Tag Info

New answers tagged

0

Number of finance books=6 number of maths books=4 - step1 consider all the different finance books as 1unit and all the different maths books as 1 unit According to your question Now we are having just two different units ( first one is finance books unit and the second one is maths books unit ) these 2 different units ...


0

Suppose the finance books are identical and so are the finance books. No matter how we order them, there are always $6!\cdot4!$ ways to reorder them so that the math books occupy the same places they did before reordering them. In other words, each order when the books are identical gives us exactly $6!\cdot 4!$ orders when the books are distinct, so this ...


2

Let the singers be (AB)(CD), we are interested in the first group We need P(both right) + P(only 1st right) + P(1st from wrong group)*P(2nd right) = $(\frac{1}{4}\cdot \frac{1}{3}) + (\frac{1}{4}\cdot\frac{2}{3}) + (\frac{1}{2}\cdot\frac{1}{3}) = \frac{5}{12}$


0

You do have a martingale! \begin{align*} \mathbb{E} [Z_t | \mathcal{F}_{t-1}] &= \mathbb{E} [X_t | \mathcal{F}_{t-1}] + \mathbb{E}[Z_{t-1}|\mathcal{F}_{t-1}] \\ &= 0 + Z_{t-1} \end{align*}


3

You almost solved the problem yourself: $\begin{align} P(A\cup B) &= P(A)+P(B)-P(A\cap B)\\ P(A)=P(B)&= \frac{1}{4}\\ P(A\cap B)&=P(A)P(B\mid A)\\ &=\frac{1}{4}\frac{1}{3}=\frac{2!}{4!}\\ P(A\cup B)&=\frac{5}{12} \end{align}$ The reason why $P(B\mid A)=\frac{1}{3}$ is simply that name used for the first guy is clearly not the one of the ...


0

Partial answer: According to a theorem (Gil-Pelaez), if $0$ is a point of continuity of $F_X$, then $$ F_X(0) = \frac{1}{2} - \frac{1}{\pi} \int_0^{\infty} \frac{\mathrm{Im}[\phi_X(t)]}{t} dt, $$ so, since a random variable $X$ is a.s. non-negative, then $F_X(0) = 0$, $\int_0^{\infty} \frac{\mathrm{Im}[\phi_X(t)]}{t} dt = \pi / 2$ is the condition for a.s. ...


2

Let $A_i$ denote the number of times number $i$ appears (each number is equally likely to appear) and $\mathcal{A}$ be the set of all possible combinations of $a\equiv(a_1,\dots,a_K)$ s.t. $\sum_{k=1}^Ka_k=N$ and each $a_k\ge 0$. Then for $a\in \mathcal{A}$ ...


0

This seems to be the kind of problem one would find at the end of a chapter on addition and multiplication rules of probability and the idea of conditional probability. Please go back and read the chapter and look at examples of these ideas. Then try to follow the outline toward an answer I have given below. (Generally speaking, probability is not a subject ...


0

The approach through the cumulative distribution function is reasonable, but there are troubles with the algebra. I am interpreting $\log$ as the natural logarithm. Minor modification will take care of things if we interpret $\log$ as logarithm to the base $10$. We have for suitable $y$ $$\Pr(Y\le y)=\Pr\left(-\frac{\log(1-X)}{\lambda}\le ...


2

Consider any $\pi\in (0,1)^I$ with $\sum_{i\in I}\pi_i=1$. Then $$\sum_{i\in I} p_i\ln p_i-\sum_{i\in I} p_i\ln \pi_i=-\sum_{i\in I} p_i\ln \frac{\pi_i}{p_i}$$ $$\ge -\sum_{i\in I} p_i\ln \frac{\pi_i}{p_i}\ge -\ln\left[\sum_{i\in I} p_i\frac{\pi_i}{p_i}\right]=-\ln\left[\sum_{i\in I}\pi_i\right]=0$$ This inequality shows that $$\max_{\pi} \sum_{i\in ...


1

$(x,y)$ is uniform so it's density is constant, say $c$. To determine $c$, note that $F(1,1)=1$ or $$\int_0^1\int_0^1cdxdy=1\Rightarrow c=1$$ It's not clear what do you mean by $P\{Y\ge x|Y\ge 1/2\}$. If $x$ is a constant then $$P\{Y\ge x|Y\ge 1/2\}=2(1-x)\wedge 1$$ because the marginal distribution of $Y$ is uniform over $[0,1]$. If you actually ...


