# Tag Info

1

Try $$f(x)=\begin{cases}\frac14,&1\le x\le 3\\\frac12,&4\le x\le5\\0,&\text{otherwise}\end{cases}$$

1

You can simply choose a piecewise constant function $f$. Let the value of $f$ on $[1, 3]$ be $a$ and the value on $[0, 1) \cup (3, 6]$ be $b$. Now you can determine the value of $a$ and $b$ with the two equations you have.

1

You can use generating functions. Let $P=p_1x+p_2x^2+p_3x^3+p_4 x^4+p_5 x^5 +p_6 x^6$ where $p_i$ is the probability of $i$ occurring when rolling the die once. Then the coefficient of $x^k$ in $P^N$ gives the probability of rolling a sum of $k$ when rolling the die $N$ times and summing. For example, suppose ...

0

It seems to me the spirit of the original context is experimental. In order to get an analytic answer, even in a simple case with only two rolls, you need to specify precisely how the die is loaded. Here is a simulation in R of an experiment with $n = 3$ rolls of a die loaded so that faces 1 to 6 appear with probabilities $(1/12, 1/12, 1/12, 3/12, 3/12, ... 0 You have$100$balls in one bag, labelled$X_1,\ldots,X_{100}$and another$100$balls in another bag, labelled$X_{101},\ldots,X_{200}, where \begin{align} X_1,\ldots,X_{100} & \sim \mathcal{N}(0,1),\\ X_{101},\ldots,X_{200} & \sim \mathcal{N}(100,1)\end{align} are mutually independent normally distributed random variables with variance1$; ... 1 For finding the distribution of the first one $$(m+n-2)\frac{S^2}{\sigma^2}$$ let$S^2_1=\frac{1}{m-1}\sum_{i=1}^m(X_i-\overline X)^2$and$S^2_2=\frac{1}{n-1}\sum_{j=1}^n(X_i-\overline X)^2$. Then $$(m+n-2)\frac{S^2}{\sigma^2}=(m-1)\frac{S_1^2}{\sigma^2}\ +\ (n-1)\frac{S_2^2}{\sigma^2}$$ But as you correctly guessed, each summand of the last equation ... 2 For the first part, (i): $$\mathbb{P}\{X \geq Y\}\} = \mathbb{E}[\mathbb{1}_{\{X \geq Y\}}] = \mathbb{E}[\mathbb{E}[\mathbb{1}_{\{X \geq Y\}}\mid Y] ].$$ Dealing with the inner conditional expectation first, $$\mathbb{E}[\mathbb{1}_{\{X \geq Y\}}\mid Y] = e^{-\lambda} \sum_{k=0}^\infty \mathbb{1}_{\{k \geq Y\}} \frac{\lambda^k}{k!} = e^{-\lambda} ... 2 For the first one, for any t>0,$$P(X\geq Y)=P(tX\geq tY)=P(t(X-Y)\geq0)=P(e^{t(X-Y)}\geq1)\leq E(e^{tX})E(e^{-tY})$$by Markov inequality and independence. Noticing that E(e^{tX}) is the moment generating function of X (and noticing the same about Y), and substituting the forms of the moment general function, we get a bound that is equal to ... 0 For the first problem, i would try this way (but i'm not sure it will work though ;) : \mathbb{P}(X \geq Y) = \sum_{k \geq 0} \mathbb{P}(X \geq k, Y=k) = e^{-3\lambda}\sum_{k \geq 0} \sum_{i=k}^{\infty} \frac{\lambda^{i+k}2^k}{i! k!}. 0 By symmetry of f(x)=e^{-\frac{x^2}{2}} you have that \int\limits_{\varepsilon}^{\infty}f(x)dx=\frac{1}{2}(\int\limits_{\varepsilon}^{\infty}f(x)dx+\int\limits_{-\infty}^{-\varepsilon}f(x)dx). Now the limit as \varepsilon\rightarrow 0 will give you the integral of f(x) over all of \mathbb{R} which is a classical result and will solve your problem. 0 This trivially follows from upper bounding the distribution of standard normal. Probability density function obeys f(x)=\frac{1}{\sqrt{2\pi}}\exp(-x^2/2) which is always upper bounded by \frac{1}{\sqrt{2\pi}}. Now to obtain your bound P(|Z|<\epsilon)=\int_{-\epsilon}^{\epsilon} f(x)dx\leq ... 