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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) = ...

3

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 ...

3

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!

3

Not knowing where exactly you are messing up I've posted the whole solution, $$S_{xx}=\sum_{i=1}^{n} (x_i-\bar{x})^2 = \sum_{i=1}^{n} (x_i^2-2x\bar{x}+\bar{x}^2)=\sum_{i=1}^{n} x_i^2 -2 \bar{x}\sum_{i=1}^{n} x_i +\bar{x}^2 \sum_{i=1}^{n} 1\\= \sum_{i=1}^{n} x_i^2 -2\frac{\sum_{i=1}^{n} x_i}{n}\sum_{i=1}^{n} x_i +\left(\frac{\sum_{i=1}^{n} x_i}{n}\right)^2 n ... 3 If M = \min(X_{1}, . . . , X_{n}), it can be shown that M has an exponential distribution with parameter n \lambda_0. So E(M)=\frac 1 {n \lambda_0} Since T=nM, it follows that E(T)=E(nM)=nE(M)=\frac 1 {\lambda_0} Details can be found here: Distribution of the minimum of exponential random variables 2 In part (1) there is no theorem that states \mathbb E(\exp Y) = \exp \mathbb E(Y). You can't move the expectation past the exponential. Instead, use the general formula:$$ \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}) = ...

2

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 $$M_X(t) = \sum_{x=0}^\infty e^{tx} e^{-u} \frac{u^x}{x!} = \sum_{x=0}^\infty e^{-u} \frac{(ue^t)^x}{x!} = e^{u(e^t-1)} ... 2 What is the probability that, if we pick a book from the first shelf, it is a hardcover? Clearly, that is \frac{5}{11}. What is the probability that, if we pick a book from the second shelf, it is a hardcover? Clearly, that is \frac{7}{11}. Now, let's say that we pick a book from the table. It is either from the first shelf or the second shelf: It's ... 2 First moment of the distribution is μ_1=E[X]=α and the first moment of the sample is m_1=\bar{X}. So, set$$μ_1=m_1 \implies α=m_1$$This is the first moment estimator for α. (Note: This method is confusing at the beginnning, because you think, ok so what? But think that m_1 is your sample mean and so it is known, it will be realized when you collect ... 2 Start with$$ E|X_n-c|\le M\cdot P[|X_n-c|>\epsilon]+\epsilon\cdot P[|X_n-c|\le\epsilon]\le M\cdot P[|X_n-c|>\epsilon]+\epsilon. $$Because X_n converges in probability to c, \lim_n P[|X_n-c|>\epsilon]=0 for each \epsilon>0. It follows from the inequality displayed above that$$ \limsup_nE|X_n-c|\le\epsilon, $$for each \epsilon>0. ... 2 The table you've linked is a pretty nonstandard format for a z-score table, but it seems to be referring to a situation like this: The area under the curve between 0 and t is the probability of a normally distributed variable falling between 0 and t. Using the fact that the curve is symmetric about 0, you can deduce the probability of a normally ... 2 The hypothesis test is only on women. So, you can go with the number of women and their data alone. 2 Edit: The original question asked about independence, and was answered by the example below. This example also settles the modified question about covariance, since \text{Cov}(XY,Z)\ne 0. Toss a fair coin twice. Let X=1 if we have head on the first toss, and 0 otherwise. Let Y=1 if we have head on the second, and 0 otherwise. Let Z=1 if the ... 2 If you have Y_1,Y_2, \dotsc, Y_n and each are independent and follow some distribution G, then you could consider each Y_i as a realization, or sample, taken from G. If you then order, then each Y_{(i)} follows a new distribution. For example, say we have Y_1, \dotsc, Y_n, where each one is independent and follows a \text{unif}(0,1). Then, ... 2 We are told the painter selects two cans at random, so it can't be that she, for example, selects can 3 twice. Hence the tuples (3, 3), (4, 4), and (5, 5) are not possibilities, leaving only six choices. Thus the probability of glossy and glossy is 6/20, and these probabilities do add to 1. 2 We know that if X_1 and X_2 are independent normal random variables, then their sum$$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.$$... 2 I will use t as the variable instead of \lambda. The required expectation is$$\left(E(\exp(\frac{tX_1}{n})\right)^n,$$and$$E(\exp(\frac{tX_1}{n}))=(1-p)+pe^{\frac{t}{n}}.$$2 For (1): show you can have, for instance, \operatorname{var} Y = 0 and \operatorname{var} X >0, or \operatorname{var} Y >0 and \operatorname{var} X = 0. For (2): Show you can have \mathbb{E} X = 0 and \mathbb{E} Y > 0, or \mathbb{E} X = 0 and \mathbb{E} Y < 0. For (3): You should see the relation with (1). For (4): Show you ... 2 The answer you got - 0.246 is the probability of getting 'exactly' 5 heads. Your intuition gives the 'Expectation' E(x). When 10 coins are tossed, the Expectation is that you get 5 heads. Your intuition makes an average of all the cases while the solution takes only the cases where number of heads is 'exactly' 5. Let me give you another problem- What is the ... 1 Let's imagine a case with 4 rolls where we desire 2 heads (MathJax diagrams only go so far...)$$\newcommand{\mychoose}[2]{\bigl({{#1}\atop#2}\bigr)}\begin{array}{ccccccccccc} & & & & & & & H & & & & & \\ & & & & & & \swarrow & & \searrow & \\ & ...

