New answers tagged

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Comments: This result is true only for normal data. The idea is to do an orthogonal transformation in which one of the new random variables is a function of $\bar X$ and the other $n-1$ are a function of $S^2$. For a multivariate normal, orthogonality is equivalent to independence. You might start with $n = 2$ where the proof is especially simple. Let $Y_1 ...


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I suggest watching Ted Shrifin's lectures from Math 3500 at University of Georgia - Atlanta: Inconsistency and Normal Equations Then Projections and Max/Min


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This is necessarily true: $P(No |Present)=1-P(Yes|Present)$ $P(Yes|Absent )=1-P(No |Absent )$ This is not necessarily true: $P(Yes|Absent )=1-P(Yes|Present)$ $P(No |Present)=1-P(No |Absent )$ So your table should look like this: $ \begin{array}{c|c} 0.919 & 0.081\\ \hline 0.726 & 0.274\\ \end{array} $


1

Use rule of complement of conditional probability which states: $P(A^C/B) = 1-P(A/B)$ Thus $P(Y/SN)+P(N/SN) = 1$ $P(Y/N)+P(N/N) = 1$ You have $P(Y/SN) = 0.919$ Thus $P(N/SN) = 1-0.919 = 0.081$ You have $P(N/N) = 0.274$ Thus $P(Y/N) = 1-0.274 = 0.726$


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If $A$ and $B$ had $80$ votes each already, with five states ($20$ votes each) still undecided, there would be a total of $260$ votes to be cast in this election. To receive a majority of the votes would require at least $131$ votes, meaning it would not be enough to win two states; a candidate would have to win three. If the only thing you change is that ...


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Each individual CI has its own confidence level (and hence error probability). When considering the error probability arising from several CIs considered simultaneously, the error probabilties can "gang up" to be larger than the error probability of any one CI. What you are seeking here is a 'family' error rate. The explanation in the NIST handbook may be ...


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$X_1,\ldots,X_n$ are i.i.d. $\mathcal N(\mu,\sigma^2)$ random variables and $\bar X=n^{-1}\sum_{i=1}^nX_i$. The distributions and confidence intervals are as follow. (a) $$ \frac1{\sigma^2}\sum_{i=1}^n(X_i-\bar X)^2\sim\chi_{n-1}^2 $$ and $$ \Pr\biggl(\frac{\sum_{i=1}^n(X_i-\bar X)^2}{\chi_{n-1,\alpha/2}^2}\le\sigma^2\le\frac{\sum_{i=1}^n(X_i-\bar ...


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If $P(k) = \frac{a}{2^k}$ for $0\leq k \leq 20$, then the fact that $\sum_{k=0}^{20} P(k)=1$ implies that $$ 1 = a\sum_{k=0}^{20} \frac{1}{2^k} = a\frac{1-\frac{1}{2^{21}}}{1-\frac{1}{2}} = 2a\left( 1-\frac{1}{2^{21}}\right) $$ i.e. $$ a = \frac{1}{2\left( 1-\frac{1}{2^{21}}\right)}. $$ You can check that this necessary condition is sufficient for $P$ to be ...


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The other responder has given an elegant solution, but here is another way by enumerating it a little bit. Probability for any one case could be calculated such as this P(Picking a Head from the Table)*P(2 books being Head)*P(1 book being a head) for the first case. The three books could be H H , H $$= 1.\dfrac{{5\choose2}}{{11\choose2}}.\frac{7}{11} ...


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Let $H$ denote the event that the book turns out to be a hardcover. Let $S_{i}$ denote the event that the book comes from shelf $i$. Then: $$P\left(H\right)=P\left(H\mid S_{1}\right)P\left(S_{1}\right)+P\left(H\mid S_{2}\right)P\left(S_{2}\right)$$ Can you work this out? Edit: See the comment of @Noble to this answer. It leads ...


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What is the probability that, if we pick a book from the first shelf, it is a hardcover? Clearly, that is $\frac{5}{11}$. What is the probability that, if we pick a book from the second shelf, it is a hardcover? Clearly, that is $\frac{7}{11}$. Now, let's say that we pick a book from the table. It is either from the first shelf or the second shelf: It's ...


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I think I have it (feel free to correct me if I am mistaken): I'm going to write $c$ equal to $c(i)$, assuming that it is a function of $i$ (it could be a constant function of $i$). $$\frac{\pi_j}{c(i)} = \frac{p(i,j)}{p(j,i)} = \frac{p(i,k)p(k,j)}{p(j,k)p(k,i)}$$ for any $k$. Then$$\frac{\pi_j}{c(i)} = \frac{\pi_j}{c(k)} \frac{\pi_k}{c(i)}$$ for any $k$, ...


