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6

This problem seems simple...but its not. For example, see here for a rather complex analysis for the prima facie simple case of ratios of normal rv and ratios of sums of uniforms. In general, if your pairs are not from a bivariate gaussian, there is no nice formula for $E[R^2]$. Note: $$R_n=\frac{n\sum x_iy_i-\sum x_i\sum y_i}{n^2s_Xs_Y}$$ This mess ...


4

According to Kendall's Advanced Theory of Statistics (Exercise 16.17 in the 5th edition of Volume 1), Pitman (1937) showed the sample correlation coefficient $r$ has zero mean and variance or second moment of $$\sigma^2_{r}=E[r^2] = {1 \over {n-1}}$$ for any sample of size $n$ where $x$ and $y$ are independent continuous variates. Checking the reference, ...


4

Your answer is correct as explained and written. Completing the square gives $$0 = x^2 - 2ax + b = x^2 - 2ax + a^2 + (b-a^2) = (x-a)^2 + (b-a^2),$$ and since no real square is negative, then the equation has no solution if $b-a^2 > 0$; i.e., $b > a^2$. The desired probability is therefore $$\Pr[b > a^2] = 1 - \Pr[b < a^2] = 1 - \int_{u=0}^h ...


4

First, you probably need to switch around the probabilities on the right. If $a<b$, then $P(x<a) \leq P(x<b)$, so the right-hand side of your expression could be negative. It's probably more intuitive if you rewrite it as $$ P(x < b) = P(x < a) + P(a<x<b)$$ Since $x<a$ is disjoint from $a<x<b$. For this to work you need to ...


3

Correlation may have many meanings, but from the question, you are using the specific definition of the Pearson product-moment correlation coefficient. You are calling variables "correlated" when $\rho \neq 0$. That is solely when $\textrm{Cov}(X, Y) \neq 0$. In the case of $Y = aX$, regardless of how $X$ is distributed, we can state the following: $$ ...


3

1) Draw the square $[-1,1]\times [-1,1]$. in the cartesian plane. The value of $X$ is on the $x$ axis, $Y$ on the $Y$ axis. 2) Asking the probability of $Z\leq s$ is equivalent to finding the area under the the line $x-y=s$, bounded by the square. 3) Find this area using simple geometry.


3

Let $Z=E[X\ | \ Y]$. Your equation states: $E[X \ | \ Z]=Z$. This follows from the following fact. Tower Property of Conditional Expectation: $$E[E[X\ | \ \mathcal{F}]\ | \ \mathcal{G}]=E[X\ | \ \mathcal{G}],\text{ whenever }\mathcal{G}\subset \mathcal{F}.$$ Proof of your equation: We apply the tower property with $\mathcal{G}=\sigma(Z)$ and ...


3

We want to prove $EX^2 P(X > \lambda) \geq (EX - \lambda)^2$, i.e. $EX^2 E1_{X > \lambda} \geq (EX - \lambda)^2$. Using Cauchy-Schwartz inequality, we get $EX^2 E1_{X > \lambda} \geq (EX1_{X > \lambda})^2$ Since both $EX1_{X > \lambda}$ and $EX - \lambda$ are positive, we only need to prove $EX1_{X > \lambda} \geq EX - \lambda $, i.e. ...


3

The idea is that if you divide the entire probability space into some pieces $B_n$, and then you consider another event $A$, you can also divide up $A$ into pieces based on how the even overlaps with the $B_n$. Perhaps a picture would help. The red lines split up the space of the black box into pieces. The blue oval stands for the event $A$. Notice how we an ...


3

Because a characteristic function (Fourier transform of a real function) is hermitian $\phi_X(-t)=\phi_X(t)^*$ (complex conjugate). Hence $\phi_X(t)\phi_X(-t) = |\phi_X(t)|^2 \ge 0$. But $\sin(t)/t$ is negative for some values of $t$. Hence...


3

With countable additivity, it is easy to prove continuity of measure from below and from above. That is, the measure of an increasing countable union is the limit of the measures of the finite unions. The same holds for decreasing intersections provided one of the sets has finite measure (which holds for probability of course). This would prevent your case, ...


2

The probability measure defined by the Kolmogorov axioms (which include countable additivity) are set-continuous: the probability of a limit is the limit of the probabilities. If one denies countable additivity and accepts only finite additivity, then this set-continuity property is no longer applicable. Consider the following notion from Calculus 101 ...


2

here is a FREE pdf of Robert Ash's Probability and Measure theory. It has answers in the back.


2

$d_i(\omega)$ is the $i$th binary digit of $\omega$. Let $S_n = \sum_{i = 1}^{n} \frac{d_i(\omega)}{2^i}$, and let $S = \sum_{i = 1}^{+\infty} \frac{d_i(\omega)}{2^i}$. Then $$S - S_n = \sum_{i = n+1}^{+\infty} \frac{d_i(\omega)}{2^i} \leq \sum_{i = n+1}^{+\infty} \frac{1}{2^i} \leq \frac{1}{2^n}.$$ It follows from this that $$ S_n \leq S \leq S_n + ...


2

Hint: Since $$\sum_{n \geq 1} \mathbb{P}(X_n \neq -1) < \infty$$ it follows from the Borel-Cantelli lemma that for almost all $\omega \in \Omega$ there exists $N = N(\omega)$ such that $$X_n(\omega)=-1 \qquad \text{for all} \, n \geq N.$$


2

Suppose that $$\limsup_{n \to \infty} \frac{|S_n(\omega)|}{n}<\infty$$ for some $\omega \in \Omega$. Then $$X_n = S_n-S_{n-1}$$ implies $$\begin{align*} \limsup_{n \to \infty} \frac{|X_n(\omega)|}{n} &\leq \limsup_{n \to \infty} \frac{|S_n(\omega)|}{n} + \limsup_{n \to \infty} \frac{|S_{n-1}(\omega)|}{n} \\ &= \limsup_{n \to \infty} ...


