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$\newcommand{\cov}{\operatorname{cov}}\newcommand{\E}{\operatorname{E}}$I'm going to assume that where you wrote "multinormal" you meant "multinomial" and not multivariate normal. The probability that you wrote above should be $$ P(N_1=n_1,N_2=n_2,\ldots,N_r=n_r)=\frac{n!}{n_1!n_2!\cdots n_r!} p_1^{n_1} p_2^{n_2} \cdots p_r^{n_r}. $$ You have $n$ ...


1

Let $X,X_1,X_2,\dots$ be constant random variables: $X(\omega)=x$ and $X_n(\omega)=x_n$ for each $\omega\in\Omega$. If $x_n$ converges to $x$ then for each $\epsilon>0$: $$\lim_{n\rightarrow\infty}P(|X_n-X|>\epsilon)=0$$showing that $X_n$ converges in probability to $X$. However for each $n$ with $x_n\neq x$ we have $P(|X_n-X|>0)=1$.


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For this specific problem, $\text{E}(X_i) = \lambda$ and $\text{Var}(X_i)=\lambda$ for $i=1,\ldots,n$. This is can be calculated from the Poisson distribution. Furthermore, the sum of $n$ independent Poisson$(\lambda)$ random variables is $$ Z = \sum_{i=1}^n X_i \sim \text{Poisson}(n\lambda). $$ Hence, $\text{E}\left(Z\right) = \text{Var}\left(Z\right) = ...


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I am not sure how relevant this my answers is; but there has some literature which suggests that the act of confirminga parameter value parameter value using hypothesis testing on frequency data, can actually falsify that parameter. Its called the goodness of fit paradox; mentioned in "Falsification of Propensity Models by Statistical Tests and the ...


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According to http://www.insidescience.org/content/are-7-game-world-series-more-common-expected/681 the respective probabilities of 4, 5, 6, and 7 games are 0.125,0.25,0.3125 and 0.3125, for evenly matched teams (assuming independently played games). Perhaps knowing the answers, you can straighten out your combinatorial methods. [For example, to have a four ...


1

$\phi^{t-1} w_1 + \phi^{t-2} w_2 + \ldots+ w_t = \sum_{i=1}^{t-1} \phi^{t-i} w_i$ This equation is not correct: On the left-hand side there appears a "$w_t$"-term, but on the right-hand side it doesn't. So, instead it should read $$\phi^{t-1} w_1 + \phi^{t-2} w_2 + \ldots+ w_t = \sum_{i=1}^{\color{red}{t}} \phi^{t-i} w_i$$ Except for that your ...


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Best way to view this is to view it as TWO dependent events. a) simply use r studio and use dbinom(2, 15, 0.02) b) Use a tree diagram P(S) = 0.9 P(NS) = 0.1 P(disease) = 0.02 P(nodisease) = 0.98 therfore to get a false positive (0.1 X 0.98) and for the second section of b is just (0.01 X 0.02) Let me know if there is any question because i am doing ...


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Assuming your calculations of the expectation and variance of the estimator $$\hat e = p \sum_{i=1}^n (X_i - \bar X)^2$$ are correct, then you have $$MSE = \operatorname{E}[\hat e - \sigma^2]^2 + \operatorname{Var}[\hat e] = (p(n-1) - 1)^2 (\sigma^2)^2 + 2p^2 \sigma^4 (n-1),$$ or $$MSE = \sigma^4\left((n^2-1)p^2 - 2(n-1)p + 1\right).$$ Then take the ...


4

Just a guess, but perhaps $f$ is the expectation operator and $x$ is a random variable? If you're trying to understand why $$E[(X-E(X))^2] = E(X^2)-E(X)^2\;,$$ then this post might help. Another derivation can be found here.


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There are $100(100^2) = 10^6$ cm$^2$ in the full area of $100$ m$^2$. Therefore, in $10$ minutes, an average of one raindrop fell in each cm$^2$, and the rate is $0.1$ raindrop/cm$^2$/minute. You answer to part (a) is correct: The probability that at least one raindrop falls in any given cm$^2$ is $$ P(X >= 1) = 1-P(0) = 1-\frac{(10 \cdot ...


1

Recall that $ \hat \beta_0$ and $ \hat \beta_1$ are determined by minimizing the term of the right-hand side. Thus, we have the pair of equalities $$\sum_{i=1}^n (y_i-\hat \beta_0-\hat \beta_1 x_i)=0$$ $$\sum_{i=1}^n (y_i-\hat \beta_0-\hat \beta_1 x_i)x_i=0$$ Using these relationships we see that $$\begin{align} \sum_{i=1}^n (y_i-\hat \beta_0-\hat ...