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Edit: The original question asked about independence, and was answered by the example below. This example also settles the modified question about covariance, since $\text{Cov}(XY,Z)\ne 0$. Toss a fair coin twice. Let $X=1$ if we have head on the first toss, and $0$ otherwise. Let $Y=1$ if we have head on the second, and $0$ otherwise. Let $Z=1$ if the ...


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$$f'(\lambda) = \frac{pe^{\frac{\lambda}{n}}}{1-p+pe^{\frac{\lambda}{n}}}$$ $$f''(\lambda) = \frac{\frac{1}{n}(1-p)pe^{\frac{\lambda}{n}}}{(1-p+pe^{\frac{\lambda}{n}})^2}$$ $$f'''(\lambda) = \frac{1}{n}(1-p)p \frac{\frac{1}{n}e^{\frac{\lambda}{n}}(1-p -pe^{\frac{\lambda}{n}})}{(1-p+pe^{\frac{\lambda}{n}})^3}$$ Use GM-AM inequality, $$ f''(\lambda) \le ...


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To get exactly 5 heads and 5 tails, the score after 9 tosses has to be 5-4 one way or the other. In this case, there is a $50\%$ chance you get to 5-5. On the other hand if you aren't at 5-4 (one way or the other) after 9 flips, there is no chance of getting a 5-5 split. So certainly the chance of 5-5 is less than $0.5$.


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There are several methods to solve this, but I like the following one: I assume that you meant to say exactly 5 heads. Then what does such a toss look like? For instance, it could be: $$HHTHTTTHTH$$ where H means you tossed a head, and T a tail. So now all the possibilities you have are the permutations of this sequence, hence $\frac{10!}{5!5!} = 252$ ...


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Let's imagine a case with $4$ rolls where we desire $2$ heads (MathJax diagrams only go so far...) $$\newcommand{\mychoose}[2]{\bigl({{#1}\atop#2}\bigr)}$$ $$\begin{array}{ccccccccccc} & & & & & & & H & & & & & \\ & & & & & & \swarrow & & \searrow & \\ & ...


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can also be expressed as a binomial distribution which simplifies to the same but is ${10\choose 5}({1\over 2})^5({1\over 2})^5$. but this way if the probability was not $1\over 2$ could still calculate.


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I will use $t$ as the variable instead of $\lambda$. The required expectation is $$\left(E(\exp(\frac{tX_1}{n})\right)^n,$$ and $$E(\exp(\frac{tX_1}{n}))=(1-p)+pe^{\frac{t}{n}}.$$


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The answer you got - 0.246 is the probability of getting 'exactly' 5 heads. Your intuition gives the 'Expectation' E(x). When 10 coins are tossed, the Expectation is that you get 5 heads. Your intuition makes an average of all the cases while the solution takes only the cases where number of heads is 'exactly' 5. Let me give you another problem- What is the ...


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You can test your intuition by considering extreme versions of the original problem. If you had to toss a coin two million times, would your intution suggest that the chance is $1/2$ that you'll get exactly a million heads? No, because in many tosses, there is 'chance variation' that prevents you from getting exactly half heads and half tails. Also, if the ...


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Let's assign a measure $m$ to Borel subsets of the half-open interval $[0,\infty)$ by specifying that the measure of every open interval is its length and $m(\{0\})=1$, and measures of all other Borel sets are accordingly determined. Let $f$ by a probability density with respect to the measure $m$, so that \begin{align} & \int_{[0,\infty)} f(x)\,dm(x) = ...


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The hypothesis test is only on women. So, you can go with the number of women and their data alone.


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If there are no additional assumptions, then the smallest possible value of the variance of their mean is $0$ (suppose that both of the random variables are equal to the same constant almost surely). We have that $$ ...


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I don't think there's enough information in your model to work out a satisfactory answer. If we draw up a table of scores like this: |0 1 2 3 -+------- 0|0 1 2 3 1|1 2 3 4 2|2 3 4 5 then I can populate it with the number of students like this (assuming only 10 students): |1 3 4 2 -+------- 2|1 1 - - 5|- 2 3 - 3|- - 1 2 Each column and row adds up to ...


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Information geometry is introduced in Amari's book. A concise treatment of the relation between information theory and statistics can be found in "Information Theory and Statistics : A Tutorial by Csiszar and Shields. Particularly section 3 and 4 are dedicated to information geometry. They are credible researchers on this topic. There is this chapter of ...


