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There is much more information in a joint distribution than can be captured by its marginal distributions. It is one thing to be told that a joint distribution can't be constructed from marginals in a unique way. It is another to have some examples. Here are a few. Discrete distributions. Consider four different joint distributions with the same marginals. ...


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Yes, $\mathsf P(X=x_i, Y=y_j)$ is taken as equivalent to $\mathsf P(X=x_i \cap Y=y_j)$. It's easier to write and read lists with commas than lots of $\cap$ symbols so we often use them for lists of conjunctions. (Not always, but often.) More specifically it should be something like: $\mathsf P(\{\omega: X(\omega)=x_i\} \cap \{\omega: Y(\omega)=y_j\})$ , ...


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This is really a long, possibly useful, comment. Not a straightforward answer. If the distribution of $X_i$ allows you to find $f_T$ easily, and if $T$ has a continuous distribution with support $S$, then it will likely be easier to integrate to find $E[(1-T)^{-1}]$ by evaluating $\int_S \frac{1}{1+t}f_T(t)\,dt.$ (Not to quibble, but note the slight changes ...


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Obviously, it only works for large $n$. Heuristically, the explanation is that although $X_n$ and $Y_n$ are not actually independent, after a large number of steps $n$, the information about the individual horizontal components of $X_n$ is essentially lost, so that little can be deduced about the distribution of $Y_n$, except that $n$ and $X_n$ together ...


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So the characteristic function of $\text{B}(n,\lambda/n)$ is $$ ((1-\lambda/n)+\lambda/n e^{it})^{n} = \left( 1 + \frac{1}{n} \lambda\left( e^{it}-1 \right) \right)^n. $$ Now use that $$ \lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n = e^x. $$ Then the convergence and uniqueness theorems for characteristic functions imply that the distribution is ...


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Perhaps use of the letter $p$ for the Poisson average daily rate is confusing you. This is not a probability. Many books use $\lambda$ (lower-case Greek 'lambda') for this rate. You may not be able to replace $p$ with a number in your final answer, but that may be a worthwhile suggestion to get started understanding the problem. Suppose the average rate ...


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I'm not sure there's a simple formulation for the distribution $P_n(k)$. Let $P_n(k)$ have a $z$-transform defined by $$ Q_n(z) = \sum_{k=0}^\infty P_n(k) z^k $$ Let $\gamma = 1-\alpha-\beta$. Then the distribution $P_{n+1}(k)$ is given by $$ P_{n+1}(k) = \alpha P_n(k-1) + \beta P_n(k) + \gamma P_n(k+1), \qquad k \geq 1 $$ $$ P_{n+1}(0) = (\beta+\gamma) ...


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Hint: for calculating $p(x = 2)$, you don't need to work out $p(AC), p(BC), p(DC)$ separately, you just need to work out $p((\text{not-}C)C)$, which is $\frac{8}{10}\frac{2}{9}$. Can you see how to generalise this? Hint for part 2: work out the conditional probability $p(\text{the fish of type D gets taken out}|x=r)$, then use the law of total probability ...


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Notice that $X$ has distribution $Exp(0.5ln(2))$. Hence: $P(X<3|X>1) = 1 - P(X>=3|x>1) = 1 - P(X>3|X>1) = 1 - P(X > 2+1|X>1) = 1 - P(X>2)$ As per the memorylessness of the exponential distribution.


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I assume you want this as a function of $p$ as well, and that $p$ is a constant in $(0,1)$? (if $p=p_n$ or $p=0$, then it'll be tricky or hopeless). You can use generalize the following fact (here for $p=1/2$) with the appropriate constants: $$ \mathbb{P}\left\{ X \in \left[\frac{n}{2} - c\sqrt{n}, \frac{n}{2}+c\sqrt{n} \right]\right\} \in (1\pm ...


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For a random variable $X$ of binomial distribution of parameters $n,p$ the possible values taken are $0,1,2, \cdots, n$ with probabilities$$P(X=k)={n \choose k}p^k(1-p)^{n-k}.$$ If $a$ is a real constant and $Y=aX$ then the valueas taken by $Y$ are $1a,2a,\cdots , na$ with probabilities $$P(Y=ak)=P(X=k).$$ I don't know if this answers the question. EDITED ...


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You seem to have got the first two right. The last one is just the probability that he gets 3 or more orders, so you need $$ 1-e^{-\lambda} - e^{-\lambda} \lambda -\frac{e^{-\lambda} \lambda^2}{2!} $$ and you know $\lambda$.


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You are correct that $aX$ is also Gaussian if $X \sim \text{Gaussian}$, with mean $a\mu$ and variance $a^2\sigma^2$. As far as I am aware there is not a binomial equivalent in terms of the name, but there is the same scaling of mean and variance.


