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0

I'm not sure I understand completely your question but did you notice that for any event $A$, $$ \mathbb{P}(A) = \mathbb{E}(\mathbb{1}\{A\}), $$ where $1$ is the indicator function ?


0

You know that $$\lambda(x)=\frac{sup_{\Theta_0}L(\theta_0|x)}{sup_{\Theta}L(\theta|x)}$$ and $$-2\log \lambda(x)\rightarrow^{Distribution}_{n\rightarrow\infty} \chi^2_1$$


1

If you want to estimate some functional of the original distribution, then you can always apply the same functional to an estimate of that distribution (the "plug-in principle"). In the case of moments, this would just amount to integrating the EDF, or treating the jumps as point masses and calculating the moments of this discrete distribution.


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Hint: Define the event $$ E_n = \lbrace X_n = 0, X_{n+1}=0, X_{n+1}=1 \rbrace $$ Then find $\mathbb P(E_n)$ and look at $$\sum^\infty \mathbb P(E_n)$$ and use Borel-cantelli


0

Are you looking for a good way to use the standard deviation to go from 0% to 100% so that a layperson would understand it better, or are you looking for something to use instead of the standard deviation that a layperson would better understand? In the first case, I would recommend dividing the standard deviation of "Data 1; Data 2; Data 3; ..." by the ...


0

An important question is how did you get the sets? Are they observations of some physical phemomena? And why are they distinct sets as opposed to one large set if they are observations? In general an overdetermined system is no problem assuming the results are consistent you just get some equations which end up being superfluous. The issue you might have ...


3

You seem to be quite confused about the use of the median. Typically, you find the median of a set of numbers, not a single number. Your initial example has you finding the median of four apples, which doesn't really make sense. Imagine, instead, you have a set of buckets containing apples. Suppose you have four buckets, containing 1, 2, 3, and 4 apples. The ...


3

I'm not sure I understand what you are trying to say. There is no medium in stats. A medium might be a size of your soft drink at a fast food place or possibly someone who can communicate with the spirit world, but I can't think of any standard meaning it would have in statistics. Median on the other hand does have a definition but seems to have little to ...


1

The issue is that you confuse two problems. You have $4$ apples, but can you choose to have $0$ apple or not? In this case, the value you can choose from are $\left\{0,1,2,3,4\right\}$. In the other case, you can choose from $\left\{1,2,3,4\right\}$...


1

A A A A ^ | At that midpoint, how many complete A's are on the left? 2. On the right 2. You do not have 0.5 A's at the midpoint. ie | A | A | A | A | +---+---+---+---+ 0 1 2 3 4 The line is now measuring (the width of) the A's. The midpoint is at 2. | A | A | A | A | --+---+---+---+ 1 2 3 4 This is NOT how you should draw the ...


0

Confidence intervals for mean of a highly skewed population using a large sample. This seems to be a followup from a previous post in which you started asking about CIs for the population mean in addition to your original question (about retrieving data from confidence intervals for the frequencies of various values). Here the values $y$ and their ...


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 = ...


0

Does this make the scenario simpler? Scenario: A person “X” selects at random a number from range “r”. If X selects a number “n”, then “X” makes “n” more attempts to select numbers from the same range “r”. As an example, If “X” first selects “3” from range [0 - 5] then “X” makes 3 more attempts to select random numbers from [0 - 5]. So, 1: Given “y” ...


7

If we set $$I := \int_{\mathbb{R}} \exp \left(- \frac{x^2}{2} \right) \, dx,$$ then $$I^2 = \int_{\mathbb{R}} \int_{\mathbb{R}} \exp \left( - \frac{x^2+y^2}{2} \right) \, dx \, dy.$$ Introducting polar coordinates, i.e. $$\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} r \cos \varphi \\ r \sin \varphi \end{pmatrix},$$ yields $$I^2 = ...


0

Hint: Change variables into the integral $\displaystyle\int_{-\infty}^{+\infty} e^{-x^2}\,dx$, square that, then transform into a polar integration.


2

The "easiest" way is to use a change of variables to change your integral into a multiple of $$\int_{-\infty}^{+\infty}e^{-u^2}\,du$$ and use the famous fact that that last integral equals $\sqrt{\pi}$.