0 This is just the definitions. Distribution \mu is absolute continuous wrt distribution \nu means for any (measurable) set A, \nu(A)=0 implies \mu(A)=0. Does that hold for your examples? No, A=[3,5] has measure zero under Unif[0,3] but 1/2 under Unif[1,5]. \mu is singular wrt \nu means for any set A, \mu(A)>0 implies \nu(A)=0. ... 2 I will consider the integral without the limit process.$$I(a,b)=\int\limits_0^\infty e^{-a x^2}\cos(b x) dx=\sum_{n=0}^{\infty}\frac{(-1)^n.b^{2n}}{(2n)!}\int_0^{\infty}x^{2n}e^{-ax^2}dx$$by expanding the cosine. The integrals in the above sum are the familiar Gaussian Integrals defined by$$I_m=\int_0^{\infty}x^me^{-ax^2}dx$$for non negative integral m. ... 1 Using Euler's identity, we get:$$ \int\limits_0^\infty e^{-a x^2}\cos(b x) dx=Re \left( \int\limits_0^\infty e^{-a x^2} e^{ibx} dx \right)  \int\limits_0^\infty e^{-a x^2} e^{ibx} dx = \int\limits_0^\infty e^{-a x^2+ibx} dx $$Let's forget about imaginary unit and take ib=\beta for simplicity:$$ -ax^2+\beta x=-a ... 1 Formally, there is a probability space$Ω$(with a$σ$-algebra$\mathcal F$and a probability measure$P$) and the random variables$X_n$map$Ω$to$\mathbb R$(with$\mathcal B(\mathbb R)) as follows \begin{align}X_n:Ω&\mapsto \mathbb R\\[0.2cm]ω &\to \{0,1\} \end{align} withP(\{ω:X_n(ω)=1\})=P(\{ω:X_n(ω)=1\})=1/2$. Now, take an$ω \in Ω$and ... 2 Bayes' Rule is applicable (or just the definition of conditional probability): $$f_{Y}(y\mid Y\geqslant R) ~=~ \dfrac{f_Y(y)~\mathbf 1_{y\geqslant R}}{1-F_Y(R)}$$ 2 Note that the$X_n$are bounded by$1$, so the absolute convergence follows from the convergence of the sum$\sum \limits_{n = 1}^\infty \frac{2}{3^n}$. 0 I don't think there's enough information in your model to work out a satisfactory answer. If we draw up a table of scores like this: |0 1 2 3 -+------- 0|0 1 2 3 1|1 2 3 4 2|2 3 4 5 then I can populate it with the number of students like this (assuming only 10 students): |1 3 4 2 -+------- 2|1 1 - - 5|- 2 3 - 3|- - 1 2 Each column and row adds up to ... -1 Refer to this? Also $$M_{\Theta}(t) = E[\exp(t\Theta)]$$ $$= E[\exp(t\lim \frac{B_n + 1}{n+2})]$$ $$= E[\lim\exp(t \frac{B_n + 1}{n+2})]$$ $$= \lim E[\exp(t \frac{B_n + 1}{n+2})]$$ $$= \lim \frac{1}{n+1}[\exp(t \frac{1}{n+2}) + \exp(t \frac{2}{n+2}) + ... + \exp(t \frac{n+1}{n+2})]$$ $$= \lim \frac{a(n)}{(n+1)(1-a(n))} (1-a(n)^{n+1}), \ \text{where} ... 2 This is not a correct MGF, because M_X(0) always exists and is equal to 1 but here$$M_X(0)=-0(1-e^0)=0\neq1$$1 No distribution has such a moment generating function, for the trivial reason that M_X(0) = 0, which would imply that$$\operatorname{E}[e^{0X}] = \operatorname{E}[1] = 0,$$a contradiction. 0 An answer with hidden text, in case you want to try it out again before looking at the solution. Using your definition of Y_n, you should have used the central limit theorem to observe$$ \sqrt{n}(Y_n/n - 1)\stackrel{d}{\longrightarrow}N(0,2)\tag{1} $$So, using the delta method, with the function x\mapsto\sqrt{x} and (1), gives Therefore, ... 2 Hint: if n is even, you can group the terms using Gauss summation:$$y_1 + \cdots + y_n = (y_1+y_n) + (y_2+y_{n-1}) + \cdots.$$Can you say anything about these pairs? Try using the symmetry of the inverse CDF: If n is odd, you can first show that the (n+1)/2-th element is zero, and use the same trick as above to establish your result. 