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Hint: Think about different variations of what X and Y are allowed to be given our assumptions. In particular, compare cases when $X$ is either identically zero or uniformly distributed, and $Y$ is identically $-2$ or $+2$. See if you can generate counter examples by using these.

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To construct counterexamples: Let $Y\equiv 2$ and $X \sim U(-1,1)$. Then $Var(X)>0=Var(Y)$. Let $Y \equiv -2$ and $X\sim U(-1,1)$ or simply $X\equiv 0$. Then $E[X]=0>-2=E[Y]$. Let $Y\in\{-2,2\}$ with $P(Y=-2)=p=1-P(Y=2)$ and $p>0$ and $X\equiv 0$. Then $Var(Y)>0=Var(X)$. Let $Y\equiv2$ and $X\sim U(-1,1)$. Then median $Y=2$ and median $X=0$. ...

1

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 $$P(\mbox{child is ... 1 I formulated it a bit different, but you should be able to follow along as all of these calculations will be identical to what you should be getting. I used page 30 as a guide. We have m = 2, n = 5 and the data (x_i, y_i) pairs are:$$\text{data}=\left( \begin{array}{cc} 1.08 & 0 \\ 1.07 & 0.0659232 \\ 0.97 & 0.169517 \\ 0.77 & ...

1

With the problem as described, you are correct that the sample space consists of the outcomes where you observe any number of failures followed by a single success. Given that, all of your answers are correct: you're looking for the set of "Some number of Fs, followed by a single S" that matches the given requirements: (i) any 5 such strings, (ii) strings ...

1

Toss two coins. Let $Y_i=1$ if the i-th coin is a head and $Y_i=0$ if it is a tail. Then we have $\{Y_1, Y_2\}$, which are two independent and identically distributed Bernoulli random variables. So what does the distribution of $\{Y_{(1)}, Y_{(2)}\}$ look like? $$\begin{array}{l|llll:l} ~ & TT & HT & TH & HH \\ \hline Y_1 & 0 & 1 ... 1 It is correct, but can be further simplified by considering the order statistics of the sample. That is to say, if$$\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,$$the first and last order statistics which are equal to the smallest and largest observations in the sample, respectively. ... 1 Here are results from a simulation in R of 100,000 ten-throw sessions, baed on @lulu's analysis. The simulation is based on a matrix MAT with m = 100,000 rows and n = 10 columns. At the end of the simulation, the matrix has a 1 in cell (i,j) if the jth throw in the ith session was a success. The m-vector x contains the number of successes in ... 1 If P(k) = \frac{a}{2^k} for 0\leq k \leq 20, then the fact that \sum_{k=0}^{20} P(k)=1 implies that$$ 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)}. $$You can check that this necessary condition is sufficient for P to be ... 1 For finding the distribution of the first one$$(m+n-2)\frac{S^2}{\sigma^2}$$let S^2_1=\frac{1}{m-1}\sum_{i=1}^m(X_i-\overline X)^2 and S^2_2=\frac{1}{n-1}\sum_{j=1}^n(X_i-\overline X)^2. Then$$(m+n-2)\frac{S^2}{\sigma^2}=(m-1)\frac{S_1^2}{\sigma^2}\ +\ (n-1)\frac{S_2^2}{\sigma^2} But as you correctly guessed, each summand of the last equation ...

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