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2

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: $$ ...


1

One approach to intuition is, since Y is always positive, the positiveness of X has no impact on the positiveness of Y. Therefore, X being more positive has as much impact on Y being larger than X being more negative. This is the antithesis of high correlation. I think that is all there is to it.


1

I can't help you with the first case. What is your problem here? But to the second one: If two random variables are uncorrelated (i.e. covariance is zero), they are not necessarily independent. The example you have is the standard example to demonstrate this. Only if $X$ and $Y$ have a joint bivariate normal distribution, from $cov(X,Y)=0$ follows that ...


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 $0$.


0

The error covariance matrix is \begin{equation} \frac{1}{p+2} A^{-1}. \end{equation}


0

Your explanation of a is incoherent. You seem to be assuming that the first four cards must be of different suits, which is not required. The easiest approach to this is inclusion-exclusion. Take the total number of hands, subtract the number with a void, notice that you have subtracted the number with two voids twice, so add them back in once, then think ...


0

Assuming that at least one of $E$ or $F$ has positive probability, then with probability $1$, one of them will occur eventually. Then $P(E|(E\cup F))=\frac{P(E\cap(E\cup F))}{P(E\cup F)}=\frac{P(E)}{P(E)+P(F)}$


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...


0

I do not see an easy approach, so I tried a simulation in R using the following code, which sees whether you lose in the first million games, repeated $100000$ times: set.seed(2015) subgames <- 10 supergames <- 100000 # maximum games is supergames*subgames cases <- 100000 gain <- function(x, y=10){ ifelse(y<10, y, y + ...


0

Let $A$ be the case where the observer saw the advertisement, and let $B$ represent the case that the observer bough the item. What we know is the following: $P(B|A) = .56$, the probability that an observer buys the item given that they saw the ad is 56% $P(B|\overline{A}) = .08$, the probability that an observer buys the item given that they did not see ...


1

Nice question, in a way, but I'm afraid that the answer is probably going to be "no, you can't apply this to a job interview". Firstly, to make use of Bernoulli trials you want a sequence of trials. OK, it's possible for the "sequence" to consist of just one trial, but that's generally a very uninteresting case. You could perhaps look at a geometric ...


0

By the definition of covariance, $$\text{Cov}(X,Y)=E[XY]-E[X]E[Y]$$ In this case, we take $X=1_{A_1\leq a}$ and $Y=1_{A_2\leq a}$. Then $XY=1_{A_1\leq a,A_2\leq a}$ and your result follows.


2

To summarize: $P($Both Stocks Go Up$) = 38\%$ $P($Both Stocks Go Down$) = 11\%$ $P($A Down, B Up$) = 16\%$ $P($A Up, B Down$) = ?$ Notice that these 4 events make up the entire set of possible combinations of the movement of stocks $A,B$. Because of this, their probabilities add to $1$, and we can conclude $P($A Up, B Down$) = 1 - (.38 + .11 + .16) = 1 - ...


3

The 4 Events $E_1$ = Both decrease $E_2$ = Both increase $E_3$ = A increases, B decreases $E_4$ = A Decreases, B increases compliment each other in the sense that exactly one of these events will happen. Therefore we have $$P(E_i\cap E_j)=0, i\neq j$$ and $$P(E_1)+P(E_2)+P(E_3)+P(E_4)=1$$ What is the probability that the stock from Company A will ...


3

The sum of the probabilities for all possible cases need to add to one. Since the stocks must go up or down (not stay the same) there are four possible outcomes for two stocks: {$A\uparrow B \uparrow, A\uparrow B \downarrow, A\downarrow B \uparrow, A\downarrow B \downarrow $}. If you're given three (disjoint) probabilities of the four, subtract from one to ...


