# Tag Info

8

These are two different phenomena: $t$-statistic Holding all else constant, if $N$ increases the $t$-value must increase as a simple matter of arithmetic. Consider the fraction in the denominator, $\hat\sigma/\sqrt{n}$, if $n$ gets bigger, then $\sqrt n$ will get bigger as well (albeit more slowly), because the square root is a monotonic ...

8

While you should be able to argue that the sd of (b) should be less than that of (a), you can't coherently argue, based on the given information, that it will be less than that of (c); it may be, but it also may not. Here's an example which should suggest to you that the answer should not be the general statement implied by choosing (b). Here are histograms ...

7

You have $$EX=\int_0^{\infty}xdF(x)$$ Notice that $dF(x)=-d(1-F(x))$ and that $P(X>t)=1-F(t)$ and use integration by parts. Now show that for monotone decreasing positive function $$\sum_{n=0}^\infty f(n)\ge\int_0^{\infty} f(t) dt$$ Combine these two results and you get your desired result. Hint for the second, recall Riemman sums.

5

All you need to do is to consider a partition like $\bigtriangleup_a=\dfrac{a}{n}$ for $n>0$ and $a_k=k\bigtriangleup_a$ for $k=0,...,n-1$. Here $B(t)$ is a continuous function so you can approximate it by a Riemann integral as $Y_a=\bigtriangleup_a\sum_{k=0}^{n-1}B(a_k)$. The normality distribution of $Y_a$ comes from the fact that you have a linear ...

5

I like January's answer. May I suggest a way to write down the series so that the eye catches the rearrangement more easily (this is the way I like to write it on the blackboard)? $$\begin{eqnarray} \sum_{k=1}^\infty P(X\geq k) &=& \quad P(X\geq 1) \quad=\quad P(X=1)&+&P(X=2)&+&P(X=3) &+& \;\dots\\ &+& \quad P(X\geq ... 5 There are a variety of ways you might find that particular value. Since you seem to be using tables, that's what I'll discuss. Tables: If you're supplied with standard normal tables, they may be arranged in any number of different ways. One common way is to give a table of the cdf; other tables might give areas above 0.5 instead, or they might give upper ... 4 Is (X, X+Y) normal? Yes! It is a linear combination of independent univariate normal distributions. Means: the mean of X is \mu_1, and the mean of X+Y is the sum of the means because they are independent, so \mu_1+\mu_2. Variance-covariance. The variance of the sum of two independent random variables is the sum of their variance. So, the variance ... 4 Simply write R_n=R_{n-1}+X_n and take the conditional expectations with respect to \mathcal{F}_n. Then exploit the fact that$$E[g(X_n)f(R_{n-1})|\mathcal{F}_n]=f(R_{n-1})Eg(X_n),$$for any measurable functions f and g (assuming that the expectations exist). Also recall the formula$$\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta$$. 4 Definition of E(X) for discrete X is E(X) = \sum_i x_i \cdot P(X = x_i). P( X \ge i ) = P( X = i ) + P( X = i + 1 ) + \dots So \sum_i P( X \ge i ) = P( X \ge 1 ) + P( X \ge 2 ) + \dots =  = P( X = 1 ) + P( X = 2 ) + P( X = 3 ) + ... + P(X = 2 ) + P( X = 3 ) + ... = (we rearange the terms in the last expression) = 1 \cdot P( X = 1 ) + 2 ... 4 Ok. The model is, in matrix notation and conformable dimensions$$\mathbf y = \mathbf X\beta + \mathbf u $$The OLS estimator is$$\hat \beta = (\mathbf X'\mathbf X)^{-1}\mathbf X' \mathbf y = (\mathbf X'\mathbf X)^{-1}\mathbf X' (\mathbf X\beta + \mathbf u) = (\mathbf X'\mathbf X)^{-1}\mathbf X' \mathbf X\beta + (\mathbf X'\mathbf X)^{-1}\mathbf ...

4

You don't specify anything about $X$. Is it the general case/what is its support? If it is for discrete r.v.s, can you say something about the relationship between $\sum_{n = 0}^{\infty} n P(X = n)$ and $\sum_{n = 0}^{\infty} P(X > n)$? Consider: \begin{eqnarray} &0 P(0)& +\, 1& P(1)& +\, 2 &P(2)& +\, 3 &P(3)& +& ...

4

There are an infinite number of ways for a distribution to be slightly different from a Poisson distribution; you can't identify that data is from a Poisson distribution. What you're talking about there by checking those three criteria isn't checking that the data come from a Poisson distribution by statistical means (i.e. by looking at data), but by ...

4

I will try to explain how rejection sampling works "in general". If I am clear enough, you should then be able to answer your own question. I won't give proofs but intuitive (or I hope so) "facts". Fact 1 If $X$ is drawn in a distribution of density $g(x)$ and $U$ in a uniform $U(0,1)$, then the point $(X, U\times M\times g(X))$ is drawn uniformly in the ...

