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Algebraically, the only difference between a density and a likelihood is that a density integrates to $1$ over its support. A density is a likelihood, but a likelihood is not always a density. Semantically, a likelihood is a function of the parameters of the model for some observed data. A density is a function of the observations for some fixed (but not ...

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Sample means and variances from data in frequency-Value format. If there are $K$ different values $v_j,$ each having frequency $f_j,$ then the sample mean can be expressed as $$\bar X = \frac{\sum_{j=1}^K f_jv_j}{\sum_{j=1}^K f_j} = \sum_{j=1}^K r_jv_j ,$$ where $n = \sum_{j=1}^K f_j$ and $r_j = f_j/n,$ for $j = 1,\dots,K.$ In this notation, the sample ...

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Simply put, yes, the cdf (evaluated at $x$) is the integral of the pdf from $-\infty$ to $x$. Another way to put it is that the pdf $f(x)$ is the derivative of the cdf $F(x)$. These definitions assume that the cdf is differentiable everywhere. I mention this not to make the definitions more complicated, but to reduce the factor of surprise later when you ...

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Yes. That's correct. A PDF is a probability density function. It is stating the probability of a particular value coming out. Taking this analogy to a discrete distribution, the PDF of a 6-sided die is: $[x<1:0,x=1:\frac{1}{6},x=2:\frac{1}{6},x=3:\frac{1}{6},x=4:\frac{1}{6},x=5:\frac{1}{6},x=6:\frac{1}{6},x>6:0]$. For a continuous probability ...

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The cumulative distribution function $F_X$ of any random variable $X$ is defined by $$F_X(x)=P(X\le x)$$ for $x\in\mathbb R$. If $X$ is a continuous random variable, then the cumulative distribution function $F_X$ can be expressed as $$F_X(x)=\int_{-\infty}^xf(x)\mathrm dx$$ for $x\in\mathbb R$, where $f$ is the probability density function of a random ...

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$E \overline \Theta = \Theta$ is the definition of an unbiased estimator. If $X_1,\dots,X_n$ are i.i.d. normally distributed variables, then $\bar{X}$ is also normally distributed, as a linear combination of the components of a gaussian vector. Thus you have $$E(\bar{X})=E\left(\frac{1}{n}\sum_{i=1}^nX_i\right)=\frac{1}{n}\sum_{i=1}^nE(X_i)=E(X_1)$$ ...

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The answer is no. Example. Let $\Omega=[0,1]$. $\cal{F}=\{Y : Y \subseteq [0,1] \& Y \mbox{~is~countable ~or}~ [0,1]\setminus Y~\mbox{is~countable}\}$. Let $(\mu_x)_{x \in [0,1]}$ be the family of Dirac probability measures on $(\Omega,\cal{F})$. Then $\cal{N}_M=\{ \emptyset\}$. Hence $\sigma(\cal{F},\cal{N}_M)=\sigma(\cal{F},\{ \emptyset\})=\cal{F}$. ...

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The margin of error of a public opinion poll (95% confidence) is $1.96\sqrt{\hat p(1-\hat p)/n}.$ Where $n = 1111$ and $\hat p = X/n$ is the ratio of the number $X$ of people favoring a particular position to the number $n$ of people interviewed (and the estimate of the population proportion $p$ holding that opinion). Your difficulty is that you do not know ...

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You are on the right track (except for typos in the notation now corrected, see @Henry's Comment and my response). Express binomial coefficients in terms of factorials. Some factorials will cancel exactly. Others will have factors in common: for example, 10!/9! = 10. For a simple start, you might try the case where $p = 1/2.$ Answer: The mode is at the ...

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Look up the continuous mapping theorem.

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Consider an experiment where one outcome ("success") has probability $p$. According to the Strong Law of Large Numbers, if this experiment is repeated indefinitely (and independently), the ratio (number of successes)/(number of trials) almost surely approaches the limit $p$ as the number of trials goes to $\infty$. To a "frequentist", this is the ...