1 Suppose U, V \sim \text{Poisson}(\lambda). Then if X = \alpha U, Z = \beta V (\alpha \neq 0, \beta \neq 0),$$\mathbb{E}[K] = \mathbb{E}[X+Z] = \mathbb{E}[X]+\mathbb{E}[Z]=\alpha\mathbb{E}[U]+\beta\mathbb{E}[V] = \alpha\lambda + \beta\lambda = \lambda(\alpha+\beta)\text{.}$$1 Yes, as you noted, when \phi is not invertable we have to modify the transformation function. Such as, for instance, when \phi(x)=x^2, which is a fold mapping two intervals into one (the negative reals and non-negative reals to the non-negative reals), we modify the transformation formula to account for the fact that we have two "inverse" functions: ... 3 Through the Spitzer identity, it is possible to find some kind of transform of the distribution of M_n. Well, not exactly. The Spitzer identity involves the expressions M^+_n = \max_{0\le k\le n} S_k, where S_0 = 0, S_k = X_1 + \dots + X_k, k\ge 1. So this translates to the positive part of expression you are interested in. But it is possible to ... 3 When dealing with ceilings, the proper way to proceed is using probability intervals.$$ P(Z=z)=P(z-1<\beta X <=z)=P(\frac{z-1}{\beta}<X<=\frac{z}{\beta})=F_X(\frac{z}{\beta})-F_X(\frac{z-1}{\beta}) $$Where F_X is the cumulative distribution of X. Substituting:$$ ... 0 When$y \in [0,1]$you need integrate over x in$ [0,1]$, when$y>1$you have integrate over x in$[y-1,1] $Hope this address your question. 0 We need to draw a picture. Draw the line$y=x+1$. If we look at the conditions, we can see that the joint density lives on the trapezoid bounded by the$y$-axis, the$x$-axis, the vertical line$x=1$, and the line$y=x+1$. We want to "integrate out"$x$. Look at the picture. If$0\le y\le 1$, then$x$travels freely from$0$to$1$, and hence the density ... 1 You've neglected the possibility of ties and are over counting events where multiple dice equal$y$. You wish to calculate the probability that two dice are$x$and$y$and none of the remaining die are higher than$y$. There are two cases to consider. When$x=y$and when$x>y$When$x=y$you want the probability that all dice are at most$x$, ... 1 We have $$F_Y(y)=\Pr(Y\le y)=\Pr\left(\frac{X}{1-X}\le y\right).$$ This is$\Pr(X\le y(1-X))$, which is$\Pr(X(1+y)\le y)$. Finally, for$y$positive, which is the only interesting part, we have $$F_Y(y)=\Pr\left(X\le \frac{y}{1+y}\right)=\frac{y}{1+y}.$$ Elsewhere, we have$F_Y(y)=0$. For the density function, differentiate. The second problem is ... 0 You have the right pgf, though you could get there easier: $$G_Z(t) = E(t^Z) = E\left(t^{\alpha X}\right) = E\left(\left(t^\alpha\right)^X\right) = G_X\left(t^\alpha\right) = e^{\lambda\left(t^{\alpha}-1\right)}$$ since$X\sim$Poisson$(\lambda)$so we know that$G_X\left(t\right) = e^{\lambda\left(t-1\right)}$. In checking its validity, perhaps you are ... 1 An observation (to replace a previous wrong answer). Note that $$M_2=X_1+X_2^+$$ where$X_2^+=\max\{0,X_2\}$. So, $$E[M_2]=E[X_1]+E[X_2^+]=0+\frac{1}{σ\sqrt{2\pi}}$$ Similarly $$M_3=X_1+\max\{0,X_2,X_2+X_3\}=X_1+\left(X_2+\max\{0,X_3\}\right)^+=X_1+\left(X_2+X_3^+\right)^+$$ with$E[M_3]>E[X_2^+]=E[M_1]$. And two links here and here that might ... 0 As convergence in distribution only cares about the distribution of the random variables, one must we very careful when applying "pointwise" operation like division. Consider$\Omega = \{1,2\}$with the uniform distribution, and$X \colon \Omega \to \mathbf R$the identity embedding. Then$X$'s distribution is$\mathbf P_X = \frac 12 \delta_1 + \frac ...

