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5

First, $$P(X_1+X_2 \leq a, X_2+X_3 \leq b, X_3 +X_4 > c)=E[P(X_1+X_2 \leq a, X_2+X_3 \leq b, X_3 +X_4 > c|X_2,X_3)].$$ As far as the conditional probability: $$P(X_1+X_2 \leq a, X_2+X_3 \leq b, X_3 +X_4 > c|X_2=x_2,X_3=x_3)=$$ $$=P(X_1 \leq a-x_2, x_2+x_3 \leq b, X_4 > c-x_3|X_2=x_2,X_3=x_3)=$$ $$ =\begin{cases} 0, & \text{ if } & ...


3

Recall that if $X\sim\mathrm{Bin}(n,p)$, then $\mathbb E[X] = np$ and $\mathrm{Var}(X)=np(1-p)$. Given $\mathbb E[X] = 4$ and $\mathrm{Var}(X) = 3$, we have $np = 4$ and $np(1-p)=3$. Hence $n=16$, $p=\frac14$. So the distribution of $X$ is given by $$\mathbb P(X=k) = \binom {16}k \left(\frac14\right)^k\left(\frac34\right)^{16-k}, k=0,1,\ldots,16.$$ The ...


3

Hint: you can rewrite $$ \operatorname{Var}[X^2] = \mathbb{E}[X^4] - \mathbb{E}[X^2]^2 $$ and use the known expressions of the (raw) moments of a Poisson-distributed random variable (or recompute them — you already can get $\mathbb{E}[X^2]$ from the variance and the expectation).


2

The moment generating function of a Poisson random variable $X$ with mean $\lambda$ is given by: $$ f(t) = \mathbb{E}[e^{tX}] = e^{\lambda(e^t-1)},\tag{1}$$ hence: $$\begin{eqnarray*}\text{Var}[X^2]=\mathbb{E}[X^4]-\mathbb{E}[X^2]^2 &=& f^{(4)}(0)-f''(0)^2\\&=&(\lambda +7 \lambda ^2+6 \lambda ^3+\lambda ...


2

It is also the Pearson distributions. In the original article the change of the probabilities (at $k+1$ wrt $k$) of hypergeometric distribution divided by the current probability is a rational function of the index $k$. Pearson noted that if the $k+1$ and k go close, then at the limit this is a derivative of the pdf devided by pdf and it is equal to the ...


2

Let $(\Omega,\mathcal A,P)$ be a probability space and let $\mathcal B$ denote the Borel $\sigma$-algebra on $\mathbb R$. Any random variable $X:\Omega\rightarrow\mathbb R$ induces a probability $P_X$ on measurable space $(\mathbb R,\mathcal B)$. This by: $$P_X(B)=P(\{X\in B\})$$ for $B\in\mathcal B$. This $P_X$ is the distribution of $X$ and is ...


2

At stationarity, one asks that $$X\stackrel{\text{law}}{=}X\cdot\mathbf 1_{X<\theta}+\delta,$$ where the random variables $\delta$ and $X$ in the RHS are independent. Using the PDF $f$ and $g$ of $X$ and $\delta$, this translates as the condition that, for every $x>0$, $$\ \qquad \qquad f(x)=P(X>\theta)\cdot g(x)+\int_0^{\min(\theta,x)} ...


2

Let $U$ be standard normal. Let $V=RU$, where $R$ is a Rademacher random variable that takes values $-1$ and $1$, each with probability $1/2$. Suppose $U$ and $R$ are independent. Let $X=\frac{U+V}{2}$ and $Y=\frac{U-V}{2}$. Then $X+Y$ and $X-Y$ are each normal. But $X$ is not normal, for it takes on value $0$ with probability $\frac{1}{2}$. Remark: ...


2

We can show and use the following Lemma. Let $(X_n)_{n\geqslant 1}$ be a sequence of random variables such that $X_n\to X$ in probability and the cumulative distribution function of $X$ is continuous. Then for each $t\in\mathbf R$, the following convergence holds: $$\lim_{n\to\infty}\mathbb P(X_n\leqslant t)=\mathbb P(X\leqslant t). $$


2

No, in general the covariance does not converge to $0$. Just consider $([-1,1],\mathcal{B}([-1,1]))$ endowed with the Lebesgue measure and $$X_n(\omega) := Y_n(\omega) := -n 1_{[-1/n,0)}(\omega) + n 1_{(0,1/n]}(\omega), \qquad \omega \in [-1,1].$$ Since $X_n \to X:=0$ almost surely and $Y_n \to Y := 0$ almost surely, we have in particular $X_n \to 0$ and ...


1

So in general I know that this does not hold, because your expression is exactly the myerson virtual value, and there is a long literature about suitable regularity conditions to make it increasing. If you assume log-concavity of your distribution for example then you are done. But this is too strong in that you can actually assume -.5 concavity. I refer you ...


