# Tag Info

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You could use induction but maybe we can appeal to a counting argument: Write $P= (x+1)(x+1)(x+1)\cdots (x+1)$ where there are $m$ instances of $x+1$. Now, when we expand this, if we respect the order of the multiplication, then an arbitrary term will be a product of $1's$ and $x's$ that satisfies: the number of instances of $1$+ the number of ...

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Hint: The sum of normally distributed random variables is a normally distributed random variable with (1) its mean being the sum of the means, (2) its variance being the sum of the variances. You want to find $n$ such that $\mathsf P((\sum_{k=1}^n B_k) - C \geq 0)\gt 0.2$, where $\{B_k\}$ are the weights of boxes and C is the capacity of the elevator; ...

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We will adapt a non-inductive, combinatorial proof of the binomial theorem. We seek the expansion of \begin{align*} (x+1)^m &= (x+1)\cdots(x+1) \\ &= a_m x^m + a_{m-1} x^{m-1} + \cdots + a_1 x + a_0 \end{align*} Each integer coefficient can be treated as the number of times the variable of the term appears in a full expansion. Note that each term in ...

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Proceed by induction on $m$. For $m = 0$, the statement holds; now assume true for all values $\leq m$. Then for $m + 1$, we have \begin{align} (x+1)^{m+1} &= (x + 1) (x + 1)^m \\ &= (x + 1)\sum\limits_{k = 0}^m \binom{m}{k}x^k &\text{ by the inductive hypothesis} \\ &=\sum\limits_{k = 0}^m \binom{m}{k}x^{k + 1} + \sum\limits_{k = 0}^m ...

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The probability that any question is answered correctly is $p=0.2$. If the number of questions is $N=180$, the number you get right is $n$ and the maximum score $S_{max}=2$ your score is $s=\frac{n}{N}\times S_{max}$. Now $n\sim B(N,p)$, that is the number of correct answers has a binomial distribution with the given parameters. Then using the Normal ...

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Without seeing your data, it seems to me you might think they are lognormal (that means they would be normally distributed if you took their logs). You might try taking taking logs of a few of your samples and making histograms to see if you get something like a normal shape. A sample of several hundred would be better than several dozen because it takes a ...

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There is excellent open source software to which you could link or which you might use as a guide. For large df there are good approximations. Much of your approach will be governed by whether you need only tail probabilities and how many places of accuracy you need. Google 'approximate CDF of chi-squared' and similar phrases to tap into rich literature on ...

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One idea is you can make a histogram of the values and report the median.

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It depends on the problem and what you are trying to achieve. I could see this going either way, without further information on the problem. Just guessing, I would say hunger is the dependent variable (the more you dance, the more calories you burn, and the hungrier you get). Hope this helps!

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So two things: One would be that, given $n$, you would expect $np$ successes, so if you were to "invert" this process, I suppose you might say you expect you had $\frac{k}{p}$ trials. However, I have my doubts as to this reasoning being legitimate. The other is that you could analyze this numerically - i'm quite certain there wouldn't be a closed-form ...

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Just do some in-your-head approximations, no table-lookups: $95\,\%$ is about $2\sigma$, here $\sigma=\sqrt{npq}=5$, so anything between $40$ and $60$ heads will not raise your suspicion at this confidence level

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Your question is not well posed. When you specify $X \sim \text{Binomial}(n,p)$ you are telling us what $n$ is. It is a fixed constant, not a random variable. In that sense, $E[n] = E[n|\text{anything}] = n$.

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No $m$ need not be equal to $g(n)$. In fact $g(n) = m + f(n)$. The construction is concerned with the sequences of lenght $n$ that follow $x_1 \ldots x_m 1\ldots 1$ we are not appending $n - g(n)$ digits (as far as I understand). For example $m = 2$ $n = 4$ $g(n)=2$ $2^{n-g(n)}-1=2$. The sequences we have are $$x_1 x_2 1 1\\ x_1x_2 10\\ x_1x_2 01\\$$ note ...