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Edit: correction of a typo: exchange between $3/2$ and $2$ on line 9. @Tankonetoone: On an experimental basis, I am convinced that the assertion is NOT TRUE, under hypothesis $\sum \lambda_k \leq \sum \beta_k < \infty $. BUT, it LOOKS TRUE (for all the cases I have tried...), if one assumes the inequality $\sum \beta_k \leq \sum \lambda_k < \infty $. ...


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An answer with hidden text, in case you want to try it out again before looking at the solution. Using your definition of $Y_n$, you should have used the central limit theorem to observe $$ \sqrt{n}(Y_n/n - 1)\stackrel{d}{\longrightarrow}N(0,2)\tag{1} $$ So, using the delta method, with the function $x\mapsto\sqrt{x}$ and $(1)$, gives Therefore, ...


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A standard 95% CI for $\lambda$ for a particular time period based on counting $X$ events in that time period is $X \pm 1.96\sqrt{X}.$ This is sometimes called a Wald interval; it uses a normal approximation, and so works best for fairly large $X$. A slight improvement is the CI $X + 2 \pm 1.96\sqrt{X + 1}.$ Its coverage probability tends to be closer to ...


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Note that the support iof $X$ is dependent on the unknown parameter $\theta$, so $X_{(n)} \le \theta$, and $\hat{\theta}_n = X_{(n)}$. For consistency you have to show that $$ \lim_{n\to\infty}P(|X_{(n)} - \theta| \ge \epsilon) = 0, $$ due to the fact that $X_{(n)} \le \theta$, you have \begin{align} \lim_{n\to\infty}P(|X_{(n)} - \theta|\ge\epsilon) &= ...


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There are many ways in which the assumption of IID variables can be broken in real-life applications. Let me illustrate how it can break down in your case of hypertension. I'm not a medical doctor or epidemiologist, so I'll make up some assumptions that are at least plausible to a non-medical doctor. Suppose people who work hard and are financially ...


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I guess that we can agree on the premises that hypertensive status affected or at least correlated with factors like age, genetics, diet, sport habbits etc. Assume that you have a clinic and you are measuring BP to each person that comes to your clinic (from whatever reason). Assuming, that the measured values are absolutely independent is unrealistic. As ...


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The table you've linked is a pretty nonstandard format for a z-score table, but it seems to be referring to a situation like this: The area under the curve between $0$ and $t$ is the probability of a normally distributed variable falling between $0$ and $t$. Using the fact that the curve is symmetric about $0$, you can deduce the probability of a normally ...


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If you're allowed to take that sample repeatedly, it's basically bootstrapping. Procedure: Draw 100 points Calculate standard deviation Repeat Steps 1 & 2 a lot of times (empirically, I've found 5-10,000 to be enough), keeping track of the results of step 2. Examine the distribution of estimates from Step 2 with whatever tools you'd like -- ...


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If you want to find out the uncertainty or standard error (SE) in the standard deviation of a chosen sample, then you can simply use $SE(\sigma) = \frac{\sigma}{\sqrt{2N - 2}}$, where $N$ is the number of data points in your sample. Hope that helps!


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Note that $$\operatorname{E}[\hat \theta_1] = \operatorname{E}[e^{-X}] = M_X(-1),$$ where $M_X(t) = \operatorname{E}[e^{tX}]$ is the moment generating function of $X$. For $X \sim \operatorname{Poisson}(u)$, we can compute $$M_X(t) = \sum_{x=0}^\infty e^{tx} e^{-u} \frac{u^x}{x!} = \sum_{x=0}^\infty e^{-u} \frac{(ue^t)^x}{x!} = e^{u(e^t-1)} ...


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Your expression for likelihood is not completely correct. It should be $$L(\theta) = \exp(n\theta) \exp(-\sum_1^n x_k) 1_{\theta \leq X_{(1)}}$$ where $X_{(1)}$ is the smallest of all $X_i$'s. We need the extra term because the likelihood will be zero if any $X_i$ is smaller than $\theta$. Now, we know that exponential is increasing function and we want ...


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In part (1) there is no theorem that states $\mathbb E(\exp Y) = \exp \mathbb E(Y)$. You can't move the expectation past the exponential. Instead, use the general formula: $$ \mathbb E(g(X))=\sum_{k=0}^\infty g(k)P(X=k), $$ which is valid for any function $g$ when $X$ takes values $0, 1, 2,\ldots$. For part (1) the formula gives $$ \mathbb E(e^{-X}) = ...


