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0

The second limit (2) just represents $\Pr(x\in\mathbb{R}$. For the first limit (1), define $G(x) = 1 - F(-x)$ and note that $G$ is also a cumulative distribution function. Then using (2), we then have $$\lim_{x\rightarrow -\infty}F(x)= \lim_{y\rightarrow\infty}F(-y)=1-\lim_{y\rightarrow\infty}(1-F(y))=1-\lim_{y\rightarrow\infty}G(y)=1-1=0$$


1

Remember this following result: Theorem A probability $P$ on $(\Omega,\mathcal{A})$ and a converging sequence of sets $A_n \in \mathcal{A}$ with limit $A \in \mathcal{A}$. Then $\lim_{n \to \infty}P(A_n) = P(A)$ In particular you have that if $A_n \to \Omega \implies P(A_n) \to 1$ and $A_n \to \emptyset \implies P(A_n) \to 0$. A CDF of a probability ...


-1

My approach: $$F(x)=P(X\leqslant x)$$ We know that $\forall x\in X$, if $x_1< x_2 \Rightarrow F(x_1)\leqslant F(x_2)$ Since the probability function is defined as ( $T$ is the sample space) $$P:T \to [0,1]$$ We can conclude that $\max (F(x))=1$ and $\min(F(x))=0$. Because of those bounds and $P$ is monotonic, wanted limits follow.


0

We have $X_1,..,X_4~iid$ samples, therefore we can use the formula for the k-th order statistic $$\mathbb{P}(Y_{(k)}\le t) = \sum\limits_{m=k}^n \binom{n}{m}F(t)^m(1-F(t))^{n-m}$$ where F is the distribution function for $X_i\sim U(0,2)$ and n is the sample size. If you haven't seen this formula, yet, you can relatively easy proof it by looking at ...


3

Hints: Your first line is $P(\hat p=k/n)$, not $P(\hat p=k)$ In full generality, $E(\hat p(1-\hat p))=\sum\limits_k(k/n)(1-(k/n))P(\hat p=k/n)$


0

Another method (@satish's) for computing the expected number of turns before a head. Let $E(N)$ be the expected length of time $N$ for the first head to occur with probability p. Then I can write $$ E(N) = p + (1-p)\left(1+E(N)\right),\tag{1} $$ here I could get the head on the first go (the first term in Eq. (1)) or I could lose a turn getting a tails ...


1

The following is not a rigorous derivation (a derivation would require a lot of assumptions about what makes one estimator better than another), but is an attempt to "make sense" of the formula so that you can more easily remember and use it. Consider a bar graph with a bar for each of the classes of data. Then $f_1$ is the height of the bar of the modal ...


0

Lets formulate the optimization problem you are trying to solve: $\min\limits_{c_i} \Sigma_i(c_i^2\sigma_i^2)$ s.t. $\sum_i c_i = 1$ (since unbiasedness requries $E[\sum_i c_iX_i] = \mu\sum_i c_i = \mu$) Lets vectorize this formulation. Let $c^2=(c_1^2,c_2^2...c_n^2)$ and $\sigma^2=(\sigma_1^2,\sigma_2^2....\sigma_n^2)$ Then we get a formulatoin that ...


3

I'm going to call the uniformly distributed random variable capital $P$ and the use lower-case $p$ for the argument to the density function. So $P$ is uniformly distributed on the interval $[0,1]$ and $$(X_1,X_2,X_3,\ldots\mid P)\sim\mathrm{i.i.d.\ Bernoulli}(P).$$ Then let $N=\min\{n\in\{1,2,3,\ldots\}\,:\, X_n=1\}$. We seek the conditional distribution ...


0

Your condition does hold in the case where $x/n=1/2$, due to the symmetry in $f(p)$. That is, $$f(1/2 + h)=f(1/2 - h).$$ In the general case, I don't think there is a nice condition, since the growth/decay on one side of $x/n$ will be different than the decay/growth on the other side.


0

Revised per OP comments Your hypothesis test appears to be $H_0: p\leq 0.01$ vs. $H_a:p> 0.01$ There are two general outcomes: You run out the test without getting a head. I.e., you get M tails (T), for some $M=$ max number of trials. You get a head (H) on the $n^{th}$ toss $(X_n)$. Under (1) $P(X_1=T,X_2=T...X_M=T|p> 0.01) \leq ...


0

Here are two approaches, with properties and intuition described below. I am not claiming optimality. There may be better ways to treat this problem. Let $\mathcal{A}$ be the support of $\theta$. Assume that each $\mu_i(\theta)$ function is differentiable and is either strictly increasing or strictly decreasing in $\theta$. Without loss of generality, ...