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$$\text{Boy, High income} =4\\ \text{girl, High income} = 6\\ \text{Boy, low income} = 6\\ \text{Girl, low income} = x$$ $$P(\text{Male - M}) = \frac{10}{16+ x}\\ P(\text{High income - H}) = \frac{10}{16+ x}\\ P(\text{Male High income}- MH) =\frac{4}{16+x}$$ Independence means that $P(M \text{and} H) = P(M) P(H)$ $$\frac{10}{16+x}\frac{10}{16+x} = ... 0 Suppose our model is a distribution based on categories A_1, A_2, and A_3, with probabilities P(A_1) = \theta_1 = 1/4, P(A_2) = \theta_2 = 1/4, and P(A_3) = \theta_3 = 1/2. This multinomial model is our null hypothesis. Also suppose we have n observations with observed counts X_1, X_2, X_3 in the respective categories. As an approximation, we ... 0 Would have rather posted this as a comment but I don't have any reputation. Anyway I run http://kappa.ws The original idea behind it was some kind of interactive trading platform. I came up with the idea right around the time that twitchplayspokemon started. The twitch name was kappamarket where it would stream current rates and show balances/trades. The ... 0 Cribbing from the answer by Brian B., assume your covariance matrix is Σ and let D = sqrt(diag(Σ)), a vector of square roots of the diagonal of Σ. then the correlation matrix is given by ϱ = D-inverse Σ D-inverse-prime D, here, is a p x 1 vector (from the diagonal of Σ) and its inverse is the item by item inverse of D -- i.e., vector of {one over ... 1 I don't think there's an elegant solution but there is a correct solution. Note that (N_i,N_j,N-N_i-N_j) follows a multinomial distribution with parameters (N,p_i,p_j,1-p_i-p_j). Therefore the probability that N_i > N_j is$$P(N_i > N_j) = {\sum_{k=1}^N {\sum_{h=0}^{k-1}{ N! \over {k! h! (N-k-h)!}} p_i^{k} p_j^{h} (1-p_i-p_j)^{N-k-h}}}$$1 Pst: The density function for the M largest ordered statistics is:$$f_{{X}_{(L)}, {X}_{(L-1)},\ldots,{X}_{(L-M+1)}} ({x}_1,{x}_2,\ldots,{x}_M)$$This is the probability that: M of the results have the given values \langle x_1,x_2,\ldots,x_M\rangle$${f}_X({x}_1)\cdot {f}_X({x}_2)\cdots {f}_X({x}_M)$$There's no particular placement of these ... 0 To get you started: You have been given the events, which are presumably pairwise independent: A :- attendance is marked. \mathsf P(A) = 0.40 B :- Bob is punctual. \mathsf P(B) = 0.90 C :- Marty is not punctual. \mathsf P(\neg C) = 0.20 (NB: Assuming that to be marked anything, attendance has to be taken. Obv.) You wish to find: (a) ... 0 Let f\geq 0 be a measurable function on X with \int_X f = 0. We claim that f is zero a.e. Suppose that this is not the case. For each \epsilon > 0, the set U_\epsilon = f^{-1}(\epsilon, \infty) is measurable. Since f^{-1}(0, \infty) = \bigcup_{n=1}^\infty U_{1/n}, some U_\epsilon must have measure \mu > 0. Thus \int_X f \geq ... 1 The density of \pi = e^{-\epsilon} where \epsilon \sim \operatorname{Gamma}(a,b) is parametrized by rate, is given by$$f_\pi(p) = \frac{b^a (-\log p)^{a-1} p^{b-1}}{\Gamma(a)}, \quad 0 < x < 1.$$Therefore$$\begin{align*} \Pr[X = x] &= \int_{p=0}^1 \Pr[X = x \mid \pi = p] f_\pi(p) \, dp \\ &= \binom{n}{x} \frac{b^a}{\Gamma(a)} ...

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Your questions are quite simple to answer. In case of Chi-Square test, the p-value is computed as $1 - CDF_{\chi^{2}_{df}}(ts)$, where $CDF_{\chi^{2}_{df}}$ is a cumulative distributive function of Chi-Square distribution with $df$ degrees of freedom and $ts$ is the value of your test statistic. (Try to figure out why is p-value calculated this way - it is ...

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I would suggest looking at "On the Identification of Variances and Adaptive Kalman Filtering" by R. Mehra as a starting point. This paper is highly cited and the methods are not difficult to implement.

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In the case of $n$ persons the graph of possible friendships is made of ${n \choose 2}$ pairs. For each edge $e =\{a,b\}$ of this graph we consider a random variable $X_e$ if $X_e = 1$ this means that they are friends. If $X_e=0$ they are not friends. ($a,b$ are two individuals of our population). We assume that $X_e$ are independent random variables and ...

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$P(A_1\cup A_2 \cup A_3)= P(A_1\cup A_2) + P(A_3) - P((A_1\cup A_2)\cap A_3 )$ $=P(A_1\cup A_2) + P(A_3) - P(A_1\cup A_2).P(A_3 )$ ..since $(A_1\cup A_2)$ and $(A_3)$ are independent Similarly, $P(A_1\cup A_2)=P(A_1)+P(A_2)-P(A_1\cap A_2)=P(A_1)+P(A_2) - P(A_1).P(A_2)$ Now you can just do the calculation. I guess the final answer is 3/4

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If the $\chi^2$ test David suggested fails $H_0$ , i.e., that all means are equal, you can then use, either Fisher's LSD test or, for non-parametric, you can use Kruskal-Wallis , to decide which two ( or more ) means are not equal ( at the given choice of significance).