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You could put your paradox in a simpler form: We know that for continuous variables it can happen that the differential entropy is negative (say, $h(X)<0$) but still we have $I(X;Y)\ge 0$ always. Now, the relation $I(X;Y)=h(X)-h(X|Y)$ (differential entropies) holds for any variables $X,Y$. Then, if we take $Y=X$ we get $$0 \le I(X;X)= h(X)-h(X|X)=h(X)$$ ...

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One way to look at this is to assume you have a very, very large population you are sampling from to get height or whatever. Then assuming independent uniform random sampling of individuals from the population, to a very good approximation, each sample $j$ can be treated as an independent identically distributed (i.i.d.) random variable $X_j$ sample value ...

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The question you refer to in the link seems to have been the subject of some bickering. Anyway, I gather that you're asking how to find the mode of a distribution in general, not just for the binomial distribution. The mode is the outcome(s) which arises most frequently. This is easy to understand in a discrete random variable. If $X$ is a random variable ...

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As you are aware the mode is the place of maximum likelihood within a density or mass function. For continuous distributions, this equates to taking the derivative of the pmf/pdf function and finding the zero of it's derivative. In the case of the binomial distribution, this is a discrete distribution and a derivative is not available. Because of this you ...

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The best way is to experiment, as it depends so much on the actual shape of the data. You can get a sinusoidal type graph with a lop-sided cycle by considering a function in the form of a fraction consisting of $a\sin(nt+\epsilon)$ in the numerator and and $b\sin(mt+\delta)+k$ where $k$ is only just large enough to ensure the denominator remains strictly ...

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Hint: the expected mean of the sample is the same as the mean of the parent distribution. The variance of the mean goes down as 1/n. So how many standard deviations high would the sample have to be?

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If $X$ is a random variable, we say that a realized value $x$ is 'in the $k$th percentile' if the following is true: $$P(X < x) = k$$ Treating your test score as a realization of a random variable, we see that being in the $74$th percentile implies that you outperformed 74% of the students.

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If your grade falls in the 74th percentile, it simply means you outperformed 74% of the students.

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You have $H_0: p=0.5$ $H_A: p > 0.5$ $T=\frac{\frac{57}{99}-0.5}{\sqrt{0.5\cdot 0.5}}\cdot \sqrt{99}$ If $|T|>z_{1-\alpha }$, then you reject $H_0$. $\alpha$ is the significance level. The value of $z_{1-\alpha }=z_p$ can be found at a table of the standard normal distribution, e.g here

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Regardless of any independence assumptions, $f_{(X,Y)}(x,y)=f_{X|Y}(x|y) f_Y(y)$. This is just the definition of the conditional density. It looks just like the definition of conditional probability. Now the marginal density of $X$ is $f_X(x)=\int_{-\infty}^\infty f_{(X,Y)}(x,y) dy$.