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Suppose there are $2n$ elements of $A$ and $2m$ elements of $B$. Then there are $n$ elements of $A$ which exceed (or equal) $10$ and $m$ elements of $B$ that exceed (or equal) $20$. of course the latter implies that there are at least $m$ elements of $B$ which exceed (or equal) $10$. Combining those we see that there are at least $n+m$ elements of $C$ which ...


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We know that if $X_1$ and $X_2$ are independent normal random variables, then their sum $$X_1 + X_2 \sim \operatorname{Normal}(\mu_1 + \mu_2, \sigma_1^2 + \sigma_2^2).$$ Therefore, $$\mu_Y = \operatorname{E}[Y] = \operatorname{E}[(X_1 + X_2)^2] = \operatorname{Var}[X_1 + X_2] + \operatorname{E}[X_1+X_2]^2 = \sigma_1^2 + \sigma_2^2 + (\mu_1 + \mu_2)^2.$$ ...


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When $y \in [0,1]$ you need integrate over x in $ [0,1]$, when $y>1$ you have integrate over x in $[y-1,1] $ Hope this address your question.


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Using the fact that the density must integrate to one over $[0,1]$, then proportional means $$1 = \int_0^1cf(x)\,dx = \int_0^1\frac{c}{\sqrt x}\, dx = c\left[2\sqrt x\right]_0^1 = 2c.$$ Thus $c = 1/2$. If you now try to compute the expectation, you will find $$\int_0^1x\cdot \frac{1/2}{\sqrt x} = \frac{1}{3},$$ which is option 3.


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We need to draw a picture. Draw the line $y=x+1$. If we look at the conditions, we can see that the joint density lives on the trapezoid bounded by the $y$-axis, the $x$-axis, the vertical line $x=1$, and the line $y=x+1$. We want to "integrate out" $x$. Look at the picture. If $0\le y\le 1$, then $x$ travels freely from $0$ to $1$, and hence the density ...


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We are told the painter selects two cans at random, so it can't be that she, for example, selects can 3 twice. Hence the tuples (3, 3), (4, 4), and (5, 5) are not possibilities, leaving only six choices. Thus the probability of glossy and glossy is 6/20, and these probabilities do add to 1.


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$\newcommand{\Var}{\operatorname{Var}}\newcommand{\Corr}{\operatorname{Corr}}\newcommand{\Cov}{\operatorname{Cov}}$ Using the known properties of covariance gives \begin{align*} \Corr(X,Z) &=\frac{\Cov(X,Z)}{\sqrt{\Var(X)\Var(Z)}}\\ &=\frac{\Cov(-10Y+10,Z)}{\sqrt{\Var(-10Y+10)\Var(z)}}\\ &=\frac{-10\Cov(Y,Z)}{\sqrt{100\Var(Y)\Var(Z)}}\\ ...


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If in doubt, compute. We have $$\text{Cov}(X,Z)=\text{Cov}(-10Y+10,Z)=-10\text{Cov}(Y,Z).$$ Also, $\text{Var}(X)=100\text{Var}(Y)$. It follows that $$\rho(X,Z)=\frac{\text{Cov}(X,Z)}{\sigma_X\sigma_Z}=\frac{-10\text{Cov}(Y,Z)}{10\sigma_Y\sigma_Z},$$ and therefore $r_1=-r_2$. Informally, scaling by some positive factor does not change the degree of linear ...


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It's not Bayes theorem; it's just the definition of conditional probability. The desired prob, by definition of conditional probability, is $${P(\mbox{child is heterozygote & child has brown eyes & parents have brown eyes})\over P(\mbox{child has brown eyes & parents have brown eyes})}.$$ But the numerator simplifies to $$P(\mbox{child is ...


1

It is correct, but can be further simplified by considering the order statistics of the sample. That is to say, if $$\boldsymbol x = (x_1, \ldots, x_n)$$ is the sample, then consider $$x_{(1)} = \min_i x_i, \quad x_{(n)} = \max_i x_i,$$ the first and last order statistics which are equal to the smallest and largest observations in the sample, respectively. ...


1

In general, let $X_1,\dotsc, X_n$ be iid eponential distribution with mean $1/\lambda$. Then the distribution of the minimum $M$ is $$P(M\leq m) = 1-P(M>m) = 1-(e^{-\lambda m})^{n} = 1-e^{-\lambda nm}.$$ Notice that this shows that $M$ follows an exponential distribution with mean $\frac{1}{\lambda n}$. In our case, $\lambda = 1/22.5, n =4$, and so $$E[M] ...



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