0

This is correct, but you need to be careful about notation. $\mathrm{erf}^{-1}(z):= \{x:\mathrm{erf}(x)=z\}$ Therefore, we know the value of $\mathrm{erf}$, but not the corresponding value of $x$. Lets use this fact to derive your answer. $\Phi(x) = {1 \over 2}[1+\mathrm{erf}({x \over \sqrt 2})]\rightarrow \Phi(\sqrt{2}x) = {1 \over ...


0

Unfortunatly it's a lot more complicated than it seems to be. I know what you would like to know but it wouldn't be really useful in the reality (of the game). Actually it would be quite useful in arena mode. Let's list a few facts or things you haven't said : Each player have 30 health points (HP), the game finishes when a player reach 0. You (your hero) ...


0

The likelihood function, assuming the samples are independent, is $$L(\alpha) = \prod_{i=1}^N p_X(x_i)= \prod_{i=1}^N {m \choose {x_i} } \alpha^{x_i} (1-\alpha)^{m-x_i}.$$ The next step is to take the logarithm of the likelihood function $$ \log L(\alpha) = \sum_{i=1}^N\log{m \choose {x_i} }+\sum_{i=1}^Nx_i \log(\alpha) +\sum_{i=1}^N(m-x_i) \log(1-\alpha), ...


1

probability of tossing a head = p Expected Number of trials = average number of trials till you get the first head E = $p(1+2q+3q^2+4q^3+....)$ S = $(1+2q+3q^2+4q^3+....)$ qS = $q+2q^2+3q^3+....)$ S-qS = $1+q+q^2+q^3+\cdots$ S(1-q) = $\frac{1}{(1-q)}$ S = $\frac{1}{(1-q)^2}$ E =$\frac{p}{(1-q)^2}$ E = $\frac{1}{p}$ E = $\frac{1}{.01} = 100$ ...


0

If $p=\frac1{100}$, you do not stop because of head with $(1-p)^{100}\approx \frac1e$, so even with $p$ slightly below $0.01$, you may stop because of head most of the time.


0

Let $c$ be the number of necessary channels and let $X \sim \text{Binomial}(12, 0.2)$ be the total number of subscribers that want to use the link at the same time during peak hours. Then we want to find the minimum $c$ such that: $$ \text{Pr}(X \leq c) \geq 0.8 $$ Using this formula: $$ \text{Pr}(X = c) = \binom{12}{c}(0.2)^c(0.8)^{12 - c} $$ we guess and ...


2

Yes, using $\widehat p=\frac{Y}{n}$ instead of $p_0$ is valid (but your treacher's reason I think is incorrect). The reason that it is valid to use $\widehat p$ instead of $p_0$ is because the asymptotics are the same i.e. the hypothesis test is based on $$\sqrt{n}(\widehat p - p_0) \stackrel{d}{\rightarrow}N(0,p_0 (1-p_0))$$ By the continuous mapping ...


0

The SD of the sample mean is $1/\sqrt{36}=1/6$. The $z$ value corresponding to $0.95$ probability is $\approx 1.645$. So the answer to question a is $9 \pm 1.645/6 \approx 8.725-9.275$. For question b, we must compute $P(8.725 < x < 9.275)$ taking into account that x is normally distributed with mean $8.9$ and SD $1$. The cumulative probabilities ...


0

Just calculate the two-sided probability of a standard normal variable being at least $z_i$ standard deviations from the mean (i.e., $2(1-\Phi(z_i)$)


0

I like Devore and Berk. It assumes univariate and some multivariate, but its substantial enough to be useful after you graduate (not a lot of overly theoretical minutae) and they use real data sets.


1

In general, yes, your interpretation is correct. Bayesian models support a generative interpretation, unlike classical frequentist models (a grey area is mixed-effects modeling).


1

When one is constructing a hypothesis test, you have to find a way to balance Type I error and Type II error the best. The problem is that Type I error and Type II error are usually inversely related (as in the case of this problem) which means that if you want decrease probability of Type I error in test you are going to increase Type II error. Thus the way ...


3

Broad question! All I can offer is my own experiences regarding the above. How closely related are math and statistics? First of all, you need to make the distinction between applied math/stats and theoretical math/stats. I've found the applied/theoretical divide to be more substantial then the subject-level division per se. For example, take a look at a ...


1

You are performing a multinomial experiment. The wikipedia article has formulas for the distribution of the frequencies for a given (M,N). I will not reproduce them here. Specifically, you are looking at a multinomial distribution over M trials with N objects, with $p_i = \frac{1}{N}\;\; i\in \{1....N\}$. Note that when you generalize to more than one ...


