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0

If the $X_n$ are independent, then $Y_n$ is Poisson distributed with parameter $\Lambda_n=\lambda_1+\ldots+\lambda_n$ for every $n$. This makes it clear that without additional restrictions on $(\lambda_n)$, the convergence of $Y_n/n$ is hopeless even in this simple case. The characteristic function of $Y_n/n$ is ...


1

If each result is independent of the others, the probability to see some given result of probability $p$ at least once in $n$ results is $$1-(1-p)^n.$$ If $p=1/N$ with $N$ large, after $n\approx cN$ results, this is roughly $$1-\mathrm e^{-c}.$$ Thus, to get the result with probability at least $x$, one needs a number of results roughly $$-\log(1-x)\cdot ...


0

The mode is the most likely value to occur. Poisson takes only discrete values, so you are given that $$P(X=1)=P(X=2)$$ which implies that $$e^{-λ}\frac{λ^1}{1!}=e^{-λ}\frac{λ^2}{2!} \iff λ=\frac{λ^2}{2} \iff λ^2-2λ=0 \iff λ(λ-2)=0$$ so $λ=0$ which is rejected (since $λ>0$) or $λ=2$ which is acceptable. Thus $X \sim$ Poisson$(λ=2)$ and therefore ...


1

As often, it is difficult to answer this question because the OP says nothing about their background. Anyway, a standard approach to determine the recurrence/transience of a Markov chain (its type) is to compute $P_1(T_0\,\text{infinite})$ as the limit of $P_1(T_n\lt T_0)$ when $n\to\infty$, since each $P_1(T_n\lt T_0)$ involves only a finite Markov chain. ...


0

Your approach is fine, except that you have to be more careful with the domain of the new random variable. Before you start ask yourself what is the domain of $$Y=\frac{1}{1+U}$$ in order to discriminate cases for $y$ in a proper way. Since $U$ takes values in $[0,1]$, obviously $$\frac12\le Y\le 1$$ Therefore $$P(Y\le y)=\begin{cases}0, ...


0

Define $I:=\left\{ \left(r,s\right)\in\mathbb{N}\times\mathbb{N}\mid0\leq r\leq s\leq5\right\} $. For $\left(r,s\right)\in T$ let $E_{r,s}$ denote the event that before the first $6$ appears exactly $r$ distinct numbers $\neq6$ appear, that before the second $6$ appears exactly $s$ distinct numbers $\neq6$ appear, and that before the first $6$ appears ...


1

I would say M - the number of minutes / day = random variable $m_i=i*25\,$, i = 0, 1, 2, 3, 4, 5 $P(M = m_i)=p_i$ Then expected value $\displaystyle E(M)=\sum_{i=0}^{5}m_ip_i=\cdots$


1

What type of distribution is this? I do not recognize it, but the wording of the question seems to imply it is identifiable. Indeed it is! Hint: The distribution puts weight $p_k$ on each integer $k$ between $0$ and $n$, where $$p_k=\binom{n}{k}a^k(1-a)^{n-k},\qquad a=\frac{t}{t+s}.$$ Does this ring a bell? Second hint: The LaTeX encoding of the first ...


1

Consider some nonnegative integers $i$ and $j$. To reach $i+j$ starting from $0$, one must first reach $i$ starting from $0$ then reach $i+j$ starting from $i$. The probability of the first event is $d_i$. By the Markov property and the invariance of the dynamics by the translations of $\mathbb Z$, the property of the second event conditionally on the ...


0

I don't know regarding the distribution, but support is easy to check: you know that $X_t$ can take any value in $\Bbb N = \{0,1,2,\dots\}$ unconditionally. You also know that $X_t$ is non-decreasing, so if $X_{t+s} = n$ then $X_t\in [0;n]$ almost surely. I'm pretty sure that you've used this fact while computing the conditional distribution. An educated ...


0

The probability that any birth is a boy or a girl is NOT 1:1 as many people believe; in actuality 105 boys are born for every 100 girls. This ratio of 1.05 is known as the "secondary sex ratio." Given these real world statistics, one must give 1.05 weight to the four scenarios that include one boy and 1.00 weight to the all-girl possibility. Therefore, the ...


1

So far, so good. (Note though, that the sample size $n$ is $5$.) Integrate to find the order statistics' cumulative distribution functions. $$\begin{align} F_{Y_{(1)}}(0.6) & = \int_0^{0.6}n[1-F_Y(y)]^{n-1}f_Y(y)\operatorname d y \\[1ex] & = \int_9^{0.6} 10y(1-y^2)^4\operatorname d y & \because f_Y(y)=2y, F_Y(y)=y^2, n=5 \\[1ex] & = ...