1

You can do it directly: let $W = \rho Y$ where $Y$ is distributed as described. Then $$\Pr[W \le w] = \Pr[\rho Y \le w] = \Pr[Y \le w/\rho] = G(w/\rho).$$ Then we have $$G(w/\rho) = \mathfrak{S}(\gamma,cw/\rho) = \mathfrak{S}(\gamma,(c/\rho)w).$$ Thus $W \sim \Gamma(\gamma,c/\rho)$. All of this is just a fancy way of showing that $c$ is a scale ...


1

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 ...


1

Let $E$ be the event you described ($6$th correct answer is on last question) As you have stated, you need the probability of $5$ correct out of the previous $7$ given by $${ 7 \choose 5 } \left(\frac{1}{4} \right)^5\left(\frac{3}{4}\right)^2$$. Let's call this event A. His ability to guess the answer to the last question properly is independent of his ...


1

Choose $5$ out of the first $7$ questions to be correct. Choose the remaining $2$ questions to be incorrect. Choose the last question to be correct. $$\binom75\cdot\left(\frac14\right)^5\cdot\binom22\cdot\left(\frac34\right)^2\cdot\binom11\cdot\left(\frac14\right)^1=\frac{189}{65536}$$


0

You can think of it like this: The envelopes have fixed positions $1, 2, 3, 4, 5, 6, 7, 8$. Now which letter gets which envelope is described by an 8-tuple. The 8-tuple $(2, 1, 3, 4, 5, 6, 7, 8)$ means that he gets every letter right, except for letter 1 and 2. Your sample space is the set of all permutation of the number 1, ... 8: $$\Omega = \{(x_1, ...


1

Note: The following is not an answer, but merely some thoughts which might or might not be helpful to you. First note that you confused(?) your inequality signs. I think you want $$ \gamma_{n}\left(\left\{ x\in\mathbb{R}^{n}\,\mid\,\left\Vert x\right\Vert ^{2}\geq\frac{n}{1-\varepsilon}\right\} \right){\color{red}\leq}e^{-\varepsilon n/4} $$ and $$ ...


2

Since January has $31$ days, the most days a month can have, and $\frac1{12}= 0.0833\ldots $, there is no obvious way to get a figure as low as $0.0828$. Either it is a trick question or you have spotted an error.


1

I'll offer two classical ways to predict a value: Plug-in estimate: let $\hat \mu$ be your estimate of $\mu$ (e.g., the MLE), then model $X\sim Poi(\hat \mu)$. Use the upper and lower 2.5th percentiles of this estimated distribution as your predction interval. It won't be exact, but you should be reasonably close to 95% coverage. Percentile bounds with ...


1

If exactly 6 of 8 are right, then exactly 2 of them have been swapped. So choose two of then to exchange. $\binom{8}{2} = 28$.


0

Hint: Let $X$ denote the sum of eyes by throwing two dice and let $E$ be some event. Then (*) $P\left(E\text{ and } X=4\right)=P\left(E\mid X=4\right)P\left(X=4\right)$ So finding $P\left(E\text{ and }X=4 \right)$ and $P\left(X=4\right)$ leads to finding $P\left(E\mid X=4\right)$. Apply this on the events mentioned in a), b) and c). Let $D_{1}$ be the ...


1

You have $3$ possible events in your probability-space: First Die | Second Die -----------|------------ 1 | 3 -----------|------------ 2 | 2 -----------|------------ 3 | 1 Out of these $3$ events: There is $1$ event in which the first die gave 3, hence the probability is $\frac13$ There are $2$ events in ...


1

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} ...


0

Your idea is correct, and so is your answer $\frac23$. However, there is a route that avoids calculating infinite sums and is in my view more elegant. Let $W$ denote the event that you win and let $X$ denote the sum at the first toss. Then: $$\mathbb{P}\left(W\right)=\mathbb{P}\left(W\mid X\in\left\{ 7,11\right\} \right)\mathbb{P}\left(X\in\left\{ ...


