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I provide two answers and you may decide if they could be considered as intuitive. The first answer is a geometrically based combinatorial proof the second is based upon Egorychev's formal residual calculus for power series (and may also be considered intuitive if you like to play with generating functions). First answer using lattice pathes: ...


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Let $C$ be the event a randomly chosen patient is critical, and $D$ be the event that the event that a randomly chosen patient dies. We want $\Pr(C|D)$. By the usual formula for conditional probability, we have $$\Pr(C|D)=\frac{\Pr(C\cap D)}{\Pr(D)}.\tag{1}$$ We want to find the probabilities on the right of (1). The probability $\Pr(C\cap D)$ is, as you ...


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The probability of A given B is equal to P(A|B) = $\frac {P(B | A)\, P(A)}{P(B)}$. In your problem, B = "Given that a patient dies", and A = "the probability that the patient was classified as critical". This is known as Baye's Theorem. I suggest you read up on it as it is very important in probability.


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Let $A=$ the event that the patient is in critical condition, $B=$ the event that the patient is in serious condition, and $C=$ the event that the patient is in stable condition. Let $D=$ the event that the patient dies. The we know $P(A)=0.2, P(B)=0.3, P(C)=0.5, P(D|A)=0.3, P(D|B)=0.1, P(D|C)=0.01$. Then ...


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You want linearity of differentiation and the chain rule: $$ \begin{align} \frac{\partial \sum \ln (1 + \theta x_i)}{\partial \theta} &= \sum \frac{1}{1 + \theta x_i} \frac{\partial (1 + \theta x_i)}{\partial \theta} \\ &= \sum \frac{x_i}{1 + \theta x_i} = 0 \end{align} $$ I can't see a solution to that though... I would use the Newton-Rhapson ...


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WHICH product integral? The geometric product integral: $\prod_a^b f(x)^{\operatorname d x} = {\sf e}^{\int_a^b \ln f(x) \operatorname d x}$ $\begin{align} \prod_{0}^{\theta} \left((1+x)^{-(\theta+1)/\theta}\right)^{\operatorname{d} x} & = {\sf e}^{-((\theta+1)/\theta)\int_0^\theta \ln(1+x) \operatorname{d}x } \\ & = {\sf e}^{-((\theta + ...


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If we have $ n $ sample size, then likelihood is $ L(x,\theta)= \prod f(x_i) $ So $ L(x,\theta)= (1/\theta^{2n})\prod x_i e^{-\sum x_i/\theta} $ Since $ 0<\theta <\infty $. We can use likelood as $ log L $. So $ log L = -2nlog\theta + \sum logx_i -\sum x_i /\theta $ differentiate with respect to $\theta$ and equate with $ 0$ we get $\theta_0 = \sum ...


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The likelihood function is given by: $$ L(\theta)=\prod_{i=1}^nf(\theta|x_i)=\prod_{i=1}^n\theta^{-2}x_ie^{\frac{-x_i}{\theta}}=\theta^{-2n}\left(\prod_{i=1}^nx_i\right) exp\left(\frac{-1}{\theta}\sum_{i=1}^nx_i\right) $$ Taking the log, to find the log-likelihood: $$ ...


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For (a), consider Alice, who owns two motorcycles, and Bob, who owns one. What was the prior probability that Alice would be selected for the survey? What was Bob's probability? For (b), one might think that the department could determine the total number of motorcycle owners with exactly one registered motorcycle each, and multiply the estimated fraction ...


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Presumably, the opinion $R$ of someone doesn't change when their asked multiple times, but it may depend on the number of motorcycles owned $O$. If so, then the expected value of your sample proportion $p$ is $\sum P(O=i)E(X|O=i)$. Therefore, your estimator is biased if there is substantial stratification of opinions based on number of motorcycles owned. In ...


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Why guess randomly? First off, your first three digits are nearly uniquely determined by your state of birth. This information is not hard to deduce, as most people's current state is also their state of birth. The group numbers are generated not consecutively, but in a defined order based on birthdate, so if you can get an estimate of the time of birth, you ...


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Use the Neyman Factorization Theorem: Given a sample $x=(x_1,x_2,...,x_n)$, the joint pdf of our sample is is $f(x;\theta)=\left(\frac{2}{3\theta}\right)^n\prod_{i=1}^n \left(1-\frac{x_i}{3\theta}\right)=\left(\frac{2}{3\theta}\right)^n\prod_{i=1}^n \left(\frac{3\theta-x_i}{3\theta}\right)=\left(\frac{2}{9\theta^2}\right)^n\prod_{i=1}^n ...