3

The hierarchical model seems to be $$\begin{eqnarray} X_i\mid N=n,\Theta=\theta,\Lambda=\lambda &\sim& \mathrm{Bin}(n,\theta) \qquad\qquad\qquad i=1,\dots,m \\ N\mid\Theta=\theta,\Lambda=\lambda &\sim& \mathrm{Poisson}(\lambda/\theta) \\ \Theta &\sim& \mathrm{Beta}(\alpha,\beta) \\ \Lambda &\sim& ... 3 Let me answer the questions in order. Claim 1. We can show that if the distribution of X is Cauchy distribution, then I = \{0\}. Obviously, M_X(0)=1. When t>0, M_X(t) \geq C \int_1^\infty \exp(tx)/x^2 dx = \infty. The same observation applies for t<0 case. Claim 2. Convexity of M_X Let t_1, t_2 be two points in I. for 0 \leq ... 3 Well, the short answer is that's what falls out of the math. The long answer would be to do the math^3. Instead I'll try to rephrase gung's explanation that these are two different (though related) things. You've collected a sample X_1...X_n that is normally distributed with unknown variance^4 and want to know if its average is different from some ... 3 Fix \varepsilon\gt 0. We have for each positive A,$$\{|X_n^{-1}-1|\gt \varepsilon\}=\{|X_n-1|\gt |X_n|\varepsilon\}\subset\{|X_n-1|\gt A\varepsilon\}\cup\{|X_n|\leqslant A\},$$this because we wrote S=(S\cap \{|X_n|\gt A\})\cup (S\cap\{|X_n|\leqslant A\}). Take A:= 1/2; then \{|X_n|\leqslant 1/2\}\subset\{|X_n-1|\geqslant 1/2\} (because ... 3 If the problem does not state explicitly that X and Y are independent, then it doesn't have a solution, because the marginal distributions of X and Y do not determine their joint distribution. Supposing that X and Y are independent, then e^X and Y are also independent. Proof:$$ \begin{eqnarray} P\left\{e^X \in A, Y\in B \right\} ...

3

Let $U$ be a $\mathrm{U}[0,1]$ r.v. Let $F$ be a distribution function. Remember that every distribution function is non decreasing and right continuous. Define the quantile function $$F^{-1}(u) = \inf\{x:u \leq F(x)\}.$$ Drawing a picture we see that $F^{-1}(u)\leq x$ if and only if $u\leq F(x)$. Please, make sure that you understand both ...

3

Define the sets $A_n=\{x\in \mathbb{R}:x>n\}$, for $n=0,1,2\dots$. For any fixed $\omega$, let $n_0$ be the smallest integer such that $X(\omega)\leq n_0$. Since $X(\omega)\geq 0$, we have $$X(\omega)\leq n_0 = \sum_{n=0}^{n_0} I_{A_n} (X(\omega)) = \sum_{n=0}^\infty I_{A_n} (X(\omega)) \, ,$$ yielding $$\mathrm{E}[X]\leq \sum_{n=0}^\infty ... 3 Not an answer, at least not to your question, but an example of how to use Ito's formula (http://en.wikipedia.org/wiki/It%C5%8D_calculus). Since [B]_t=t, we have$$B_t^n = \int_0^t nB_s^{n-1}dB_s + \frac{n(n-1)}{2} \int_0^t B_s^{n-2} ds$$for n\ge 2. In particular,$$B_t^3 = 3 \int_0^t B_s^2 dB_s + 3 \int_0^t B_sds$$so that$$ \int_0^t B_s^2 dB_s ...

2

You seem to have some conceptual issues. In the classical non-bayesian context (the fact that your are learning about bias, and your working example, suggest that this is your context) the parameter $\theta$ is ... a parameter, a number; which is perhaps unknown to us but which takes nonetheless some determined fixed value. In short: $\theta$ is not a ...

2

To understand the problem easily let's consider that all the five cards are of the same 'type' (either all spade, or all diamond, or whatever you wish). So the face value is different only by the 5 different symbols from the 13 symbols that a particular 'type' of card has. There are four 'types' of cards- diamonds, hearts, spades and clubs. Now consider ...

2

The exercise says that each day exactly 10 customers walk in, asking for one of two sorts of pie. The question is which total number of pies of each sort must be available to ensure with 95% probability that each customer gets what he/she requests. Let us have a look at the extreme cases first: If the total number is 4 of each sort, then a total number of 8 ...

2

Let's back up a minute and remind ourselves what the definitions of these things are. First, we have a class of models $$\{f_\theta(x): \theta \in \Theta\},$$ Let $\mathcal L_\theta(X \mid T(X))$ denote the conditional distribution of $X$ given $T(X)$ and a fixed $\theta$. Then $T(X)$ is sufficient if $\mathcal L_\theta(X \mid T(X)) = \mathcal ... 2 Wom Man Mar x y Sing w z You are mixing up the concepts of joint and conditional probability. Problem:$x = P(M,W) = 0.06x+w = P(W) = 0.3P(M|W) = ?$Solution:$P(M|W) = P(M,W) / P(W) $2 You should be able to draw and use something like the diagram below to lay out the information (write it in the spaces and margins) and then you may be able to see how to do the problem. You would write all of the values on the diagram and from them fill in probabilities for every subregion and colored region(/margin). Then you should have a clearer idea ... 2 That is the correct "long" way to do it. Essentially any combinatorics problem could be solved this way with enough ink and paper or computing power. This seems a bit like homework, so I'll get you started in the right direction without giving the answer away completely. You've got a bunch of conditional probabilities$P(A|B)$that you need to calculate. Use ... 2 You don't do tests of equality of sample proportions at all. Indeed, you can tell at a glance whether sample proportions differ! It's population proportions you test the equality of. Under$H_0\$, where the population proportions are equal, there are several possible estimators of the standard error of the difference in sample proportion. They make ...

2

You could add the tag self-study to your question. First estimate the expectation of the normal distribution by the sample mean, then the sample variance, and the square root of the sample variance. > x<-c(11,13,13,14,14,14,15,15,17,18) > mean(x); var(x) [1] 14.4 [1] 4.044444 > sqrt(var(x)) [1] 2.01108 The confidence interval is contructed ...

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