0

Here's a fact about hypothesis testing: For large sample sizes, almost any deviation from the null will be labeled "significant"..even if its slight. If a statistic is asymptotically normal, that does not ensure that the AD test will fail to reject, since it will never, in fact, be normal. Hence, with a large sample size, all you are doing is amping up the ...


2

For hypothesis testing purposes, you always use the actual null distribution, which requires no estimation. Therefore, I am suspicious of your teacher's response. However, your teacher may have been referring to how confidence intervals are formed, in which case, you do use estimates. Your case appears to look like the beginnings of the basic "textbook" CI ...


1

We use a Poisson model. If a random variable $X$ has Poisson distribution with parameter (mean) $\lambda$, then $\Pr(X=n)=e^{-\lambda}\frac{\lambda^n}{n!}$. a) The number of cyclones in a one month in-season period has Poisson distribution with parameter (mean)$\frac{12}{6}$. So the probability of exactly $5$ cyclones in a month is $e^{-2}\frac{2^5}{5!}$. ...


0

I have developed a simple voting app using Schulze algorithm. Assuming I did input your data correctly the results are as follows: \begin{equation} \mathbf{Choices}= \begin{pmatrix} KG & LJ & JS & DL & TC\\ \end{pmatrix} \end{equation} \begin{equation} \mathbf{D}= \begin{pmatrix} 0 & 4 & 3 & 2 & 3\\ 1 & 0 & 3 ...


0

Using $G_{n+2}:=\frac{(\sum_{i=1}^n G_i)+G_{n+1}}{n+1}+\varepsilon$, we have that $$\displaystyle G_{n+1}=\frac{\sum_{i=1}^n G_i+\frac{\sum_{i=1}^n G_i+G_{n+1}}{n+1}+\varepsilon}{n+1}=\frac{(n+2)\sum_{i=1}^n G_i+G_{n+1}+(n+1) \varepsilon}{(n+1)^2}$$ and then, solving for $G_{n+1}$, $$\displaystyle (n+1)^2 G_{n+1}=(n+2)\sum_{i=1}^n G_i+G_{n+1}+(n+1) ...


4

$K=1.$ $N_m=$ Binomial$(m,J)$ r.v.. The number on the counter after $m$ seconds is $min(I,N_m).$ If we collect the value on the counter after $m$ seconds, the expected value is: $$E_m=E[min(I,N_m)]$$ If $m\le I $ then $N_m\le I$ and $E_m=E(N_m)=mJ.$ The expected number collected per second is $\frac{E_m}{m+1}=\frac{mJ}{m+1}.$ Since this increases in $m,$ ...


1

In the example, there are three protocols. In the Bayesian viewpoint, the idea is that the number of events of interest is a binomial random variable whose probability of "success" is itself a random variable $p_i$. Specific realizations of the data may be drawn from realizations of the underlying random variables in the model, but that does not mean that ...


1

OK, you found the ratio correctly. Now you need to construct the critical region in the form $\Omega = \{x:\Lambda(x) > C_{\alpha}\}$, where $C_{\alpha}$ corresponds to your significance level $\alpha$. Let's vary $C$ and look what we have. If $C=0$, then our critical region consists of all possible $x$, i.e. $\Omega_1 = [0, 1.1]$. In this case the ...


0

The answer is, yes, one can prove that there is no other estimator $b$ that minimizes $E(L(\theta,b)).$ To summarize the problem, we define $$ L(\theta, a) = \begin{cases} k_1 |\theta - a| & \text{if } a \le \theta, \\ k_2 |\theta - a| & \text{if } a > \theta, \end{cases} $$ where $k_1 > 0$ and $k_2 > 0.$ We let $p = k_1/(k_1 + k_2)$ and ...


1

There are several algorithms and measures that can be used to quantify the difference/similarity between sequences. For short strings - as those reported in the question - using the Levenshtein distance could be a good choice. This is a simple measure of difference between two strings that computes the minimum number of single-character edits necessary to ...


0

Let $M$ be the mathematics branch, $M'$ be the languages branch, and $B$ signify "boys." We are given $\mathbb{P}\left(B\mid M\right) = 0.65$ and $\mathbb{P}\left(B\mid M^{\prime}\right) = 0.65$. Furthermore, for this particular class, $\mathbb{P}\left(B\right) = 0.55$. By the Law of Total Probability, $$\mathbb{P}\left(B\right) = \mathbb{P}\left(B \cap ...