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No, not quite that.   Conditional Mutual Information is: $$\begin{align} I(X;Y\mid Z) & =D_{KL} (p(X,Y,Z)\parallel p(X\mid Z)\;p(Y\mid Z)\;p(Z)) \\[2ex] & = \sum_{z\in Z} p_{_Z}(z)\; D_{KL}(p(X,Y\mid Z=z)\parallel p(X\mid Z=z)\;p(Y\mid Z=z)) \\[3ex] & = \sum_{x\in X}\sum_{y\in Y}\sum_{z\in Z} p_{_Z}(z)p_{_{X,Y\mid Z}}(x,y\mid ...


1

Mutual information is often written $I(X;Y)=D_{KL}(p(X,Y)||p(X)p(Y))$. Your instructor has provided a slightly generalised version which depends on some other variable, $Z$. You are free to take expectations over this. Note that if $Z$ is constant, then that leads to the usual definition.


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The "Note: $P(X=x^2)=P(X=x)$" is simply wrong. They should have stated: $P(X^2=x^2)=P(X=x)$   so that: $$\begin{align}E(X^2) & = \sum\limits_{x^2\in\{1,4, 9, 16\}} x^2 \;\mathsf P(X^2=x^2) & \color{silver}{= \sum\limits_{y\in\{1,4, 9, 16\}} y\; \mathsf P(X^2=y)} \\ & = \sum_{x=1}^4 x^2\; P(X=x)\end{align}$$ PS: More generally ...


1

I think this is just an error in the book. It looks to me like every occurence of $P(X = x^2)$ on the page should be $P(X^2 = x^2)$.


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Chebyshev's Inequality is true for any $c > 0$, but you are right that it only provides useful information for $c > 1$. This is actually surprisingly easy prove. Define $\mu = E(X)$ and $\sigma^2 = E((X-\mu)^2)$. Observe that for any $c \geq 0$ we have $\mathbb{1}\left\{\left|\frac{X-\mu}{\sigma}\right|\geq c\right\} \leq \frac{(X-\mu)^2}{\sigma^2 ...


1

I am having trouble making sense of this problem. In your notation, the usual regression model is $$Y_i = a + bx_i + \epsilon_i,$$ where $\epsilon_i$ are distributed $Norm(0, \sigma_\epsilon^2)$, for $i = 1, \dots, n.$ A 95% confidence interval for the slope is $$ b \pm t^* s_\epsilon \sqrt{1/S_{xx}},$$ where $t^* = 3.182$ cuts off area .025 from the upper ...


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I think by $p = 0.6$ you mean $\rho = 0.6,$ which is the correlation. If so, you have only to look at the relationship between covariance and correlation. If not, I wonder what $p$ might be.


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It looks like you are given the correlation $\rho$ of the two random variables $X$ and $Y$. There is a formula connecting the correlation to the covariance, and it is $$\rho(X,Y) = \frac{\text{Cov}(X,Y)}{\sigma_X\sigma_Y}.$$ Do you see what to do from here?


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Let $W_1,W_2$ denote the random variables for weights for the randomly selected students. Let $Y=W_1 + W_2 + 10.$ Assuming $W_1$ are $W_2$ are independent, we get: $E[Y]=E[W_1+W_2+10] = E[W_1]+E[W_2]+E[10] = 170 + 170 + 10 = \boxed{350}$ and $Var[Y]=Var(W_1 + W_2+10) = Var(W_1) + Var(W_2) + Var(10)= 10+10+0=\boxed{20}$


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Let us assume that the 1007 measurement are equally spaced in time. It follows that in most of the years ($637$) there was only one measurement. In the remaining years ($185$) there were two measurements. Now the probability that one measurement exceeds $X$ is $0.1$. For a year with two measurements the probability is easily found to be $0.19$. For a ...


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There are so many possible points of confusion here that I hardly know where to start. First, I wonder what 'time' means in your tabulation. It can hardly be seconds. If the rate is 88/sec. then how can there possibly be 908 within the first second. Second, one might suppose 'rates' refer to rates of exponential distributions. But then there is no ...