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Assuming you are treating your data as the population, rather than a sample, then for Q1 you count the number of cases which meet the criteria "stroke = yes, income = yes, exercise = no, and smoke = no" and divide by the number which meet the criteria "income = yes, exercise = no, and smoke = no". Similarly for Q2 you count the number of cases which meet ...


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I think I understand what you want. Here is a one-dimensional version. Make a sketch as follows: (a) Draw a standard normal curve (it has height about .4 in the middle, inflection points at $\pm 1$, and nearly touches the x-axis at $\pm 3.$ (b) Superimpose a rectangle with base extending between $\pm 2$ and 1 unit it height. You will generate a candidate ...


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It is not clear where to begin, it depends how the Pareto distribution was defined. I will assume that it was defined by the density function $$f_X(x)=\frac{\alpha \lambda^\alpha}{x^{\alpha+1}} \quad\text{when}\quad x\ge \lambda,$$ and $0$ elsewhere. Then the mean of $X$ is given by $$E(X)=\int_\lambda^\infty x\frac{\alpha ...


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Fortunately, you gave a frequency histogram so the sample size can be discerned (at least very nearly). I guessed at bin counts and used uniform noise to spread data through their intervals. As is, the data overwhelmingly failed a Shapiro-Wilk test for normality (P-value about 0.003). Removing the few observations in the bin from 0 to 5, I found that the ...


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Isn't a good approach to calculate the average wait time if B arrives at 2pm (50% probability) and the average wait time if B doesn't arrive at 2pm (50% probability), then average these two average times out, in 50% - 50% proportion. If B arrives exactly at 2pm, then B's average wait will be 30 minutes. If B doesn't arrive exactly at 2pm, I believe the ...


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This may or may not be what you mean by resolution 10 minutes, but it might mean that the 30 minute interval is marked off into three 10 minute intervals, and during any 10 minute interval the person is in room A,B, or C with the given probabilities of .9, .07, and .03. At the end of the first 10 minute interval the person could change rooms or remain where ...


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You can turn it into a Beta function with the substitution $e^{-y}=x$, so $-e^{-y} \, dy = dx$. Then $$ \binom{2}{k}\int_0^{\infty} (e^{-y})^{k+1} (1-e^{-y})^{2-k} \, dy = \binom{2}{k}\int_0^1 x^k (1-x)^{2-k} \, dx = \binom{2}{k} B(1+k,3-k) = \binom{2}{k} \frac{k!(2-k)!}{(3-k+k)!} = \frac{2!}{k!(2-k)!} \frac{k!(2-k)!}{3!} = \frac{1}{3}. $$ Boring ...


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$Z_i$ and $W$ aren't needed: you just need to note $\forall i, E(X_i^2)=\text{Var}(X_i)+E(X_i)^2=\alpha+\alpha^2$ and use the linearity of expectation to get: $$ E\left[\sum_{i=1}^n X_i^2\right]=\sum_{i=1}^nE(X_i^2)=n\alpha(1+\alpha). $$ Note also that the independence assumption is superfluous.


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The probability of $X=x$ tests needed is the sum over the number of students $n$ who fail the original test of: the probability of $100-n$ passes and $n$ fails in the $100$ original test multiplied by the probability of $n-1$ passes and $x-100-n$ fails in any order in the first $x-101$ retests multiplied by the probability the final retest is a pass so ...


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ok , first find the probability for at most 5 arrive in hour. sum Probabilities for k=1...5 using parameter λ for 1 hour. subtract this probability from 1.i.e 1-P(atmost 5) This gives you the probability of at least 5. Do the same for second part , except λ is calculated for 2.5 hours


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It can be shown that the integral is a time-changed Brownian Motion and from there we can deduce the distribution from looking at this BM. Define $f(u) = \sigma(T-u)$ and $X_t = \int^t_0 f(u) dW_u$. Define the stopping time $$ \tau_t = \inf\lbrace u \geq0 : [X]_t > t \rbrace $$ and look at the process $B_t := X_{\tau_t}$. We see that by definition of the ...


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You picked $10$ people from a population with half male and half female. You are now picking $2$ people from them. With replacement, every time there is $50%$ probability to get a female. So the probability is $0.5\cdot 0.5$. Without replacement, it is like tossing $10$ coins and find the probability of existence of $2$ consecutive heads. You can refer ...


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$\phi^{t-1} w_1 + \phi^{t-2} w_2 + \ldots+ w_t = \sum_{i=1}^{t-1} \phi^{t-i} w_i$ This equation is not correct: On the left-hand side there appears a "$w_t$"-term, but on the right-hand side it doesn't. So, instead it should read $$\phi^{t-1} w_1 + \phi^{t-2} w_2 + \ldots+ w_t = \sum_{i=1}^{\color{red}{t}} \phi^{t-i} w_i$$ Except for that your ...