1

Each random variable similar to $$\xi_1 - \frac{\xi_1 + \xi_2 + \xi_3}{3}$$ is normal centered with variance $\frac23\epsilon$ and they are independent hence the expectation to be computed is $\frac13\sqrt{2\epsilon}E(Z)$ with $$Z=\sqrt{X_1^2+X_2^2+X_3^2},$$ where $(X_k)$ is i.i.d. standard normal. Now, the distribution of $Z$ is well known...


0

Hint: The distributions of $(X_1,S_n)$ and $(X_k,S_n)$ coincide hence $E(X_1\mid S_n)=E(X_k\mid S_n)$ for every $1\leqslant k\leqslant n$ (this uses the fact that conditional expectations only depend on joint distributions). Summing these over $k$, one gets $$n\,E(X_1\mid S_n)=\sum_{k=1}^nE(X_k\mid S_n)=E\left(\left.\sum_{k=1}^nX_k\right| S_n\right)=\cdots$$ ...


-2

They are mutually exclusive events. How many son/daughters the couple already has is irrelevant. Hence, the probability of the last child (or rather any child) being boy is 1/2 (some would say 1/3 too).


0

$\begin{align} \mathsf P(W\leq w) & = \mathsf P(X+Y \leq w ) & \text{where}\quad w\in[0,\infty) \\[2ex] & = \int_0^w\mathsf P(Y\leq w-x\mid X=x)\;f_X(x)\operatorname d x \\[2ex] & =\int_0^w\int_0^{w-x}f_{Y\mid X}(y\mid x)f_X(x)\operatorname dy\operatorname dx \\[2ex] & =\int_0^w\int_0^{w-x}f(x,y)\operatorname dy\operatorname dx \\[2ex] ...


0

You're pretty close. Be careful with the distribution function of $X_n$, though. It should be: $$F_{X_n}(x)=\begin{cases} 0&\text{if}\ x<0,\\ \frac{k}{n+1}&\text{if}\ \tfrac{k-1}n\leq x<\tfrac{k}n,\ k=1,\ldots,n,\\ 1&\text{if}\ x\geq1. \end{cases}$$ Now we need to do an $\epsilon$-proof. Fix $\epsilon>0$ and let $N\in\mathbb{N}$ be ...


0

As mentioned above, a good spot to start is figuring out probabilities for having selected a card of a given type, then figuring out probabilities of having won having selected a given card. I would personally recommend creating a tree diagram: There is a $\frac{30}{55}$ of having picked card $A$, a $\frac{20}{55}$ chance of having picked card $B$, and a ...


0

You get $1 or 10 according to whether you obtain one heart or two hearts respectively; otherwise you get nothing. Assume that at the start of the game, there are 30 coupons of type A, 20 coupons of type B and 5 coupons of type C. Find the probability that, for just that one game, you will obtain You proceed by ignoring distracting details. The only ...


0

If the maximum is less than the median, then all points in the sample have to be less than the median. A sampled point will be less than the median with probability $1 \over 2$, so if $Y$ is the maximum and $m$ is the population median, then $$P[Y \le m] = {\left({1} \over {2} \right) }^n$$


1

You did it all right, except for the final fraction. The joint event is: $$A \cap B \equiv (1 \le X \le 3) \cap (0 \le X \le 4)\equiv (1 \le X \le 3)\equiv A$$ (Notice that this holds because here $A \subset B$) Hence $$\frac{P(A\cap B)}{P(B)}=\frac{P(A)}{P(B)}=\frac{.383}{.6826}=0.561$$


0

You have $n$ iid random variables, $\{Y_1, \ldots, Y_n\}$, and wish to know the probability that the $Y_n$ is less than all of the others. You know the probability density function $f_Y$ and have found the cumulative distribution function $F_Y$. You have almost got it. What you have is the probability that a specific value of $Y_n$ is the greatest of the ...


1

I see there is already a solution, but just to note that in a simple problem like this, you can easily get to the answer by enumerating the mutually exclusive cases: 1: Draw 1: prob = 1/5 2: Draw 2, 3, 4: prob = 3/4 3: Draw 2, 3, 4: prob = 2/3 total = 6/60 1: Draw 2, 3, 4: prob = 3/5 2: Draw 1: prob = 1/4 3: Draw 2, 3, 4: prob = 2/4 total = 6/60 1: ...


1

The favoured case happens when you draw a 1, and 2 to 4 on all other draws, of three draws( and never a 5).  When drawing without replacement this is determined by ways to select the favoured items divided by the ways to select any 3 items from 5. $$\mathsf P(F) = \frac{{1\choose 1}{3\choose 2}}{{5\choose 3}} = \dfrac{3}{10}$$


1

You need to sum over wins and losses to get $E(X)$. $$ E(X)={18\over38}\times5+{20\over38}\times-5 = -10/38=-0.26 $$ If you play 100 times then the amount is just $100\times$ that, so the average loss is \$26 (the average "win" is -\$26).