3

I'm using the definition of "discrete random variable" as a random variable taking values in a countable set (not necessarily a discrete set). The sum of two discrete r.v.'s is discrete, because the sum of two countable sets is countable. So (2) can't happen. And then (1) can't happen, because if $X + Y = Z$, $Y = Z - X$.


0

Since each 1 is distinct on the first reel, you should probably label them 1a, 1b,1c. Once you label everything distinctly, it is easier to see that you are correct. It is confusing to say 1-4-2-5-5 is different than 1-4-2-5-5. It is clear that 1a-4-2-5-5 is different than 1b-4-2-5-5


1

Integration by parts is the right idea, but let's start with the left side of $\displaystyle\int_{0}^{\infty}(1-F(x))\,dx - \int_{-\infty}^{0}F(x)\,dx = E(X)$ instead of the right side. We have $\displaystyle\int_{0}^{\infty}(1-F(x))\,dx = \underbrace{\left[x(1-F(x))\right]_{0}^{\infty}}_{0}-\int_{0}^{\infty}-xF'(x)\,dx = \int_{0}^{\infty}xf(x)\,dx$, and ...


1

I don't know if this counts as an elegant solution in your book, but I think it's cute. Let's say the "frequency state" of a deck is the number of cards of each face value remaining. A full deck, for example, has the frequency state "4 aces, 4 twos, 4 threes...," while an empty deck has the frequency state "0 aces, 0 twos, 0 threes...." There are $5^{13}$ ...


0

Assume I have to pay $r$ dollars per pull. I shall pull until I obtain the black ball. The expected number of pulls for that is $5.5$. Therefore the expected revenue per game will be $5.5(10-r)+90$ dollars. If this is positive the game is in my favor. The break-even point is $r_0={145\over5.5}\doteq 26.35$ dollars.


1

Since the bag contains 190 dollars, I'll eagerly pay $19 per pull. At the worst, I'll break even by continuing till the end ! Since the $100 bill can come equi-probably on any draw, if the operator wants to break even, I will get $\frac{1}{10}\cdot [100 +110 +120 + 130 + ...+190] = 145$ dollars against $\frac{1}{10}\cdot[x+2x+3x+4x+5x+6x+7x+8x+9x+10x] = ...


0

Define a random vector $Z$ which is a result of concatenating $X$ and $Y$. $H(X,Y) := H(Z)$.


1

Denoting the current day as index $i$ (so that yesterday is $i-1$ and the day before yesterday is $i-2$), the previous state comprises two elements, $State_{n-1}=\{S_{i-1},S_{i-2}\}$, where $S_k$ is the state of the weather for day $k$. Thus the current state is $State_{n}=\{S_{i},S_{i-1}\}$ Lettting $R$ be the occurrence of rain on a particular day, so ...


1

So you have a $9/10$ chance of winning $10$ and a $1/10$ chance of winning $100$. Thus your expected outcome is $$\frac{9}{10} \cdot 10 + \frac{1}{10}\cdot 100 = 19$$ If this is how much you expect to win per pull, you can answer how much you would be willing to pay per pull. After the first pull, you either have 8 white balls and 1 black, or 9 white and 0 ...


0

Let at time t, the first player has moved a distance R1 while second player is situated at a point in circumference and has caught the first player at the same time t. Let the distance R2 be between his initial point in the circumference to the point where he caught the first player. Let R be the radius of the circular field. The first player randomly ...


1

When someone says that a baby is born every (instert a period of time), they actually mean that this is true on average. It can happen in a small village that three babies are born on the same day. Many of these deviations from the average get washed out if you have enough data. But some things don't. Several things affect when people are born. A mother's ...


3

The cited statistic, "every second 3 babies are born" is just a rough average. In truth, human birth rates are extremely seasonal (the weather having a dramatic effect on human behavior). Here is a reference: http://www.sas.upenn.edu/~valeggia/pdf%20papers/birth_seasonality.pdf


0

The most convinient way to get concentration results on the median is isometric inequality. Basically it states that among all geometrical shapes with equal area, the circle has the smallest perimeter. But when this result is generalized to a high-dimension metric space with a probability measure, you may get some counter-intuitive results called "measure ...


0

To me both proposed results seem wrong. With $P_{1/2}=:p'$, $P_1=:p\>$ I argue as follows: The probability that neither of the two rods fails is $(1-p')^2$. Therefore with probability $2p'-p'^2$ we shall see the failure of a first rod, and then the second rod will fail as well with probability $p$. The probability $F$ that the beam will fall to the floor ...


