New answers tagged

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 variance $1$; ...


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 \begin{equation} 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} with $P(\{ω:X_n(ω)=1\})=P(\{ω:X_n(ω)=1\})=1/2$. 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} ...


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 $$μ_1=m_1 \implies α=m_1$$ This 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 ...



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