1

The number of independent Bernouli trials until a success is a geometric distributed random variable. The expected number of trials so required is then the mean of the distribution. $$N\sim\mathcal{Geo}(1/1000) \implies \mathsf E[N] = \boxed{\qquad?}$$ Do you know what is the mean of a geometric distribution?


0

The probability that the highest order statistic in a sample of $n$ iid continuos random variables is below the median $m$ is: $$\mathsf P(\max\{X_i\}<m) = \mathsf F_X(m)^n$$ Here, $F_X$ is the cumulative probability function for the distribution. We cannot proceed any further unless we know what actually is the distribution. What is the median, and ...


0

The probability that the girl is in the room directly below you depends on a number of factors you have not considered. How many floors are there in the accommodation? You first have to consider the probability that you are on the floor directly above her. You also have to consider bias in assigning guests to floors and rooms. Are higher level floors ...


2

I would say. When 4 cards are not aces, then remaining 48 cards are 4 ace. So pull of 48 cards 2 cards without replacement --> hypergeometric probability distribution. The probability that between these two cards is not ace will: $\displaystyle P=\frac{{4 \choose 0}{48 - 4\choose 2 - 0}}{{48 \choose 2}}=\frac{473}{564}$ Edit: I see that I reached the ...


4

Rajkumar's direct solution works nicely. You mentioned that the problem could be considered as conditional probability, and I'll show here that that method gives the same answer. As I mentioned in a comment, the main reason for the inaccuracy of your original answer is overcounting the number of ordered $6$-card hands with the first four being non-aces, and ...


0

From what you have said, you need to solve $$(1-p)+p(1-p) \ge \frac12$$ subject to the usual condition for any probability that $0 \le p \le 1$. This is at worst a quadratic which I will leave for you to solve. Your example of $p=0.5$ has an error: the probability of reaching state $T_2$ in one step would indeed be $0.5$ and of reaching state $T_2$ ...


2

We know that given 4 cards are not aces. So remaining 48 cards = 4 ace cards + 44 non-ace cards. Probability of getting 5th card as non-ace card = 44/48 Probability of getting 6th card as non-ace card(given 5th is non-ace card) = 43/47 So probability of getting no-ace card at all is = 44/48 times 43/47.


0

you can write all possible remainders by divisibility of $3$ these are $0,1,2$ and you will get $$a^3+b^3 \equiv \mod 0+0;1+0;0+1;1+1;2+0;0+2;2+1;1+2;2+2, 3$$ now we will find $$0+0;2+1;1+2\equiv 0 \mod 3$$


1

1) We are permuting two distinct letters of $26$, so we have $26!/24!$. We then multiply (by rule of product) for permuting four digits of $10$ to get a total count of: $26!/24! * 10!/6!$. 2) To see the letters, notice a pattern. If we choose $A$ as our first character, then there are $25$ options for the second character. If we choose a $B$ first, we have ...


0

Hint: For every real number $t$, define $X_t=\max(X,t)-t$ and $Y_t=\max(Y,t)-t$, then $X_t$ and $Y_t$ are nonnegative random variables and $X_t$ is stochastically larger than $Y_t$ (can you show this?) hence $E(\max(X,t))\geqslant E(\max(Y,t))$. Consider the limit when $t\to-\infty$, assuming that $X$ and $Y$ are integrable.


2

You need to conduct a hypothis test for the variance not for the mean. The confidence interval $9\pm 1.5$ is a confidence interval for the mean and will not help you to draw conclusions about the variance! Accordingly, you do not need the $t$-distribution but the $\chi^2$-distribution. Formally In order to conclude whether $σ^2$ has been reduced below $1.0$ ...


0

Picking up Robert Israel's suggestion: If a couple keeps having children until they have at least a boy and a girl, and ignoring twins, ... it's difficult. If they had children until they reached their goal, the probability is 1/16 each that they stop at 3 boys and 1 girl, at 1 boy and 3 girls, 3/4 that they stopped with 2 or 3 children, 1/8 that they had ...


2

We have $0 \le x_1<x_2< \cdots < x_N\le M$ (with $N=10$, $M=20$). Define $$\begin{array}{} y_0&=& x_1 \\ y_1&=&x_2-x_1-1\\ y_2&=&x_3-x_2-1\\ &\cdots&\\ y_{N-1}&=&x_N-x_{N-1}-1\\ y_N&=&M-x_N \end{array}$$ Then any $\{y_i\}_{i=0\cdots N}$ defines a valid configuration (and viceversa) iff $y_i\ge0$ ...