1

There are two answers depending on how restrictive your setting is. As the OP asks for a general principle I will consider the following easier example: Let $\Omega =\{0,1,2\}$, $P = (\delta_0+2\delta_1+\delta_2)/4$ and $Z:\Omega \to \mathbb{R}$ with $Z(\omega ) = \omega$. We know that we may obtain $Z \stackrel{d}{=} Z'$ with $Z' = X'+Y'$ where $X'$ and ...


1

Assume you have pocket aces and are looking for the conditional probability of flopping a full house of aces over some other rank given that you have the aces. Let's look at the chance of getting aces over twos. There are 2 aces in the deck and 4 twos. So the chance of flopping a boat of aces over twos is $${{{2 \choose 1} {4 \choose 2 } } \over {50 ...


0

If everyone is 5 feet tall in country A and 6 feet tall in country B, then each individual country has mean 5,6 height respectively, with zero variance. However if you want to compare the two, you need to know the population size. For example if there are $n_A,n_B$ people in both countries, then the mean height amongst both is: $m=\frac{5n_A+6n_B}{n_A+n_B}$. ...


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 ...


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 ...


2

Suppose $g(x) = \sum_{n \leq x} f(n)$ denotes the partial sums of the arithmetic function $f(x)$. Then the heuristic in the post you mention follows from the intuitive idea behind the derivative. That is, consider $g(M) - g(N)$, the sum of the last $(M-N)$ elements in the partial sum of $f(n)$. If we're interested in the average size of $f(n)$, then we are ...


0

Intro statistics focuses on the PDF as the description of the population, but in fact it is the CDF (cumulative density function) that gives you a functional understanding of the population, as points on the CDF denote probabilities over a relevant range of measures. If you look at all stats from this perspective, then the PDF is just the description of ...


1

What is the probability that the sun will rise tomorrow? The sunrise problem is such an open statistics question. As far as I know, there are 22 unsolved problems in statistics. Difficulties in identifying problems have delayed statistics far more than difficulties in solving problems.


0

As others have said in the comments, you almost certainly have enough data to warrant using a normal approximation for the distribution of the sample mean (using a plug-in estimate of population variance, suitably scaled). If you want to double-check, then do bootstrap sampling with $n=19593$ samples per bootstrap trial. Compare the distribution of the ...


1

0) I think you have some typos in the question. Let us make things more clear by assuming $Y$ has standard deviation $\left(\sqrt{\lambda}\right) \sigma$. 1) Draw a picture and compare the midpoints of the two line segments. 2) Define "centered densities": \begin{align} \tilde{f}_X(x) &= f_X(x+\mu)\\ \tilde{f}_Y(x) &= f_Y(x + \mu) \end{align} ...


1

Your terminology is a bit vague, confusing "plan to get grade A" with "she gets A", and confusing the probability of one particular person with the probability of group that person is in. But assuming you mean what you seem to mean, .... The meaning of the $75\%$ seems to be the probability of being from Berlin, or getting grade A, or both. Let $A$ be ...


1

Rewrite the thesis as $$E[u(X)] = \int u(x) dP^X \ge \int u(x + y) dP^X \otimes dP^Y = E[u(X + Y)]$$ Now let's concentrate on the right hand side; in particular we can use fubini tonelli to write $$\int u(x + y) dP^X \otimes dP^Y = \int \left(\int u(x+y)dP^Y\right) dP^X$$ The inner integral is simply $E[u(x + Y)]$ for some constant $x$. Since $u$ is ...


1

The naive approach is first to choose $k$ of the $m$ letters, then each location in the string has $k$ possibilities. This fails because if the string only has $k-1$ different letters in it, you will count it $m-k+1$ times, once for each different letter as the missing one. You need to use the inclusion exclusion principle to account for this.


1

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


0

Roughly, yes. For example, take the CI for the proportion of $4$'s. It is centered at 0.053. If these are the 'usual' kind of CIs used for a sample of size $n=1000,$ then the number of $4$'s in the sample must have been $0.053(1000) = 53$. In some cases, because the endpoints are rounded, you will not get an integer answer, but you can come very close to ...


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

Yes, there are many ways to incorporate the number of users in your rating scheme. For example, suppose N is the number of raters you think a thread should have and A is the number of raters the thread actually has. Then multiply the average rating by max(1,A/N).