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I asked a very similar question only yesterday. If your samples are homoscedastic (can be assumed if subjects are randomly chosen) you can estimate the variance to behave reciprocally to sample size. $\sigma^2 \propto \frac{1}{N} \Rightarrow N \propto \frac{1}{\sigma^2} \Rightarrow N_1=\frac{\frac{1}{1.07^2}}{\frac{1}{1.07^2} + \frac{1}{1.32^2}} * 583 ... 0 Take x as your free throw probability. To find the "x" or free throw probability, you need to find out what how many times x goes into 1.72. To do that, first find what 1.72 is equal to. The percentage times he would make 2 points would be: 2x^2 Two points would be his free throw percentage (in decimal form) squared because he has to make his free throw ... 1 Let$f : \mathcal{M} \to \{0,1\}$be defined as $$f(m) = \begin{cases}1 &\text{if } \mathrm{value}(m) \neq 0 \\ 0 &\text{otherwise}\end{cases},$$ then your average is $$\frac{\sum_{m \in \mathcal{M}}\mathrm{value}(m)}{\sum_{m \in \mathcal{M}}f(m)}.$$ When you write it like below (which gives the same result because of how$f$is defined): ... 1 Let$m_i$be non-negative integers such that$1 \leq i leq 5$. Let$P$be the set of$m_i$such that$m_i > 0$i.e.$P = \{ m_i | m_i > 0 \}$. The average can be written as $$A(m) = \frac{1}{|P|} \sum_{m_i \in P} m_i$$ where$A(m)$is defined to be something for the empty set. 1 Since an average is a quotient of a sum and the number of samples (which we can itself write as sum), one way to encode this to incorporate the condition into the sum. For example, if we denote the month by$a$and the value for the month$a$by$v_a$, we can write $$\text{(average)} = \frac{\sum_{v_a \neq 0} v_a}{\sum_{v_a \neq 0} 1}.$$ The denominator here ... 0 This is not really an answer, but please bear with me. Baye's Rule is about conditional probability, and probability trees. There are a lot of factors which affect the price of oil. Therefore, it seems likely that the Baye's rule won't be practical. If you assume that there are really only a handful of factors: production, consumption, wars in the Middle ... 1$\newcommand{\E}{\operatorname{E}}$I'm going to GUESS that what's intended is that this is a discrete uniform distribution on the set$\{0,1,2,\ldots,N\}. Then \begin{align} \Pr(X=x) & = \E(\Pr(X=x\mid N)) = \sum_{n=0}^\infty \Pr(N=n)\Pr(X=x\mid N=n) \\[10pt] & = \sum_{n=x}^\infty \Pr(N=n)\Pr(X=x\mid N=n) \quad (n=x\text{ replaced }n=0.) \\[10pt] ... 0 Suppose you have say5$red balls and$8$green balls, and some one draws them all at random, without replacement, giving some ordering of all the reds and greens. If the person tells you that the first ball was red, then from your point of view, the probability that the second ball is red is$\frac{4}{12}$. On the other hand, if the person tells you ... 0 Since we need more 1-dollar coins than 2-dollar coins in the payment, we can investigate solutions using only 1s and 3s, then split the 3s into 2+1 for more solutions. Counts will be given as$(x,y,z)$where$x+2y+3z=20$. So$(11,0,3)$splits into 3 additional solutions that use \$2 coins: $(12,1,2), (13,2,1), (14,3,0)$ - and in general if we start with ...

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Discrete positive solution for the plane x+2y+3z=20 where x>y always. The count comes to be 28.

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1 way with 6x \$3 coins (\$2 with \$1 and \$2, so must be 0 \$2 coins) 2 ways with 5x \$3 coins (\$5 with \$1 and \$2, so between 0 and 1 \$2 coins) 3 ways with 4x \$3 coins (\$8 with \$1 and \$2, so between 0 and 2 \$2 coins) 4 ways with 3x \$3 coins (\$11 with \$1 and \$2, so between 0 and 3 \$2 coins) 5 ways with 2x \$3 coins (\$14 with \$1 and \$2, ...

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There are two different but commonly used parametrizations of a geometric random variable. One version states that $$\Pr[X = x] = (1-p)^{x-1} p, \quad x = 1, 2, 3, \ldots.$$ The other has $$\Pr[X = x] = (1-p)^x p, \quad x = 0, 1, 2, 3, \ldots.$$ Note that they are computationally the same, but the second one allows $X = 0$ whereas the first does not. You ...

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$\mathbb E[X] = \frac{1-p}p$ as you computed above. Here is a trick to make the computation of $\mathrm{Var}(X)$ easier: $$\mathrm{Var}(X) = \mathbb E[X^2] - \mathbb E[X]^2 = \mathbb E[X(X-1)] + \mathbb E[X] - \mathbb E[X]^2.$$ Since the generating function of $X$ is \begin{align*} P(s) &:= \mathbb E\left[s^X\right]\\ &= \sum_{n=0}^\infty (1-p)^n ... 0 No answer to your question but a suggestion to follow an alternative route (too much for a comment). Let S denote the event that the first experiment is a succes and let F denote the event that the first experiment is a failure. Then make use of:\mathbb EX^n=\mathbb E(X^n|S)P(S)+\mathbb E(X^n|F)P(F)=\mathbb E(1+X)^nq$$This for n=1 and n=2 ... 0 You can first note that \sum_{i=0}^\infty iq^i = \sum_{i=1}^\infty iq^i because the first term is zero. Then this equals q\sum_{i=1}^\infty iq^{i-1} = q\frac{d}{dq}\sum_{i=1}^\infty q^i. Now you have a term that looks like:$$pq\frac{d}{dq}\left(q\frac{d}{dq}\sum_{i=1}^\infty q^i\right)$$You can complete the derivation from there. 4 It's one possibility out of 2^6 equally likely possibilities. So \frac 1 {2^6} = \frac 1 {64}. The first calculation you did is correct because the probabilities are independent.  In more detail:$$\Pr[\text{first four children are boys and rest are girls}] = \\\Pr[\text{second to fourth children are boys and rest are girls} \mid \text{first ...