2

No: let $\Omega=\{a,b\}$, $\mathcal{F}=2^\Omega$ and suppose that your filtration is $\mathcal{F}_1=\{\emptyset,\Omega\}$ and, for any $k\ge 2$, $\mathcal{F}_k=2^\Omega$. Let $\mathbb{P}$ be given by $\mathbb{P}(\{a\})=\mathbb{P}(\{b\})=\frac{1}{2}$. Let $X_1\equiv 1$ and for $k\ge 2$ $X_k(a)=2$, $X_k(b)=0$. Then $(X_k)$ is a nonnegative converging ...


2

As you've observed, the method of moments estimator for the mean of a Poisson is just the sample mean, $2.3$ in your case. (The maximum likelihood estimator - which I suspect you're likely to come across soon, if you're learning about the method of moments - is the same for the Poisson but is not necessarily the same in general.) A Poisson random variable ...


2

$Var(X)=E[X^2]$ is the variance (measures spread around mean), which for a standard normal distribution is $1$. $E[X^3]$ measures the skewness of the density and since a normal distribution is symmetric the skewness is $0$.


2

Consider the probability space $((0,1),\mathcal{B}(0,1))$ endowed with the Lebesgue measure $\lambda$ and the random variables $$X(\omega) := 1_{(0,1/2)}(\omega) \qquad \qquad Y(\omega) := 1_{(1/2,1)}(\omega), \qquad \omega \in (0,1).$$ Then $X \sim Y$. Set $X_n(\omega) := Y(\omega)$ for all $n \in \mathbb{N}, \omega \in (0,1)$. $X_n \to X$ in ...


2

Here is an answer that doesn't (explicitly) use normed vector spaces. Nevertheless, I'd like to mention that, for a given probability space $(\Omega, \mathcal F, P),$ we have the Hilbert space $L^2(\Omega) := \{Y:\Omega\rightarrow\mathbb R\ |\ Y\ \mathcal F$-measurable and $E(Y^2) < \infty \}$ with norm $\|Y\| = \bigl(E(Y^2)\bigr)^\frac12$ and scalar ...


2

For $p\geqslant 1$, we have $\lVert X_n\rVert_p^p=2/n$, hence $X_n\to 0$ in $\mathbb L^p$. Since (in general) convergence in $\mathbb L^1$ implies convergences in probability, which implies in turn convergence in distribution, we can answer questions 2), 3) and 4). Note that for these question, we did not use independence. Question 1) can be solved using ...


2

Let $A$ be the $\sigma$-algebra generated by the intervals $(a,b]$ for $a,b \in (0,1]$ with $a < b$, and let $B$ be the Borel $\sigma$-algebra. To show $A \subseteq B$, it is enough to show that each interval $(a,b]$ as above is a Borel set. If $b=1$, this is clear, because $(a,1]$ is open. If $b < 1$, then $(a,b]$ is the intersection of the countable ...


2

If you want a probability book that uses real analysis but not measure theory, then you want an older book back before measure theory became so central to the subject. I recommend Feller's two volume work, An Introduction to Probability and Its Applications.


1

Consider a Markov chain on the integers where it moves right with probability 1. The limit distribution is all zeros, but this doesn't qualify as an invariant distribution because it needs at least one positive entry.


1

Ash's book on Probability Theory is fantastic and covers what you want, although at a bit higher level, with some measure theory: http://www.amazon.com/Probability-Measure-Theory-Second-Robert/dp/0120652021/ref=asap_bc?ie=UTF8


1

Unfortunately, the difference is not negligible (though of course it depends on how much one is actually willing to neglect). Take for example the sets $X,Y=\{0,100\}$. Then over a uniform distribution, $$E[\min(x,y)]=25$$ while $$\min(E[X],E[Y])=50$$ One is double the other! To see $E[\min(x,y)]=25$, we note that there are $4$ possibilities which ...


1

You said: $${\rm P}(\text{no male in the 4 people sample})={2\over 3}.{{2n\over 3}-1\over n}.{{2n\over 3}-2\over n}.{{2n\over 3}-3\over n}$$ But it should be: $${\rm P}(\text{no male in the 4 people sample})={2\over 3}.{{2n\over 3}-1\over n-1}.{{2n\over 3}-2\over n-2}.{{2n\over 3}-3\over n-3}$$Because after choosing a person, the total choices also decreases ...


1

No, $T_n<\infty$ does not hold true. Proof: Suppose that $T_n<\infty$ almost surely. Since $(X_t)_{t \geq 0}$ is a martingale, $X_0 = 1$, the optional stopping theorem yields $$\mathbb{E}X_{T_n \wedge t} = 1$$ for any $t \geq 0$. Since $|X_{T_n \wedge t}| \leq n$ for all $t$, we get $$\mathbb{E}X_{T_n}=1 \tag{1}$$ by the dominated convergence ...


1

See "Testing statistical hypotheses" by E.L. Lehmann (196), Chap. 6 regarding regularity conditions. If you don't have access to that, here is a link to them. See p. 6-12 of the pdf. To summarize, you need to demonstrate a bit more than just consistency of the mode. Specifically, you need to show: $\lim_{n \to\infty} E(X_n)=\lim_{n \to\infty} \int xf_n(x) ...



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