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

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 $$X_n(\omega)=-1 \qquad \text{for all} \, n \geq N.$$


1

If $X\sim\mathrm{Poisson}(\lambda)$ then $\mathbb E[X]=\lambda$. From the method of moments, equating the sample mean and population mean we get $\hat\lambda=\overline X$. So your reasoning is correct. Another estimator can be obtained by $$\tilde \lambda = \frac1n\sum_{i=1}^nX_i^2 -\left(\frac1n\sum_{i=1}^nX_i\right)^2, $$ as also ...


0

This is how you can attack such problems: First rewrite the inequality as $$ \frac{\binom{40}{10-k}\binom{10}{k}}{\binom{50}{10}} > 0.5 $$ By using the definition of the binomial coefficient and some algebra, rewrite this as $$ 2\frac{40! 10! 10!}{41\cdot 42\dots 50} > (10-k)! (30+k)! k! (10-k)! =F(k) $$ On the last line we have defined a new ...


0

The probability that $1$ accident occurs on a given day is $\frac17$ The probability that $7$ accidents occur on a given day is $(\frac17)^7$ The probability that $7$ accidents occur on the same day is $(\frac17)^7\cdot7$


0

Here is the answer to the question, based on the answer manual (it was worded slightly differently in my textbook, hence why I did not find it). My part 1 was correct. The textbook question is as follows -- I'm not sure the cells versus boxes thing is supposed to make it be any different. >If $n$ balls are placed at random into $n$ cells, find the ...


1

There are seven days and seven accidents (which are distinguishabla from one another), so the number of possible outcomes is $7^7$. However, only seven of them fulfil our criteria: all accidents on Monday, all on Tuesday, ..., all on Saturday. So the answer is $1/7^6$.


1

For a fixed $t$, taking logarithm gives $\log \phi_{Y_n}(t) = \sum_{k=1}^n \log \cos(kt \sqrt{\dfrac{3}{n^3}})$ Using $$1-\cos(x) = \dfrac{x^2}{2} - O(x^4), x\to 0$$ $$\log(1-x) = -x + O(x^2), x\to 0$$ we have $$1- \cos(kt \sqrt{\dfrac{3}{n^3}}) = \dfrac{3k^2t^2}{2n^3} + O(\dfrac{9k^4t^4}{n^6}) = \dfrac{3k^2t^2}{2n^3} + O(\dfrac{k^4}{n^6})$$ and ...


0

Note: The following is a two-step approach. I will start with a simpler problem in order to be well prepared for the more challenging problem stated by OP. We also take a look at small examples as simple plausability check. We will see that the resulting generating function is the one stated by @RusMay. The following is based upon section I.4.1 and ...


0

I doubt $Z$ has a simple distribution. But it might be possible to describe its moments. For example if you have a random variable $X_i$ with a Poisson distribution with mean $\lambda_i$ then its variance is also $\lambda_i$ and its second moment is $\lambda_i+\lambda_i^2$. So if you have a random variable $Y_i$ which is $X_i$ with probability $p$ and $0$ ...


0

The simple solution would be to sit and wait until 5 loses come. This will take statistically 32 turns. Then use this betting system (Assuming you lose) £5 £12 £30 £60 £70. At this point you have a 1/1024 chance of losing. £150 £200 £300 £500 £700. At this point you have a seriously low chance of this happening and when it does you'd be extremely likely to ...


0

This seems to be an old question, but I find it interesting hence a treatment: We want to show that $\displaystyle X_n=\frac{\sum x_i}{s_n} \sim N(0,1)$ for an appropriate choice of $\displaystyle s_n \rightarrow \infty $. To do so recall the Lindeberg's condition that $\displaystyle lim_{n \rightarrow \infty} \sum \int_{|z|\geq s_n ...


0

Because the bus interval arrival time is 30-minute or 15-minute with equal probability, the probability your arrival falls into a 30-minute interval is $\frac{30}{30+15}=\frac{2}{3}$ the probability your arrival falls into a 15-minute interval is $\frac{15}{30+15}=\frac{1}{3}$ if your arrival falls into a 30-minute interval, the expected waiting time is ...