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Best way to view this is to view it as TWO dependent events. a) simply use r studio and use dbinom(2, 15, 0.02) b) Use a tree diagram P(S) = 0.9 P(NS) = 0.1 P(disease) = 0.02 P(nodisease) = 0.98 therfore to get a false positive (0.1 X 0.98) and for the second section of b is just (0.01 X 0.02) Let me know if there is any question because i am doing ...


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If two random variables, X and Y, are independent, then $Var(dX+ eY)=Var(dX)+Var(eY)$. $Var(cX+(1-c)Y)=Var(cX)+Var((1-c)Y)$ $=c^2\cdot Var(X)+(1-c)^2Var(Y)=c^2\cdot a+(1-c)^2b$ Calculating the derivative w.r.t c $2ca-2(1-c)b=0$ Now you can solve for c.


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First of all, note that if $\mathbf{P}_Z$ is the law of $Z$, then you can check that $\mathbf{P}_Z$ is absolutely continuous with respect to $\lambda$ (the Lebesgue measure), so there will be a density function. Now to find it, use the cumulative distribution function. Let $t \in \mathbb{R}$. $$F(t) = \mathbf{P}(Z \leq t) = \mathbf{P}(e^{-3t}\leq|X|)$$ ...


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No, the joint probability distribution does not have to be stationary w.r.t. time. Just imagine a precessing top with no friction. The probability distribution for $l^2$ is just a delta peaked at some particular value. But the joint probability distribution of $(l_1,l_2,l_3)$ is a delta peak that moves around in the angular momentum space.


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In calculating $F_Y$, you did wrong. Note that $X^2$ and $X$ are not independent. But $X^2 \le X$ iff $X \le 1$. We have for $t \in [0,1]$: \begin{align*} \def\P{\mathbf P}\P(Y \le t) &= \P(X \le t)\\ &= \frac 12 t\\ \end{align*} and for $t \in [1,4]$: \begin{align*} \P(Y \le t) &= \P(Y \le 1) + \P(1 < Y \le t)\\ ...


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By definition, the pair of random variables $\langle{X,Y}\rangle$ has a bivariate normal distribution, only if the (marginal) distribution of any linear combination of the variables is a univariate normal distribution. $X$ is a linear combination of the two random variables; witness: $X=1\cdot X+0\cdot Y$.   Thus it has a univariate normal ...


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Chebyshev's Inequality is true for any $c > 0$, but you are right that it only provides useful information for $c > 1$. This is actually surprisingly easy prove. Define $\mu = E(X)$ and $\sigma^2 = E((X-\mu)^2)$. Observe that for any $c \geq 0$ we have $\mathbb{1}\left\{\left|\frac{X-\mu}{\sigma}\right|\geq c\right\} \leq \frac{(X-\mu)^2}{\sigma^2 ...


1

You can start by computing the number of ways to place $k$ balls into $n$ slots such that no two consecutive slots are filled (this will be $n!$ times the probability that no two consecutive numbers are drawn). Call this $N_p(n,k)$. To do that, note that either the last slot on the right is filled or not. In either case, group each of the balls (except ...


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After looking through some textbooks from fluid mechanics I found a similar and problem and it seems that this type of problem is solved by adopting a certain type of PDF and performing a little calculus involving Dirac's delta. PDF definition Firsltly, we assume we assume that the deterministic version of our problem has a solution: \begin{equation} ...


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Let the RV $X$ take value $n^2$ with probability $\frac{c}{n^2}$ where $c$ is a norming constant. Then $$E \left( \frac{X}{n^2+X} \right) \geq \sum_{k=n}^{\infty} \frac{c}{k^2} \frac{k^2}{n^2+k^2} \geq \sum_{k=n}^{\infty} \frac{c}{k^2} \frac{n^2}{n^2+n^2}= \frac{1}{2}\sum_{k=n}^{\infty} \frac{c}{k^2} \sim \frac{c}{2n} $$ which is not summable. Note that ...


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Three-dimensional case (n=3) The Wikipedia article on area and volume element states that the area element of the unit sphere is $\sin\phi\,\mathrm d\phi\,\mathrm d\theta$ where I use $\phi$ to denote inclination and $\theta$ to denote azimuth. If you consider your fixed vector as the zenith direction, then my $ \phi$ will be the same as yours. To turn ...


