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3

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 & \\ & ...


1

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.


1

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 $j$th throw in the $i$th 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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