0

In order to count the number of perfect loops with $n$ people, we need to count the number of distinct labelings of a cycle on $n$ vertices, say $c(n)$. While I don't have a proof yet, I think the number is given by $$ c(n)=\frac{1}{2n-3}\binom{3n-6}{n-2}. $$ On the other hand, to count the number of valid loops we want to know the number of ways to ...


0

Hint: $$ \Bbb Q = \bigcup_{q\in\Bbb Q} \{q\} $$ and $P(\bigcup A_n)\le \sum P(A_n)$ if $(A_n)$ is enumerable.


3

You are asking for the chance of a single cycle given that you have a derangement. For $n$ people, the number of derangements is the closest integer to $\frac {n!}e$ To have a cycle, person $1$ has $n-1$ choices, then that person has $n-2$ choices, then that person has $n-3$, etc. So there are $(n-1)!$ cycles. The odds are then (just about) $\frac e{n}$


0

Let $P_N$ be the probability to obtain a perfect loop with $N$ participants. $ A \rightarrow C \rightarrow B \rightarrow D \rightarrow A $ $ A \rightarrow B \rightarrow D \rightarrow C \rightarrow A $ $ C \rightarrow A \rightarrow B \rightarrow D \rightarrow C $ are perferct loops. We can see that the second and third one describe the same loop, so we can ...


0

Simply put, a set isn't indexed, it's just a collection of unique elements. Without context, I don't know why it serves the author's purposes to define this structure as opposed to just considering a set of random variables, but presumably it is helpful for an argument they are making.


0

Let $X \sim Uniform(0,1)$ with pdf $f(x)$: Let $X_n$ and $X_{n-1}$ denote the largest and second largest order statistics in a sample of size $n$. Then, the joint pdf of $(X_{n-1},X_n)$, say $g(x_{n-1},x_n)$, is: where I am using the OrderStat function from the mathStatica package for Mathematica to automate, or do it manually following: Wiki -- ...


0

Ok let's go but I will only outline the solution (using theroem 2 of planetmath.org) : Take $\mathcal{K}$ as of processes of the form $\sum_{i=1}^n C_i.1[({s_i},{t_i}]\times A_{s_i}]$ where $0\leq {s_i}<{t_i}$, $s_i$ are increasing and $A_{s_i}\in\mathcal{F}_{s_i}$ (those processes are bounded and predictable, and $\mathcal{K}$ is stable by ...


0

The probability that $n-1$ of them are less than $x$ is $nx^{n-1}(1-x)+x^n$, so the PDF of the second-highest is the derivative of that. If the second-highest is $x$, then the expected value of the sum of the top one is $(1+x)/2$, and the sum of the top two is $(1+3x)/2$. Integrate the PDF of the second-highest, multiplied by $(1+3x)/2$.


1

Let $\left(\Omega,\mathcal{A},P\right)$ be a probability space. Then a random variable on it is a function $X:\Omega\rightarrow\mathbb{R}$ such that $X^{-1}\left(B\right)=\left\{ \omega\in\Omega\mid X\left(\omega\right)\in B\right\} \in\mathcal{A}$ for each Borelset $B$. Denoting the collection of Borelsets on $\mathbb{R}$ by $\mathcal{B}$ we state that ...


1

Your proof is correct. More generally: If $E(X_n)\to C$ and $\mathrm{var}(X_n)\leqslant\epsilon_n$ where the series $\sum\limits_n\epsilon_n$ converges, then $X_n\to C$ almost surely. Hint: There exists some positive sequence $(\alpha_n)$ such that $\alpha_n\to0$ and $\sum\limits_n\frac{\epsilon_n}{\alpha_n^2}$ converges. Consider the events ...


-1

To help you prove that $Y$ is a random variable, we need to know what kind of definition of random variables you are familiar with. Regarding the distribution function: $F(t) = P(X \leq t)$. Let $F_Y$ be the distribution function of $Y$. Then $$F_Y(t) = P(Y \leq t) = P(aX + b \leq t) = P\left(X \leq \frac{t-b}{a}\right) = F\left(\frac{t-b}{a}\right)$$ if $a ...


1

This is an old question, but I still like to write an answer. First note that for your c.f. it holds that $\phi_x(\omega)=\phi_x(\omega+2\pi)$, indicating that your random variable takes only value in $\mathbb{Z}$ and hence is of lattice type for which a simple inversion formula is $f_X(x)=\frac{1}{2\pi}\int_{-\pi}^{\pi} e^{-i \omega ...



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