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If you are testing fit to a specific distribution, then note that for a sample of data $X$ and hypothesized distribution $F, F^{-1}(X) \sim U(0,1)$. Therefore, you need to derive the distribution of the largest vertical difference between your transformed sample and the hypothesized CDF, which will be taken to be the standard uniform. The distribution of ...


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Add up the following: The probability that it lands on the red side exactly $4$ times is $\dbinom{6}{4}\cdot\dfrac{5^2}{6^6}=\dfrac{375}{46656}$ The probability that it lands on the red side exactly $5$ times is $\dbinom{6}{5}\cdot\dfrac{5^1}{6^6}=\dfrac{30}{46656}$ The probability that it lands on the red side exactly $6$ times is ...


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If the random variables are sampled from a density $f$ that is bounded and strictly positive in a neighborhood of zero, show that $\text{E}|\bar{X}_n^{-1}|=\infty$ for every $n$. This is wrong if the density $f$ is allowed to be zero on the left of $0$ (and bounded and strictly positive on the right of $0$): try $f$ the standard exponential density, ...


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The minimal case would be if one dice rolled a 1, another a 2 and a third a 3. The sum of these numbers is not less than six. The probability is zero.


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You will want to using the following theorem that For rv $X$ with cdf $F_{X}(t)$, $Y=F_{X}(X)\sim Uni(0,1)$ Proof: We will look at case that $F_{X}(t)$ is invertible, which you will see is true in with your example. So to find distribution of $Y$ we will see what cdf is. Thus we have ...


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You want some function $G$ so that, when $U$ is uniform on $[0,1]$, $G(U)$ has density $f$. If we let $F$ be the cumulative distribution corresponding to $f$, this means we want $P(G(U)<x)=F(x)$. But $P(G(U)<x)=P(U<G^{-1}(x))=G^{-1}(x)$, meaning $G^{-1}(x)=F(x)$, or $G(x)=F^{-1}(x)$. Thus, to find $G$, you need only compute $F$ and find its inverse. ...


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If you're having trouble finding distributions (or densities), it's usually not a bad idea to start from the cdf, $F_X(x) = P(X \leq x)$. Let $Y = |X|$, $X \sim N(0,\sigma^2)$ and let $F_X$ be the cdf of $X$. Then, for $y \geq 0$, \begin{align*} P(Y \leq y) &= P(|X| \leq y) \\ &= P(-y \leq X \leq y) \\ &= P(-y \leq X \leq 0) + P(0 < X \leq y) ...


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For a), \begin{gather*} P(\{\text{at least one ball defective}\}) = 1 - P(\{\text{no balls defective}\}), \end{gather*} while $$P(\text{{no balls defective}}) = (1-0.01)^{100}.$$ For b), \begin{gather*} P(\text{{at most one ball defective}}) = P(\text{{exactly 1 ball defective}} \cup \text{{exactly 0 balls defective}}). \end{gather*} You already have $P( ...


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a. The probability that at least one is defective is equal to: $1 -Pr$(none are defective). Probability that none are defective=$(0.99)^{100}$ So the answer is $1-(0.99)^{100}$ b. This is straight binomial probability, so the answer should be: $\binom{100}{1}(0.99)^{99}(0.01)+(0.99)^{100}$. If you don't understand that let's approach it combinatorially. ...


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Presumably $X$ and $Y$ are independent. Hint: the joint pmf for $(X,Y)$ is $$p_{X,Y}(x,y) = \dfrac{e^{-\lambda}}{1+\lambda} \dfrac{ \lambda^{x+y}}{x!} \ (x \in \mathbb N, y \in \{0,1\})$$ Do you know the factorization criterion (Fisher-Neyman factorization theorem)?


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You can calculate the marginal probability of $ Y=1$ as following: $ P (Y=1) = \int_0^1 P (Y=1|X=x) f_x (x) dx = \int_0^1 x dx = 0.5$. We used the fact that $ x $ comes from uniform distribution. Then it is $ P (Y=0) = 1-P (Y=1) = 0.5$.


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Try twiddling with the sliders on this page.


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If we assume that districts have the same number if inhabitants, the divergence between party control percentages and national popular vote percentages is substantially dependent on the variable proportion of actual voters in each district. In a theorerical situation where this proportion is constant across districts, divergence would be eliminated. So, to ...