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I suggest you calculate the cumulative distribution function of $\frac{k}{\hat{k}}$.

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These questions originated from an assignment (due tomorrow) of UNSW - course code, MATH2901. Please put this on hold. Thanks.

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My interpretation on my frequencist basis: Past and future lose its causal relationship in probability theory. Consider the following example. We have a friend who goes to the pub some evenings. Also, he is in bad mood or in good in some mornings. We observe our friend at $N$ evenings and at the following mornings. Our data is collected in a table: ...

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In the context of the question (drawing coloured balls from an urn without replacement, so with the hypergeometric distribution), the argument is that knowledge of the colour of the $k$th ball drawn affects the conditional probability of the colour of the $j$th ball drawn, whether $j \gt k$ or $j \lt k$. In a sense this is obvious, in that knowing a ball ...

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We want to obtain the distribution of the estimator $\hat{\theta} = max_{i = 1}^n{X_i}$: $$P(max_{i=1}^n{X_i} \leq x) = \prod_{i=1}^nP(X_i \leq x) = P(X_1 \leq x)^n = (\frac{x - 0}{\theta - 0})^n = \frac{x^n}{\theta^n} I_{[0, \theta]}(x)$$ Which means that, differentiating the function with respect to $x$, we obtain: $$P(max_{i_1}^n{X_i} = x) = ... 2 This is an assignment question for MATH2901, UNSW. Please do not post up solutions. Hint for 1 and 2: Work backwards as you have learnt in Disrete Mathematics. 0 We can use the given data to estimate the probability that a student passes in each of the three categories. For instance, p_1:=\frac{9500-3325}{9500} is the proportion of students who passed the mathematics exam. Similarly you can compute p_2,p_3 for the language and general exams. Since the events of passing in each exam are assumed to be ... 10 We want to estimate the probability of success when we make a prediction. We are using the estimator S/N, where N is the number of trials and S is the number of successes. In the case you describe, both estimates are \frac{1}{2}. However, the variance of the estimator when N=40 is \frac{1}{10} times the variance when the estimator is based on ... 0 By definition it is:$$\begin{align} \Bbb P \quad & = \quad \sum_{k=\lceil 3n/4\rceil}^n \binom{n}{k}p^k(1-p)^{n-k} \\[1ex] & = \quad \tfrac 1{2^n}\sum_{k=\lceil 3n/4\rceil}^n \binom{n}{k} & : p=1/2 \end{align}$$And that's why Normal approximations get used. 0 You can obtain the square root of a matrix M using the Cholesky Decomposition, M = LL'. Then compute the inverse of L 0 Not sure if this was ever totally resolved. It is a question that raises a number of basic issues. Following the terminology of @mattbiesecker, Let Y = X_1 + X_2 + \cdots X_{16}, where X_i are a random sample from Norm(0,1). Then$$E(Y) = E(X_1) + E(X_2) + \cdots + E(X_{16}) = 16(1/2) = 8$$and, by independence,$$V(Y) = V(X_1) + V(X_2) + \cdots + ...

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Well, you have largely solved it. The length of an interval $[a, b]$ is just $b-a$. So if the 95% CI for $\mu$ is $$\left[ \bar x - 1.96 \frac{\sigma}{\sqrt{100}}, \bar x + 1.96 \frac{\sigma}{\sqrt{100}} \right]$$ the length is just $$2 \cdot 1.96 \frac{\sigma}{\sqrt{100}}$$ Now for b the length of the 99% CI is going to be  2 \cdot 2.58 ...