0

The probability space for X+Y = S = {0,1,2}, through the independence assumption we can calculate the event of $P(X = 0, Y=0) = P(X=0)*P(Y=0)$ therefore the following holds: $P(X+Y = 0) = P(0 + 0 = 0) = P_x(0)*p_y(0) = 1/16$ $P(X+Y = 1) = P(1 + 0 = 1) | P(0 + 1 = 1) = P_x(1)*P_y(0) + P_x(0)*P_y(1) = 3/4*1/4 + 1/4*3/4 = 3/16 + 3/16 = 6/16 = 3/8$ $P(X+Y = 2) ...


0

$S_n^1 = \sum X_i^1, S_n^2 = \sum X_i^2$, since they are both of zero expectation $$cov(S_n^1, S_n^1) = E((\sum X_i^1)^2) = \sum E((X_i^1)^2) = n$$ $$cov(S_n^1, S_n^2) = E((\sum X_i^1)(\sum X_i^2)) = \sum E(X_i^1X_i^2) = n\rho$$ Similarly for $cov(S_n^2, S_n^2)$.


0

I'll consider the case of $K=1$ counter here. I believe Ragnar's comment is correct, that a strategy for one counter can be applied separately to more counters because of independence. Let's say that after $m$ seconds, with no collection steps, the expected value of the counter is $E_m$. Note that if $E_m\le I-1$, then $E_{m+1}=E_m+J$ (since there is a ...


0

On average the weight of a bag is only 0,5 kg (=25.5-25) over 25 kg with the new maschine-instead of 1 kg (=26-25) . The equation is $\Delta e=p \cdot \Delta w\cdot q$. $\Delta e$= difference of earnings=$5,000 $\Delta w$=difference of the weights per bag=1 kg/bag-0.5 kg/bag=0.5 kg/bag q=amount of bags p=price of a bag of cement=$0.8/kg You have to ...


0

You know that the old machine is underselling each bag by an average of $\mu _1 - 25 = 1 {\rm kg}$, and the new machine is underselling each bag by an average of $\mu _2 - 25 = 0.5{\rm kg}$. Hence the average savings per bag by switching to the new machine is $1 {\rm kg} - 0.5 {\rm kg} = 0.5 {\rm kg}$. Since the cement sells for \$0.80 a kilo, this is an ...


1

For example, $f_{X\mid T}(t\mid t)=1/(1+\mathrm e^{-2t\mu})$ if $t\ne0$ hence $f_{X\mid T}(\ \mid t)$ does depend on $\mu$ except if $t=0$.


0

You could try the following: Let $\pi_n\left( n_i | \mu _i, \mu _j\right)$ be the marginal PDF describing the probability that $n$ comparisons between items $i$ and $j$ drawn from processes with means $\mu _i$ and $\mu _j$ would yield $n_i$ instances of $i$ being favoured over $j$. You could compute this distribution easily either by using basic statistics ...


0

The assumption that $x$ and $\epsilon$ are orthogonal, that is, that $\mathrm{Cov}(\epsilon,x)=0$, does not imply that $E(\epsilon|x)=E(\epsilon)$. Thus the quote in your question is indeed faulty (even assuming that $E(\epsilon)=0$). Example: let $x$ uniform on $(-1,1)$ and $\epsilon=1-3x^2$, then $E(x)=E(\epsilon)=E(x\epsilon)=0$ hence ...


0

In linear models sometimes you want the regressors or other r.v. of interest to be uncorrelated which would imply a linear independence. $E(Y_i|X_j)=0 \Rightarrow E(Y_iX_j)=0 \text{ (orthogonality condition in econometrics) } \Rightarrow Cov(X_j,Y_i)=0$ Check Fumio Hayashi Economtrics book. P.S.: You may have better luck with an answer at ...


1

You can formulate this as a binomial test of a proportion and use either the Wilson score interval or the Clopper-Pearson interval. Specifically: let our sample space be all hostels under study and assume random sampling. Let X be a random variable that takes the value 1 if a selected hostel has at most 30 rooms and 0 otherwise. For a sample of N rooms the ...


1

How about this: If old_score*delta<0 then damp = 1 Otherwise If old_score < -90 then damp = 2+(old_score/50) If -90 < old_score < -10 then damp = 1.1+(old_score/100) If -10 <= old_score < 10 then damp = 1 If 10 <= old_score < 90 then damp = 1.1-(old_score/100) If 90 <= old_score then damp = 2-(old_score/50) ...


0

Hint If rv $X$ has PDF: $$f(x)$$ and $c>0$ then $cX$ has PDF: $$\frac{1}{c}f(\frac{x}{c})$$


1

The constraints actually give six attempts, so while the following algorithm may fail to produce the answer in just five tries, it is still guaranteed to give it in at most six. (So far I could only find two examples of codes requiring six guesses, and I think these might be the only two.) For each digit define its status as $u,r,a,g$ meaning untested, ...



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