1

Not necessary. Let A be a set of 100 humans with mean weight 70 kg, and standard deviation 20 kg. Let B be a set of 100 elephants with mean weight $8000$ kg and standard deviation 40kg. Clearly even with standard deviation doubled (for the same sample size) elephants seem to be more homogeneous in weight.


0

I would see it this way: if you have a large-enough sample-size for each, and the data is roughly-symmetric and the sample standard deviation is not too large, then you can consider $\sigma_a , \sigma_b $ to be valid measures of spread _from the underlying populations from which the data came. In this case, you may say that one population has less, or ...


0

Fitting of $$y=A(1-2e^{bx})$$ to a set of data $(x_1,y_1),(x_2,y_2), ... , (x_k,y_k), ... , (x_n,y_n)$ I agree with the method proposed by Claude Lebovici. Another approach consists in : Ranking the data in increassing order of $x_k$ Let $S_1=0$. Then, from $k=2$ to $n$ , compute $S_k=S_{k-1}+\frac{1}{2}(y_k+y_{k-1})(x_k-x_{k-1})$ Compute $b$ from : $$ ...


0

The model $$y = a\,(1-2e^{bx})$$ is nonlinear with respect to its parameters $a,b$ and nonlinear regression should be used. The problem is that you need to start with rather good estimates and that, as already said in comments, you cannot linearize the model. So, how to get estimates ? First approach Suppose that you fix $b$ at a given value. Then define ...


0

You understand more than you think. Try plotting the function when $\lambda = 1$. Note we have a nice curve. This is a probability mass function (pmf). This tell us the probability that the value $x$ will occur. We express this by saying $P(X=x)$. Consider the plot of our function, what if we want to know the probability that our random variable $X$ is ...


0

The model being intrisically nonlinear wih respect to its parameter, you will need nonlinear regression. However, the problem is to provide good estimates. One way to do it is to rewrite the model as $$y=a e^{qx}+b$$ with $q=\log(p)$. Now, a hint already provided by Yves Daoust here : choose three points $(x_1,x_2,x_3)$ such as $x_2\approx ...


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 ...


0

Just answering this for others searching (even though the current question has relatively low views). The only magic in computing an inhomogeneous Poisson process in a given time comes from computing the second moment. This is done quite easily by noting that, for some random parameter $\lambda_t$, the second moment if given by (for $N_T\sim ...


0

The result does not dependent on $n$ in the asymptotic information matrix. \begin{align*} \mathcal{I}\left(p\right)&=\underset{n\to\infty}{\mathrm{plim}}\dfrac{1}{n}\dfrac{n}{p\left(1-p\right)}\\&=\dfrac{1}{p\left(1-p\right)} \end{align*}


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 $y$ ...


0

P(Y < y) means the probability that Y is less than y. Similarly P(X > x) is the probability that X is greater than x. I hope I answered your basic question.


0

What you have calculated is the overall CV for the league in 2014. You need to find the mean and standard deviation of the points scored by each team in the IPL, then compute the CV for each team. This will give you an idea of each team's consistency. I think your second table will help.


0

To start you off, Case 1 : $\mathscr P$(a player selects a different # than $\mathcal A$) = $\frac{k-1}{k}$ $\mathscr P$(n-1 players select a different # than $\mathcal A$) = $[\frac{k-1}{k}]^{(n-1)}$ = X (say) $\mathscr P$(at least one player selects the same # as $\mathcal A$) = 1 - X Case 2 : Those who chose 2 or more (category P) can't select ...


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 ...


0

You have p=mu+0.67sigmafor the upper quartile andp=mu-0.67sigmafor the lower quartile. You are given values ofp. So just solve those two equations for mu and sigma(Please check values of 0.67 and -0.67 using z-table as I may remember those wrong)


0

Four speculative scenarios follow: Perhaps the context of the problem in the text or lectures could provide a clue which (if any) might be intended. 1) Uniform. Perhaps, somehow, "constant" is supposed to suggest a uniform distribution. To have mean 35, that would have to be $X \sim Unif(0, 70),$ so $P(X \le 45) = 45/70 = 0.6429.$ 2) Normal. Some yards may ...



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