0

We don't need to know the marginal distribution of $X_1$ to compute the variance of $X_1$. That is a good thing, since the required integral cannot be evaluated using elementary functions. By symmetry the mean of $X_1$ is $0$. So the variance is $E(X_1^2)$, which is $$\int_{x_2=0}^\infty \left(\int_{x_1=-x_2}^{x_2} ... 1 If P(X=n) = \dfrac{\lambda^{n}}{n!}e^{-\lambda} when n is a non-negative integer and 0 otherwise, and Z=\alpha X then$$P(Z=z)=\dfrac{\lambda^{z / \alpha}}{(z/\alpha)!}e^{-\lambda}$$when z/\alpha is a non-negative integer and 0 otherwise 1 If X=Y then 1,2,3,4 would be correct. In particular X-Y=0 with probability 1 If X and Y are independent then 1,2,4 are correct but 3 might not be. For example, suppose X takes the value 1 with probability 0.4 and the value 0 with probability 0.6 But if X and Y have a more complicated relationship then there is little you can say ... 2 First moment of the distribution is μ_1=E[X]=α and the first moment of the sample is m_1=\bar{X}. So, set$$μ_1=m_1 \implies α=m_1This is the first moment estimator for α. (Note: This method is confusing at the beginnning, because you think, ok so what? But think that m_1 is your sample mean and so it is known, it will be realized when you collect ... 1 The time T until the next event has continuous distribution, with density function f_T(t)=\lambda e^{-\lambda t} for t\gt 0, and f_T(t)=0 elsewhere. The distribution of T is called the exponential distribution with parameter \lambda. The expectation E(T) (mean) of T is given by E(T)=\int_0^\infty t\lambda e^{-\lambda t}\,dt. Integration ... 0 I found one approach, although we get a 1/(np) term rather than 1/n. According to [1], for any integer-valued random variable X (including a Binomial) with variance V, we have \begin{align} H(X) &\leq \frac{1}{2} \log_2 \left[ 2\pi e \left(V + \frac{1}{12}\right) \right] . \end{align} This is already very nice/useful, and probably often quite ... 0 A general keyword you could search for is "distributional similarity". My formulation: Words are similar to the extent they occur together with the same words. For this purpose, word co-occurrence statistics mean basically the number of times that other words occur together with the words of interest. First you build a "vector" that tells how often any of ... 1 The way you have written it, "\theta = probability that the psychic has ESP", \theta essentially is your prior distribution. There are only two possibilities, ESP and not-ESP, so the full statement of the prior is (I write e for ESP and \neg for negation): P(e) = \theta P(\neg e) = 1 - \theta Writing d for the observed data (3 out of 5 ... 1 First, some basic calculations. let p be the probability of guessing a card correctly. Then the probability of getting exactly 3 correct is \binom 53 p^3(1-p)^2. If p=.2 this is .0512, if p=.5 this is .3125 Let's say your prior is \theta_0. That is, before you test anything, you estimate that the "ESP probability" is \theta_0. Then, ... 1 We give a somewhat expanded version of your professor's hint, correcting some errors of transcription. For convenience of typing I will call your Y^\ast by the name W. We want an expression, preferably nice, for the probability that W=w. Note that by the defining formula for conditional probability, we have\Pr(W=w)=\Pr(Y=w\mid ...