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Since you assume $t$ to be a natural number and properties of $\cos$ it can be shown that $Z_t$ and $Z_{1}$ are equivalent. For $Z_2$, you practically have two $Z_1$ and you toss a coin which one of them to select. Anyway, you have $Z_1$ at the end. Similartly for $t>2$.


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Use LIE three times, then simply because of independence of the two coordinates w.r.t. each other. $\begin{align} \mathsf E[Z] & = \mathsf E_{X_1}\left[\mathsf E_{X_2\mid X_1}\left[\mathsf E_{Y_1\mid X_1,X_2}\left[\mathsf E_{Y_2\mid X_1,X_2,Y_1}\left[\frac{Y_1+Y_2}{X_1+X_2}\middle| X_1,X_2,Y_1\right]\middle| X_1,X_2\right]\middle| X_1\right]\right] ...


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Hint: Don't split the numerator. Use both $E[Y_1|X_1] = \theta X_1$ and $E[Y_2|X_2] = \theta X_2$.


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The variance of the average of $n$ i.i.d. random variables is $\sigma^2/n$. Solving the equation $(0.002)^2 \ge 0.0009 / n$, we get $n \ge 225$ which means that you need at least $225$ bricks.


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If we have a sample space of $N$ equally likely outcomes, and the number of outcomes that result in failure in $F$, then the probability $p$ of failure is given by $$p=\frac{F}{N}.$$ This can be rewritten as $F=Np$. Remark: The example you have in mind is that we consider the $n!$ permutations of the set $\{1,2,3,\dots,n\}$. Under the right assumptions, ...


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I would disagree with Michael's statement that "Pointing out trivial consequences like "log turns product into sum..." is very misleading." In fact, those consequences are precisely WHY the logs are used when finding the MLE for a parameter. The MLE has the value that it has in order to minimize the Kullback-Liebler divergence. However, I think this ...


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We have that $X_1$ and $X_2$ are independent $\Gamma(\alpha_i,1)$ random variables. Note that $X_1$ and $X_2$ are non-negative with values in $[0,\infty).$ Given the transformation, the joint density of $Y_1$ and $Y_2$ is: $$ f_{Y_1Y_2}(y_1,y_2)=f_{X_1X_2}[X_1(y_1,y_2),X_2(y_1,y_2)]|J|= f_{X_1}(y_1y_2)f_{X_2}(y_2-y_1y_2)y_2= ...


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Here How exactly are the beta and gamma distributions related? you can find in one of the solutions posted. X1/X1+X2 has betta distribution


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For $i=0,1,2,3,\dots$, you have found that $$\Pr(Y=i+1)=e^{-i}(1-e^{-1}).$$ Let $k=i+1$. You have found that for $k=1,2,3,\dots$, we have $$\Pr(Y=k)=(1-e^{-1})(e^{-1})^{k-1}.\tag{1}$$ That's because $e^{-i}=(e^{-1})^i=(e^{-1})^{k-1}$. Formula (1) is exactly the formula for the probability that $Y=k$, where $Y$ has geometric distribution with parameter ...


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I would say $\Phi(\frac{c-\mu}\sigma)=0.08\Rightarrow \frac{c-\mu}\sigma =\Phi^{-1}(0.08)=-1.406$ $\Phi(u)=\,\,$ distribution function of the standard normal distribution $N(0,1)$ You'll find in the tables N(0,1), in Ecxel (function "=NORMSINV(0,08)) etc.


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The Cartoon Guide to Statistics by Larry Gonick and Woollcott Smith. The Cartoon Guide to Calculus by Larry Gonick.


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Is it correct to write this for the above problem with d different variables. $|\{ \forall (L_{1},L_{2}, \cdots L_{d}) \in \mathbb{Z} : \sum_{i=1}^{d}L_{i} = L , 0 \leq L_{i} < h_{i}\}|$ where $h_i$ is the limit of each $L_i$


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A late answer, just for completeness with a different view on the thing. You might look at your data as measured in a multidimensional space, where each subject is a dimension and each item is a vector in that space from the origin towards the items' measurement over the full subject's space. Additional remark: this view of things has an additional ...


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I don't really know what the etiquette is on answering your own question (whether it's frowned upon to answer your own question), but I'll post my solution/work anyway... So, we have $\displaystyle g(\theta) = \frac{1}{n}\sum\left(\frac{X_i}{X_i+\theta Y_i} - \frac{1}{2} \right) = \frac{1}{n} \sum \left( U_i - \frac{1}{2} \right)$. We verified that ...