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The formula for a 95% confidence interval with a lower bound will be $p - z^*\sqrt{p(1-p)/n},$ where $p = .45,$ $n = 1060,$ and $z^* = 1.645$ cuts 95% from the upper tail of a standard normal distribution. This gives $0.5148181 \approx 51.5\%.$ The interpretation is we have 95% confidence that more than 51.5% of individuals in the population (from which ...


0

You need to distinguish between events and random variables. I think you mean to say random variable $X$ has the discrete uniform distribution on the integers between 100 and 400 inclusive. Then $P(X = i) = 1/301$ for any integer $i \in [100, 400].$ Also $Y is an independent random variable with the same distribution. In order to grasp what this problem ...


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Arrange ZLGEBR, then replace Z with AA.


2

Linearity of expectation is (sometimes surprisingly) always valid. If I roll two dice then the expected value of the first is $3.5$ and the exopected value of the second is $3.5$ and the expected value of the sum is $7$. Now you'll say "Sure, that's cause they are independent." But now do this: I roll the first die and then manually place the second die so ...


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Step $I$: Pick $6$ balls and weigh them $3$ on each side. Step $II$: $1$. If the balls balance each other, then weigh the remaining two balls and the heavier one will tilt the balance on its side. $2$. If the balls do not balance each other, then the heavier ball should be one of the three balls on the side the balance tilts. From these three balls, ...


0

$$P(Z < 1) = \alpha \approx 0.8413$$ $$P(Z > 1) = 1-\alpha = P(Z < -1)$$ $$P(-1 < Z < 1) = 1 - P(Z>1)-P(Z<-1)$$ $$P(-1 < Z < 1) = 1 - 2\cdot P(Z>1) = 1 - 2\cdot (1-\alpha)$$ $$P(-1 < Z < 1) = 2\alpha - 1 \approx 0.6826$$


0

You can assume some model for votes for a book, e.g. that is basically Gaussian with an unknown mean vote score $\mu$, and some global standard deviation that you calculate for example by computing the average standard deviation $\sigma$ over all books, where you take the average over all books that have more than 1 vote. Then your assumption, under a ...


1

This question has been plaguing internet ranking sites for a while, e.g. reddit. Fortunately it has a nice theoretical solution. It involves a bit of statistics; the Ruby code is: require 'statistics2' def ci_lower_bound(pos, n, confidence) if n == 0 return 0 end z = Statistics2.pnormaldist(1-(1-confidence)/2) ...


1

i.) Your first result is wrong because the question is related to the total waiting time, $X+Y$: $$P(X+Y<30)=\iint_{\{(x,y):x+y<30\}}f_{X,Y}(x,y)\ \ dxdy=$$ $$=\frac{1}{100}\int_0^{30} e^{-\frac{1}{10}x}\int_0^{30-x}e^{-\frac{1}{10}y}dydx=$$ $$=\frac{1}{10}\int_0^{30} e^{-\frac{1}{10}x}\left[1-e^{-\frac{1}{10}(30-x)} \right]dx=$$ ...


2

I will set out a solution strategy and some hints, in keeping with the site's homework policy. It's quite an effective one in general - in this case, checking the back of my envelope, I used four simple lines of algebra and three rough sketch plots (two PMFs and an overlaid contour plot) - but other approaches are available. In particular it can make life ...


2

Here's an explicit counterexample. Let $U=|X|$ and $V=|Y|$. Then $\text{Var}(X)\geq \text{Var}(Y) \Rightarrow E(|X|)\geq E(|Y|)$ is equivalent to $E(U^2)\geq E(V^2) \Rightarrow E(U)\geq E(V)$ for $U,V$ continuous in [0,1]. Consider a $\text{beta}(\alpha,\beta)$; it has mean $\mu=\frac{\alpha}{\alpha+\beta}$ and second raw moment $\mu_2'=\mu ...


0

What's the probability the first card is a king? What's the probability the second card is a king? etc. Then apply the rule of product: $$P= \frac 4 {52} \cdot \frac 3 {51} \cdot \frac 2 {50} \cdot \frac 1 {49}$$


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Assuming you're drewing four cards, there's $1$ way of drawing four kings, and there are $52 \choose 4$ ways of drawing any four cards. Hence the probability of drawing four kings is $\frac{1}{52 \choose 4}$.


0

Yeah. If your data points are $x_1,x_2,\dots,x_n$, then you can compute $n-1$ new data points $y_1,y_2,\dots,y_{n-1}$ which are the differences between consecutive $x_i$s. Specifically let $$y_i=|x_i-x_{i+1}|$$ Now you can simply compute the average of this to find the average difference from one day to the next (you can also compute standard deviation, ...


