MVN}\left(x=\{-a,-a\},\mu=\{0,0\},\Sigma=\begin{bmatrix}b^{2}+1 & 1\\1& b^{...
7
Q1: In sector 'Coefficients:' there is no 'factor(color)1' but others. Why?
Because it is included in the intercept. It's similar to a situation where you have a continuous variable - the intercept is the expected value when the variable is zero. With categorical variables we call the level which is included in the intercept the "reference level" ...
8
Your questions are already answered in several threads on the site, so I'll provide links to detailed explanations:
There is no factor(color)1 because the categorical variable is dummy coded, so one of the categories is always dropped.
Your second question is answered in Interpreting Residual and Null Deviance in GLM R.
AIC stands for Akaike Information ...
0
I think $E[XE(X_n|X)]$ should be computed as $E[X[(1-\dfrac1n)X+\dfrac{e^n}{n}]]\\=(1-\dfrac1n)E(X^2)+\dfrac{e^n}{n}E(X)\\=1-\dfrac1n$
However, I agree that $X_n$ does not converge to $X$ in quadratic mean.
4
If I understand the problem correctly, you have $X\sim\mathsf{Pois}(\lambda)$ and you seek $\lambda$ such that
$$P(X \ge 365) = 1 - P(X \le 364) = 0.8.$$
Normal approximation to Poisson. For an approximate answer (I guess what is expected) you can use
the fact that, for large $\lambda,$ one has $X \stackrel{aprx}{\sim}
\mathsf{Norm}(\mu=\lambda, \sigma=\sqrt{...
2
I see no reason to take these as paired data, from your descriptions of
the sampling. Why should the first tube from A be particularly related to the first tube from B?
x1=c(97.1, 101.3, 107.8, 101.9, 97.4, 104.5, 99.5, 95.1)
x2=c(103.5, 105.3, 106.5, 107.9, 102.1, 105.6, 109.8, 97.2)
Boxplots show different medians, perhaps significantly different at the 5%...
0
You have (a) correct! The key thing to understand here is that the effects of A and B will look totally different, because they are represented by different things in the diagrams. Moving a dot higher or lower means that that group mean is higher or lower. Factor A is split into three points along the x-axis, so a difference between the means across those ...
0
Let $x$ and $y$ denote the number of heads and tails needed to win the game by Player 1 and 2 respectively. Let $$Y\sim \operatorname{bin}(x+y-1,1/2)$$
denote the number of tails in next $x+y-1$ rounds. Without loss of generality we can ignore the fact that some of these rounds may not need to be played.
Letting $X$ denote the number of heads in the same ...
5
If $X\sim\mathcal N(\mu,1)$ and $\mu\sim\mathcal N(0,10)$, the MAP of $\theta$ is
$$\theta^\text{MAP}(x)=\dfrac{10}{11}x$$while the MAP of $\alpha=\alpha(\theta)=\exp\{\theta\}$ is
$$\alpha^\text{MAP}(x)=\exp\left\{\dfrac{10}{11}x-\frac{\sqrt{10}}{\sqrt{11}}\right\}$$
The plot of the plugged-in density estimates $\varphi(\cdot;\theta^\text{MAP}(x),1)$ versus ...
0
For anyone in the future finding this question: the 2nd variance value computed from the conditional variance formula (at the bottom of the question) is correct. The first value is incorrect.
The answer above that says "The correct answer is" shows the value of the conditional variance $Var(\bar{X^*_n}|X_1,\dots,X_n)=\frac{n-1}{n^2}S^2$. The ...
0
For your first set of problems (A and B), your attempt to mimic a classic ANOVA in R is leading you astray. If you look at the details of a mod returned by your code, you will see that the F-test is based on 3 degrees of freedom for the numerator. An F-test with 3 groups, however, should only have 2 degrees of freedom for the numerator.
What's going on?
When ...
3
On the assumption $H_0$ is correct, the probability $X_{(n)} \ge 1$ is $0$, while the probability $X_{(1)} \ge g$ is the probability all the observations are in $(g,1]$ which is $(1-g)^n$. So $\alpha = (1-g)^n$ and $g=1-\sqrt[n]{\alpha}$.
If you want the power of the test for some particular $\phi_1 > 0$, you do much the same calculation:
if $1 \le \...
-1
As my master said : specificity and sensitivity are not complement of each other , but we should say that specificity is a complement for false positive and sensitivity is a complement for false negative
0
Compared to interpolation, regression takes the uncertainty of measurements into consideration. The pairs of observed values may be noisy.
2
Hint 1: random walk is a cumulative sum of i.i.d. increments and is I(1). A cumulative sum of a random walk is a cumulative sum of an I(1) process and thus I(2). Your case is where i.i.d. is replaced by something more general, namely, an I(0) process.
I will write $z_t=a_0+a_1z_{t−1}+a_2z_{t−2}+e_t$. Is this true?
Hint 2: It is more like $z_t=y_t+y_{t−1}+...
0
For a more sophisticated approach to statistical inference, I would recommend Casella and Berger, Statistical Inference. You can skip the early chapters on probability if you are already familiar with measure-theoretic probability and move on to the later chapters that deal with the topics you mentioned. In particular, chapters on Hypothesis Testing, ...
0
I would recommend you the Data Analysis Using Regression and Multilevel/Hierarchical Models book by Andrew Gelman and Jennifer Hill. It covers a broad variety of topics in regression analysis, including hierarchical and Bayesian models. It is very readable and has many code examples. It doesn't go in-depth into theory, rather focuses on practical ...
2
$5.5$ years after posting this question, I've since taken measure-theoretic probability and can answer this question.
The very definition of a random variable $T \sim t_{\nu}$ is
$$T = \dfrac{Z}{\sqrt{V/\nu}}$$
for some $Z \sim \mathcal{N}(0, 1)$ and $V \sim \chi^2_\nu$ independent, with probability one ("almost surely").
This immediately implies ...
2
Hi: If $z_t$ is I(2) with zero mean then that means that
$(z_{t} - z_{t-1}) = (z_{t-1} - z_{t-2}) + \epsilon_{t}$.
So, if you difference the differences, that gives an I(0) process.
$(z_{t} - z_{t-1}) - (z_{t-1} - z_{t-2}) = \epsilon_{t}$
The above was for $z_t$. Keep doing the same thing for $z_{t+1},z_{t+2},\ldots, z_{t+n}$. Then put everything except $...
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