3

Intuitively, saying that two logical propositions $E$ and $F$ are independent means that learning whether $E$ is true or false tells you nothing about $F$, and vice versa. For example, suppose you flip two coins, a penny and a nickel, and consider the propositions $E$ = "The penny came up heads" $F$ = "The nickel came up heads" For ordinary coins, it's ...


3

The two concepts of "independent" and "mutually exclusive" are different. Two events are "mutually exclusive" ( that is, $$P(E \cap F) = 0$$) if they can't both happen at the same time. For example, if I roll a die and define E = "I roll a 6" and F = "I roll a 3", these can't both be true. If E happens, i know F didn't. Two events are "independent" (that ...


1

You're correct that if $E$ and $F$ are independent, then $P(E \cap F)=P(E)P(F)$. If $E$ and $F$ are independent, then it means that the result of $E$ has no influence on the result of $F$ and vice versa. For example: I toss a quarter and a dime. The result of the quarter toss has no influence on the dime toss. However, if $P(E \cap F)=0$, then they are ...


0

0.5 probability of player 2 winning means that if player 1 runs perpendicular to the line joining the initial positions of the players, then player 2 will catch him on the boundary of the circle. In that case the path of player 2 will be the hypotenuse of an isosceles right with side length $r$, the radius of the circle. the length of this path is $l=\sqrt ...


1

The whole purpose of calculating the arithmetic mean of a distribution is to minimize the expectation of the square of the error. It all makes sense if you are penalized one dollar per square of the error. So if you know that tossing 3 fair coins gives you the usual distribution of outcomes, your best guess for the number of heads to show is 1.5, even though ...


0

There are two factors here. First is, as pointed out in the comments by Shalop and Gary, average value of a random variable need not correspond to an actual value assumed by that variable. There are also issues of interpretation. The well-known joke that says a man whose feet is on fire and head on ice saying he is on the average comfortable. But I would ...


1

This is a classic situation where you calculate the odds of losing both, and subtract that from 1. The situation is of the "at least 1..." type, and is calculated by first figuring out the opposite situation which is that it never happens. So the chances that you don't win the first is $9/10$, and the chance that you don't win the second is $8/9$. So the ...


1

For a person to win, they can either win on the first round or lose the first round and win the second round. The probability that they win in the first round is $1/10$. The probability that they lose the first round is $9/10$, and the probability they win the second round is $1/9$. Thus the probability that they win is $$ ...


2

$X,Y,Z\sim_{iid}\text{Exp}(1)$ because $F_{X,Y,Z}(x,y,z)=F_X(x)F_Y(y)F_Z(z)$ for all $(x,y,z)\in\mathbb{R}^3$. Denote $S_n=\sum_{i=1}^nX_i$ where $X_i\sim_{iid}\text{Exp}(\lambda)$. Distribution of $S_n/n$ follows $\text{Gamma}(n,n\lambda)$ because ...


3

$$F_{X,Y}(x,y)=P(X\leq x\cap Y\leq y) \leq P(X\leq x)$$ $$F_{X,Y}(x,y)=P(X\leq x\cap Y\leq y) \leq P(Y\leq y)$$ $$F_{X,Y}(x,y)^2\leq P(X\leq x) P(Y\leq y)$$


0

The first step is true because $\theta_n=\theta_{n-1}+1$ when $S_n\not =S_{n-1}$ and $S_n\not =S_{n-2}$ and ... and $S_n\not =S_{1}$ and $S_n\not =S_{0}$, so if $T_i=S_n-S_{n-i}$ then $\theta_n=\theta_{n-1}+1$ when $T_1\not =0$ and $T_2\not =0$ and ... and $T_{n-1}\not =0$ and $T_n\not =0$, but the $T_i$ are just the partial sums of $X_n, X_{n-1}, ...


0

It suffices to apply $(d)$ with $\tilde{g}= \theta_{N+1} g$to the expression in $\tilde{X}_{\tilde{\theta}}$. $$\tilde{X}_{\tilde{\theta}}(t) = \exp \bigg\{\langle \theta, \alpha(t)- \alpha(0) \rangle - \int_0^t \langle \theta, b(u) \rangle \, d\xi(u) -\frac{1}{2} \int_0^t \langle \theta, a(u) \theta \rangle \, du + \\ \theta_{N+1} g(t,\alpha(t)) - ...