1

1) $\max(x_1, x_1 + x_2) \le t$ if either $x_1 \le t$ and $x_2 \le 0$, or $x_1 + x_2 \le t$ and $x_2 \ge 0$. It may help to sketch this in the $x_1-x_2$ plane. Thus if $(X_1, X_2)$ has joint density $f(x_1,x_2)$, $$P(\max(X_1, X_1 + X_2) \le t) = \int_{-\infty}^t dx_1 \int_{-\infty}^{t-x_1} dx_2 f(x_1,x_2)$$ If $X_1$ and $X_2$ are iid with density $f$ and ...

0

Let me show a method for 1-d case first: $dX = aX dN$ By Ito's formula ( In this special case, it is straightforward, I guess we don't even need Ito's formula) $ln(X(t)) = ln(X(0)) + \sum_{s\in [0,t]}(ln(X_s)-ln(X_{s-}))$ Note that, whenever there is a jump at $s$, $ln(X_s)-ln(X_{s-})=ln(1+a)$, so we have $ln(X(t)) = ln(X(0)) + N(t)ln(1+a)$ $X(t) = X(0) ... 0 Comment: In case it helps, the ECDF plot from a simulation of 10,000 points in R should be reasonably close to the exact CDF. m = 10^4; x = rnorm(m); y = rexp(m); z = x+y plot.ecdf(z, pch=".") abline(v=0, col="green3") 1 If$X,Y$are independent, then$dP_{XY}=dP_XdP_Y$, and by Tonelli's theorem (Fubini's theorem for non-negative functions),$E(f(X,Y))=\int_{\mathbb{R}^2} f(x,y)dP_{XY}=\int_{\mathbb{R}}\int_{\mathbb{R}} f(x,y)dP_YdP_X=\int_{\mathbb{R}} E(f(x,Y))dP_X=\int_0^\infty E(f(x,Y))g(x)dx$The conclusion doesn't hold for general$X,Y$.For example$X=Y$,$f(x,y)=xy$, ... 2 Partial solution here.$Y$is a quadratic form. In particular, let$\mathbf{1} \in \mathbb{R}^n$be a vector of ones, and define $$P_\mathbf{1} = \mathbf{1}\left(\mathbf{1}^{T}\mathbf{1}\right)^{-1}\mathbf{1}^{T} = \mathbf{1}\left(\dfrac{1}{n}\right)\mathbf{1}^{T}\text{.}$$ Let $$\mathbf{X}=\begin{bmatrix} X_1 \\ X_2 \\ \vdots\\ X_n \end{bmatrix}\text{.}$$ ... 2${\bf X}=(X_1,\dots, X_n)^\prime$has a multivariate normal distribution with$\mu_{\bf X}=\mu {\bf 1}$and$\Sigma_{\bf X}=\sigma^2 I$. Here${\bf 1}$is the column vector of all$1$s, while$I$is the$n\times n$identity matrix. Let${\bf e}_1=(1,0,0,\dots,0)^\prime $, and let$A$be the matrix of an orthogonal transformation that takes the vector$\bf ...

Top 50 recent answers are included