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Often it doesn't make a difference. However, in essence you are calculating the largest p-value under the null hypothesis. If the null is a single point, then its just the p-value of the null associated with that value. IF you have a range of values, then you want to find the highest p-value that can be produced by parameters in the set of null hypotheses. ...


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(I will use $L$ instead of lim.) Given integers $L$, $n$, and $m$, you are counting the number of ordered pairs $(i,j)$ such that $i+j=L$, $0\le i < n$, and $0 \le j < m$. $$\left|\{(i,j) \in \mathbb{Z}^2 : i+j=L, 0 \le i < n, 0 \le j < m\}\right|.$$ Note that equivalently you can count the number of $i$ such that $0 \le i < n$, and $0 ...


1

It seems that I need to give a more detailed answer in the post but not only refer to the references for the main technical ideas. Let me start with a general picture. Estimating the entropy, from a statistical perspective, is by no means a unique problem among the problems of estimating functionals of parameters. The reason why it has attracted so much ...


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The total number of combinations is $6\cdot6=36$ i) There are $6$ combinations with a sum of $7$: $1,6$ $2,5$ $3,4$ $4,3$ $5,2$ $6,1$ So the probability is $\dfrac{6}{36}$ ii) There are $2$ combinations with a difference of $5$: $6,1$ $1,6$ So the probability is $\dfrac{2}{36}$ iii) There are $6$ combinations with the sum being a multiple of ...


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@ Superman: for simplicity, I'll write $l=\log p(x)$ and $A=\Sigma^{-1}$. Then $$ \nabla l=\frac{\partial}{\partial x}\left[-\frac{1}{2}(x-\mu)^{T}A(x-\mu)\right]=-\frac{1}{2}(A+A^T)(x-\mu)=-A(x-\mu) $$ where the last equality is due to the symmetry of $A$. It follows that the Hessian of $l$ is $$ H=\frac{\partial}{\partial x^T}\nabla l ...


2

I suppose the sample is iid. You can find the distribution of $F_{\hat\theta}(t)$ by noticing $$F_{\hat\theta}(t)=P(\hat\theta\leq t)=P(Y\leq t)^n$$ Thus you can find $f_{\hat\theta}$, and then its expected value $$E(\hat\theta)=\int_{[0,\Theta]} t f_{\hat\theta}(t)\;\mathrm{d}t$$ Here you have $$P(Y\leq t)=\int_0^t f_Y(u)\;\mathrm{d}u=\int_0^t ...


2

This sounds like a Beta-Binoimial purchase rate model would be useful. The link has a lot of details on this, but I'll give the salient highlights: We are assuming that when a product $i$ is introduced, we are uncertain as to its actual bought rate $BR_i$. We model this uncertainty with a beta distribution on [0,1]. For example, the uniform distribution on ...


1

Calling $p$ the probability of being hit on one crossing and $q=1-p$ the probability to be safe in one crossing you can compute $p_{1000}$ (the probability of being hit at least once in $1000$ crossings) like this: $$p_{1000} = p + qp + qqp +...+q^{999}p$$ This mean: you get hit at the first attempt, or you escape the first and get hit at the second, or ...


2

When $n$ is large $$ W=\frac{\bar{X}-\mu}{\sigma/\sqrt{n}}\sim N(0,1) $$ approximately, or equivalently, $$ \bar{X}= \mu+\frac{\sigma}{\sqrt{n}}W\sim N(\mu,\sigma^2/n) $$ approximately. That is, the sample mean $\bar{X}$ is approximately normal distributed with mean $\mu$ and variance $\sigma^2/n$ when $n$ is large.


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Consider \begin{align} S_{m} = \sum_{k=0}^{m} \binom{2x}{k} \binom{2m-2x}{m-k} \end{align} for which it can be seen as the following. \begin{align} S_{m} &= \sum_{k=0}^{m} \binom{2m-2x}{m} \frac{(-m)_{k}(-2x)_{k}}{k! (m-2x+1)_{k}} \\ &= \binom{2m-2x}{m} {}_{2}F_{1}(-m, -2x; m-2x+1; 1) \\ &= \binom{2m-2x}{m} \frac{(m-2x)! (2m)!}{(2m-2x)! m!} \\ ...


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Edit I made a BIG mistake in my interpretation of the simulation - revised results have been substituted. Answer A team with an 11/5 record has an 99.97% chance of making the playoffs. Simulation In coming up with an answer for this question, I have carried out the simulation for this 10,000 times (i.e. 12 teams x 10,000 seasons = 120,000 playoff ...



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