1

Consider $F(a) = \Bbb{P}(\frac{X+Y + Z}{3} \leq a) =\Bbb{P}(X+Y + Z \leq 3a)$. Now calculate $$F(a) = \int_0^{3a} \int_0^{3a-x} \int_0^{3a - x - y} e^{-z}e^{-y}e^{-x} \, dz\,dy\,dx = \\ = \int_0^{3a} \int_0^{3a-x} (1 - e^{-3a + x + y})e^{-y}e^{-x} \,dy\,dx \\ = \int_0^{3a} \int_0^{3a-x} (e^{-y} - e^{-3a + x} )e^{-x} \,dy\,dx \\ = \int_0^{3a} (1 - ...


1

It's not quite clear how you're modelling the behaviour of the rod that holds under $G/2$ and is then supposed to carry the entire load. You didn't include the factor of $1-P_{1/2}$ for the rod holding under $G/2$ in the cases where the other one fails, but you did where the other one holds. I think the most lucid way of clearing up this confusion would be ...


1

That looks mostly right, except you probably meant $g_2^{-1}(s)$ where it says $g_1^{-1}(s)$, you didn't introduce $y$, and since the person specifying $f_R$ was finicky enough to include $I_{[0,1]}(r)$ there, they probably want you to include the characteristic function in $f_V$ and $f_S$, too. By the way, you can make those absolute value bars look much ...


1

$$\mathbb{E}X=\mathbb{E}X^+-\mathbb{E}X^-=\mathbb{E}[X\vee 0]-\mathbb{E}[-X\vee 0]$$ $$=\int_0^\infty\mathbb{P}\{X>x\}dx-\int_{0}^\infty\mathbb{P}\{-X\ge x\}dx$$


0

Bag one contains six elements $$\{1,2,3,4,5,6 \}$$ Bag two is the multiset of all possible sums of two dice rolls, of which there are 36, $$\{2,3^2,4^3,5^4,6^5,7^6,8^5,9^4,10^3,11^2,12\}$$ Considering the first bag, a $1$ or a $2$ wins $0/36$th of the time (never). A $3$ wins $1/36$th of the time. A $4$ wins $3/36$ of the time. How many combinations are ...


0

Some ideas: count the appropiate events $$\begin{align}&Left\;\; Hand&Right\;\; Hand\\{}\\ &3&(1,1)\\{}\\ &4&(1,1),\;(1,1),\,(2,1)\\{}\\ &5&(1,1),\,(1,2),\,(2,1),\,(2,2),\,(1,3),\,(3,1)\end{align}$$ and I leave to you to fill the last line and do the final mathematics.


1

The limits for $j$ in Conrado Costa's equation will be $\max(x_{1min},n-x_{2max})$ to $\min(x_{1max},n-x_{2min})$.


0

If n is from 1 to 96, it is divisible by 8 if it is of the form 8m+k where k = 0, 2, 4, 6, or 7. There are 5(96/8) = 60 from 1 to 96. Therefore the probability is 60/96 = 5/8.


1

Hint: the product is divisible by $8$ if: One factor is divisible by 8, or One factor is divisible by 4 and another is divisible by 2, or All three factors are divisible by 2. The last one is impossible in your situation (why?) What are the possible remainders when $n$ is divided by $8$ such that the first and second possibilities occur?


3

Hint: If $n+1$ is even, then both $n$ and $n+2$ are odd so the only hope is that $n+1$ is divisible by $8$. If $n+1$ is odd, then can you see why your expression is always divisible by 8?


0

Assigning weights is one way to go, and you can do this without constraints, but if you want an overall score for each question, but ideally you would probably also like to rank the questions in some non-trivial way. And the ranking will depend on the weights, and like you said you don't know what the "optimal" weights are. For this, you can compute how many ...


0

The answer is k*n times . According to probability After n chooses we expect that we choose each no. 1 time each .Similarly after 2*n chooses we expect that we choose each no. 2 times each .Similarly after n choose we expect that we choose each no. n times each .


0

THe first step is to calculate the law of the number of car that pass on the street in 5s. It can be shown that it follow a Poisson law with $\lambda = 5\times\frac{8.6}{60} \simeq 0.7166$ Then the number of car that pass on the street in 5s is $$P(X = k) = \frac{\lambda^k}{k!}e^{-\lambda}$$ So the probability that no car pass in the street in the next ...



Top 50 recent answers are included