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A $\chi^2$ test would be appropriate since we are dealing with observed frequencies from mutually exclusive groups

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Let $X$ be a random variable with density function $f_X(x)$, $g(x)$ be a monotone increasing function of $x$, and $Y = g(X)$. We seek the density function $f_Y(y)$ of $Y.$ In this instance it can be shown that $$f_Y(y) = f_X(g^{-1}(y)) \frac{dg^{-1}(y)}{dy},$$ where $g^{-1}$ is the inverse function of $g$. In your Question, $X \sim Exp(1),\;$$g(x) = ... 0 You can use the approximate confidence interval for a proportion. In general it is used for normally distributed variables. But you have 3333 identical and independent distributed (iid) random variables. So you can apply the central limit theorem. The two sided limits are \Large{\left[\hat p-z_{1-\frac{\alpha}{2}}\cdot \sqrt{\frac{\hat p\cdot (1-\hat ... 0 I think that they are almost ok but it depends on what do you exactly need. I see two potential problems among the kernel functions you mentioned (they arise because kernel functions are made for density estimation but you want something little bit different). 1) All of these functions are symetrical around 0 but you have distances that are always ... 0 First: Yes, the square root of the determinant of the i-th covariance matrix. Second: No, the inverse of the i-th covariance matrix. Yes, raising 2\pi to the D/2-th power. If D is odd, this is a half integer; you still need to use D/2, without the rounding suggested in your code. x^{D/2}=\sqrt{x^D}. 0 For part b), you are asked what the 95 percentile is. You know the mean in a normal ( or symmetric) distribution is the 50 percentile. Now you need to see what z-value will give you the additional 45% of the data between the mean and the 95 percentile. By symmetry, this should be the z-value associated with the 90 percentile. By the 1-2-3 or 68-95-99 rule ... 4 The simplest way is probably using the CDF, which is known in this case. Let X \sim Exp(1) and Y = \sqrt{X}. The CDF of X is F_X(x) = 1 - e^{-x}, for x > 0. Then$$F_Y(y) = P(Y \le y) = P(\sqrt{X} \le y) = P(X \le y^2) = 1 - e^{-y^2},$$for y > 0. Then take the derivative of the CDF of Y to get F_Y^\prime(y) = f_Y(y) = 2ye^{-y^2}, ... 1 To find (Z < 1.246), possible entry points into the table are 1.24 and 1.25. The distance between them is .01. The distance from 1.24 to 1.246 is .006. Then .006/.01 = 6/10. Here are exact values from R statistical software: pnorm(.024); pnorm(.0246); pnorm(.025) ## 0.5095737 # P(Z < .024) ## 0.509813 # P(Z < .0246) ## 0.5099725 # ... 1 The crucial point is the level of significance of the test, if you consider \epsilon = 2^{-1} then you are rejecting the null conjecture (It is a random array) in many cases. For some reason you are bound to think that many 0 may occur. So you will reject randomness at the event the first m digits are zero. You could also think that it is not very ... 1 I wrote what appears below under the impression that what you had was not \vphantom{\frac \int\int}P(X_i > e^{-x/2}) = P(X_i < 1 - e^{-x/2}) but P(X_i > e^{-x/2}) = 1 - P(X_i < e^{-x/2}). But maybe that last is what you actually need.$$ \Pr(X>x) = 1-\Pr(X\not>x)=1-\Pr(X\le x) \overset{\huge\text{?}} = 1 - \Pr(X<x). $$So is it true ... 3 The median for a random variable X is m such that P(X \le m) \ge 1/2 and P(X \ge m) \ge 1/2. In the first example the correct answer is 0: P(X \le 0) = P(X = 0) = 0.728303 and P(X \ge 0) = 1. In the second example it is 2: P(X \le 2) = 0.10 + 0.20 + 0.30 = 0.6, P(X \ge 2) = 0.30 + 0.25 + 0.15 = 0.7. Your method is completely wrong. 1 Hint: T=\sum_{k=1}^nX_k \sim Poisson(\lambda) T_1=\sum_{k=2}^nX_k\sim Poisson(\lambda-\lambda_1) and P(X_1=x_1,T=t)=P(X_1=x_1, \sum_{k=2}^nX_k=t-k)=P(X_1=x_1)P( \sum_{k=2}^nX_k=t-k) now just use Poisson formula and you will get the result. 1 You might be interested in reading about Simpson's paradox (easily found online). It can happen, for example, that A has a better average every single year than player B does, yet player B has the better combined average. For example:$$A:\;\;\;\left[\frac23,\frac{5}{10},\frac{5}{10}\right]B:\;\;\;\left[\frac{12}{20},\frac13,\frac25\right]$$Then, ... 1 First we can assume without loss of generality that$n = 2$since a sum independent Poisson r.v.'s is itself Poisson. For$0 \leq k \leq t, k \in \mathbb{N}\$ we have \begin{align} P(X_1 = k \mid X_1 + X_2 = t) &= \frac{P(X_1 = k \cap X_1 + X_2 = t)}{P(X_1 + X_2 = t)} \\ &= \frac{P(X_1 = k) P(X_2 = t - k)}{P(X_1 + X_2 = t)} \\ &= ...

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