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The function $F(t)$ is a CDF (cumulative distribution function), not a PDF (probability distribution function). It gives $$ F(t) = P(T \leq t) $$ To determine the average from a CDF, you do not multiply by $t$; you just subtract from $1$ and integrate: $$ E(T) = \int_{t=0}^\infty [1-F(t)] \, dt $$ The difference between the two questions is whether you ...


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By completing the square, we obtain the random variable $\ a\sigma^2\left(\frac{X}{\sigma}+\frac{1}{2\sigma}\frac{b}{a}\right)^2-\frac{1}{4}\frac{b^2}{a}$, where $\left(\frac{X}{\sigma}+\frac{1}{2\sigma}\frac{b}{a}\right)^2$ is a noncentral chi-squared random variable with 1-degree of freedom and non-centrality parameter ...


1

the first step is to invert the covariance matrix so that you can express the joint distribution function as $$ f(x_1,f_2) = \frac{1}{N} e^{-(A_{11} x_1^2 + A_{12}x_1 x_2+ A_{22}x_2^2)} $$ where $N$ is a normalization constant. Then by integrating from $x_1 = 0$ to infinity you find the conditional distribution of $x_2$ given that $x_1 > 0$ will be ...


0

I also know the variables $S_1$ and $S_2-S_1$ are independent. Yes, and thus: $$\begin{align} f_{S_1,S_2}(s_1, s_2) & = f_{S_1, S_2-S_1}(s_1, s_2-s_1) \\[1ex] & = f_{S_1}(s_1)\cdot f_{S_2-S_1}(s_2-s_1) \end{align}$$ Now, do you know what $\;f_{S_2-S_1}(t)\;$ is? Hint: A Poisson process is memoriless.


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There are so many possible points of confusion here that I hardly know where to start. First, I wonder what 'time' means in your tabulation. It can hardly be seconds. If the rate is 88/sec. then how can there possibly be 908 within the first second. Second, one might suppose 'rates' refer to rates of exponential distributions. But then there is no ...


0

1) The cumulative distribution function $F(x)$ is equal to the anti-derivative of the probability density function. Therefore $$ \begin{align} F(x) &= 2 \int (1-x)\\ &= 2\left(x - \frac{x^2}{2}\right) + C\\ &= 2x - x^2 + C \end{align} $$ Where $C$ is some constant. We know that the cumulative distribution function must be equal to 1 ...


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Your answers to the first two questions are correct. You can then obtain the mean as follows: $$ E(X) = \int_{x=0}^1 xf(x)\,dx $$ and the standard deviation as follows: $$ E(X^2) = \int_{x=0}^1 x^2f(x)\,dx $$ and then $\sigma = \sqrt{E(X^2)-[E(x)]^2}$.


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heropup -> Don't you think we have $\hat{\theta}\sim InvGamma(n,n\theta^{-1})$ instead of $\hat{\theta}\sim InvGamma(n,n\theta)$ ?


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The assertion is not true. A counterexample: Pick a random variable $Y$ and define $Y_n:=Y$ and $X_n:=Y+1/n$. Then $X_n$ converges in distribution to $Y$, and obviously $Y_n$ converges in distribution to $Y$. But $X_n$ and $Y_n$ are equal nowhere.


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You need to know what is the expected value of the uniformly distributed variable $w$. Let me denote the expected value by $E$, then $E(w)$ is the expected value of $w$, and in fact $E(w) = \frac{1}{2}$. This is perhaps not surprising, as $\frac{1}{2}$ is the midpoint of the interval $[0,1]$. What you really want is $E(wd)$. However, $d$ is not random, so ...


2

I will set out a solution strategy and some hints, in keeping with the site's homework policy. It's quite an effective one in general - in this case, checking the back of my envelope, I used four simple lines of algebra and three rough sketch plots (two PMFs and an overlaid contour plot) - but other approaches are available. In particular it can make life ...


2

Here's an explicit counterexample. Let $U=|X|$ and $V=|Y|$. Then $\text{Var}(X)\geq \text{Var}(Y) \Rightarrow E(|X|)\geq E(|Y|)$ is equivalent to $E(U^2)\geq E(V^2) \Rightarrow E(U)\geq E(V)$ for $U,V$ continuous in [0,1]. Consider a $\text{beta}(\alpha,\beta)$; it has mean $\mu=\frac{\alpha}{\alpha+\beta}$ and second raw moment $\mu_2'=\mu ...


-1

The probability of not drawing a red ball after $k$ turns is given by $(1-p)^k$. Thus if we want to ensure that a red ball is chosen before $k$ turns, then the probability of not drawing a red ball after $k$ turns must be $0$. That is, $(1-p)^k = 0$ Therefore, $(1-p) = 0$ Thus, $p = 1$ Therefore in order to be sure a red ball is chosen after $k$ ...



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