1

Define the event $\mathcal{H}$ according to my hint above (where $\mathcal{H}$ stands for "hint"): $$ \mathcal{H} = \{\omega'\in\Omega: G(X(\omega') ) \leq G(c)\}\cap \{\omega'\in\Omega : X(\omega')< c\} $$ We want to show that $Pr[\mathcal{H}]=0$. Your reply to my hint shows you already got most of the way: I think you agree that: $$ \mathcal{H} ...


3

Your question is unclear. If I were to answer the question you posed in your title, then the transformation $$Y = g(X) = e^X$$ is monotone, hence $$F_Y(y) = \Pr[Y \le y] = \Pr[e^X \le y] = \Pr[X \le \log y] = F_X(\log y).$$ We also have $$f_Y(y) = f_X(g^{-1}(y)) \left| \frac{dg^{-1}}{dy} \right| = f_X(\log y)/y.$$ If $X$ is exponentially ...


1

$F_Y(y)=\mathbb P(Y\le y)=\mathbb P(e^X\le y)=\mathbb P(X\le\ln y)=F_X(\ln y).$ Edit: Please consider the comment, if $Y=X^2$ then $F_Y(y)=\mathbb P(Y \le y)=\mathbb P(X^2\le y)=\mathbb P(\pm X\leq \sqrt y),$ here you would have to be more careful, possibly.


2

Suppose that $A,B$ are independent, $A\sim \mathcal{N}(\xi,\sigma^2)$, and $B\sim \mathcal{N}(\eta,\tau^2)$. Then $$ tA\sim \mathcal{N}(t\xi,(t\sigma)^2),\qquad A+B\sim \mathcal{N}(\xi+\eta,\sigma^2+\tau^2). $$ Consequently, it follows that $$ \sum_{i=1}^{n}\frac{X_i}{i}\sim ...


1

You have to go through each case: $ \begin{cases} 1 & a+b<C \\ \frac{C^2}{2 a b} & (a\geq b\lor a\geq C)\land (a<b\lor b\geq C)\land C>0 \\ \frac{C-b}{a} & a+b=C \\ -\frac{a-2 C}{2 b} & a<b\land a<C\land b\geq C \\ -\frac{b-2 C}{2 a} & a>b\land b<C\land a\geq C \\ -\frac{a^2-2 a C+(b-C)^2}{2 a b} & ...


0

Right, so far. The total number of outcomes, our total probability space, is $(6\times 2)^5$, because each of 5 people can independently have $6\times 2$ results; so we multiply that five times (ie: take it to the 5th power). $$|U|=(6\times 2)^5 = 12^5$$ Now if we want the first favoured space of nobody rolls a six, the we reduce the outcomes each person ...


1

We know that $$\begin{align}\text{Cov}(X,Y)&=E[(X-E[X])(Y-E[Y])]\\&=E[XY]-E[X]E[Y]\end{align}$$ while $$\begin{align} \widetilde{Cov}(X,Y)&=\widetilde{E}[XY]-\widetilde{E}[X]\widetilde{E}[Y]\\ &=E[MXY]-E[MX]E[MY] \end{align}$$ Assume $\Omega=A_1\cup A_2\cup A_3\cup A_4$, with $A_1,A_2,A_3,A_4$ disjoint and with the same probability ...


0

** How many outcomes are in the event where nobody rolls a six? If they can't roll a six, there are 5 other numbers to roll, and either coin-flip is still allowed. So each person has $2\times 5=10$ possible outcomes. Since there are 5 people, there are $10^5$ possible outcomes. ** How many outcomes are in the event where at least one person rolls a six? ...


0

Suppose a bag exactly 5 marbles that are either red or green, and the probability of the bag containing 0, 1, 2,...,5 red marbles is uniform (e.g., each has probability 1/6). One person draws a marble from the bag, and it is green. If this green marble is not put back into the bag, what is the probability that the next marble will be red? You have, for ...



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