0

You want the probability of picking a bad watch and then two good watches. That can be expressed as: $$\frac{1}{10} \cdot \frac{9}{9} \cdot \frac 8 8 = \frac 1 {10}$$ But that's an ordered result. You can get this to happen in three different ways (pick a bad watch first, or pick a bad watch second, or pick a bad watch third). So we multiply by three. The ...


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The easiest way to get to an answer is to think that the watch breaks after you've chosen your $3$. You want the probability that the one that decides to break is among your $3$ out of $10$ watches. Then the answer is right there, it's $\frac{3}{10}$.


0

For a problem like this, it's easier to find the probability of not getting a broken watch and then subtracting it from 1 to get the probability of the complementary set--getting at least one broken watch. The probability of getting a non-broken watch is 9/10. After one non-broken watches has been removed, the probability is reduced to 8/9. Then, 7/8. We ...


0

The answer is $$\frac{C_9^2}{C_{10}^3}.$$ The denominator is the number of all possible 3 watches combinations. The numerator is the number of all possible 3 watchers combinations containing the bad watch.


1

Your first probability is incorrect. The probability of choosing an empty box on the first try is $\frac{3}{5}$; your $\frac{2}{5}$ is the probability of choosing a box that contains an item. Your second probability is good: with only 4 boxes left and 2 that contain items you want, the probability of getting an item is $\frac{1}{2}$. Mutliply them ...


1

You are confusing two different uses of "p" in this problem. First, p = 0.07 is the proportion of defective light bulbs (the stated defect rate). It simply means that the probability of a randomly selected bulb being defected is 0.07. Second, the p-value of the hypothesis test is given as 0.087. This means that the janitor has conducted a test. Typically ...


1

This density function is for an uncorrelated bivariate normal distribution. So it is the product of two normal density functions. Recognizing that is the easy path. Otherwise transform to polar coordinates to integrate. Of course, you need the integral over the plane to be unity.


3

Given that $Y$ is Bernoulli distributed, it will have probability mass function $P(Y=0)=(1-p)$ and $P(Y=1)=p$. As Frank has helpfully commented, we can condition on $Y$. When $Y=0$, we have $P(\frac{X}{Y-X}<0|Y=0)=P(-X<0|Y=0)=P(X>0|Y=0)$ When $Y=1$, we $P(\frac{X}{Y-X}<0|Y=1)=P(\frac{X}{1-X}<0|Y=1)=P(X>1|Y=1)$ Thus, we have the ...


1

As long as a matrix is symmetric and positive-definite, it can be a covariance matrix. Yours appears to satisfy the conditions, so it is a covariance matrix.


0

Just look to the symmatry in the area of the standard normal disturbution CDF, you can see that the right area of the CDF with respect to a point of abscissa say $x$ with $x$>0 is the same area of that on the left of the point of abscissa -$x$ and hence you can say that $\psi(-x)$= $P(Z \le -x)$ = $P$($Z \ge x) \ $= $\ 1 - P(Z \le x)$ =$ 1 - \psi(x).$


1

If $a=0$, then the correlation between $x$ and $y = ax + b$ is not defined. Computation of Pearson's correlation, would involve division by 0. In the computation of Spearman's correlation, all values of $y$ are equal, and there is no meaningful way to rank the $y$s. [Note: You give one formula for Spearman's correlation. Another valid computation is to ...


0

The general principle, for an interval, is $$P(a\le Z\le b)=\Phi(b)-\Phi(a)$$ Sometimes the interval is half-infinite, e.g. $b=\infty$. Then $$P(a\le Z)=P(a\le Z\le \infty)=\Phi(\infty)-\Phi(a)=1-\Phi(a)$$ There is no inconsistency, and $\Phi(1.5)\neq \Phi(-1.5)$.


1

The CDF $\Phi(x)$ is all the area to the left of $x$. $P(x \leq Z \leq y)$ is the area which is to the left of $y$ and to the right of $x$. This excludes the area which is to the left of $x$. So you subtract that off, getting $\Phi(y)-\Phi(x)$. Visually this is the area to the left $y$ minus the area to the left of $x$. Alternately, going the other way, you ...


0

Pearson's correlation measures the linear component of association. It will be +1 if a plot of one variable against another has all points exactly on an upward-sloping line, and -1 if all points are exactly on a downward-sloping line. If there is little or no linear aspect of association then Pearson's correlation will be near 0. Spearman's correlation is ...



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