1

Let $0 \leq t_1<\ldots<t_n$. Since $(B_t)_{t \geq 0}$ is a Gaussian process, we know that $(B_{t_1},\ldots,B_{t_n})$ is Gaussian. This implies in particular that $\sum_{j=1}^n B_{t_j}$ is Gaussian. Since a Gaussian random variable is uniquely characterized by its mean and variance, it remains to calculate those two. As $\mathbb{E}B_t=0$ for any $t ...


1

$$ \hat k >x \text{ if and only if }[X_1>x\ \&\ \cdots\ \&\ X_n>x] $$ and the probability of that is $\Big(\Pr(X_1>x)\Big)^n$. $$ \Pr(X_1>x) = \int_x^\infty \frac{\alpha k^\alpha}{u^{\alpha +1}} \, du = \left(\frac k x\right)^\alpha, $$ so $$ \Pr(\hat k>x) = \left(\frac k x\right)^{n\alpha}. $$ Thus $$ \Pr(x_1<\hat k < x_2) = ...


1

These questions originated from an assignment (due tomorrow) of UNSW - course code, MATH2901. Please put this on hold. Thanks.


1

For my comfort, I will change the notation. Let our two independent uniforms be $X$ and $Y$, and let $W=\frac{\max(X,Y)}{\min(X,Y)}$. We want to find the cdf $F_W(w)$ of $W$, that is, $\Pr(W\le w)$. This is $0$ if $w\le 1$. So from now on we can suppose that $w\gt 1$. We have $\Pr(W\le w)=\Pr\left(\frac{Y}{X}\le w|Y\ge X\right)$. The required conditional ...


1

The Berry–Esseen theorem gives an estimate for the normal approximation of the binomial distribution: $$\sup_{x\in\mathbb R} \left|P\left(\frac{B(n,p)-np}{\sqrt{np(1-p)}} \le x\right)-\Phi(x)\right| \le \frac{C\rho}{\sqrt{n}}$$ with $C < 0.4748$ and $\rho=\frac{p^2+q^2}{\sqrt{pq}}$.


1

A picture can help: \begin{eqnarray*} F_{X_3}(z) &=& P(X_2/X_1^2 \leq z) \\ &=& P(X_2 \leq zX_1^2). \end{eqnarray*} It should be obvious that $F_{X_3}(z)=0$ if $z\lt 0$. For $0\lt z\lt 1$, the first picture applies and for $z\gt 1$ the second picture applies. In both cases, because $X_1,X_2$ are uniform and independent, we need the area ...


1

The "universal probability space" in the sense of your second question is $([0,1],\mathcal{B}([0,1]),m)$ where $m$ is Lebesgue measure. The procedure for making the identification is the procedure that is commonly used for numerical sampling of a random variable given its CDF. Specifically, we uniformly choose $p \in [0,1]$, then choose the smallest $x$ such ...


1

It matters what your definition is of a probability measure but here is a start and it addresses the situation where $\Omega$ is a subset of the reals. For the first question. There is an equivalence between CDFs and probability measures on the space. As was suggested I will give that correspondance: $$\mu([-\infty, x]) = F(x)$$ Then $\mu$ can be ...


1

To be a lot simpler than the other answers, just use the fact that the sum of two Gaussian random variables is gaussian and the fact that the scalar multiple of a Gaussian random variable is also Gaussian. The first of these facts can be worked out by computing the convolution of two arbitrary Gaussian distributions. The second requires some thinking about ...


1

Yes. Write this in matrix form, so you have $\mathbf {AX} = \mathbf Z$ where $\mathbf Z$ is bivariate normal. Then $\mathbf X = \mathbf {A^{-1}Z}$. The RHS is a linear combination of Gaussians, therefore the LHS is Gaussian.


1

It seems your fundamental difficulty is with definitions of probability concepts, not so much with the process of integration. If $X \sim Unif(50, 150),$ then the density function $f_X(x)$ has three parts: (i) For $x < 50,\;f_X(x) = 0;\;$; (ii) for $50 \le x \le 150,\;f_X(x) = 1/100 = 0.01;$ and (iii) for $x > 150,\;f_X(x) = 0.$ If you want to ...


1

You can determine the sde that $X_t^2$ satisfies using Ito's lemma. Then you can find the pdf by solving the forward Kolmogorov equations (Fokker-Planck).


1

I've attached a picture of various beta densities from the wiki on the Beta Distribution. It really matters what your density looks like beyond the first two moments you specified, but in general, this will ensure the values are restricted to a bounded interval. If you aren't concerned with some density living outside your interval you might consider ...


1

A stochastic matrix is, by definition, any matrix for which each row, column or both, sums to $1$. Mathematics does not concern itself with what mathematical objects "represent". It only cares about what their properties are. And what you described is a stochastic matrix. In fact, every stochastic matrix will be a transition matrix for some Markov chain, ...


1

your vector has $P(n) = P(x \ge n)$ so $P(3) = P(x \in \{3,4,5,6\} ) $ So $P(x \in \{4,5\}) = P(4) - P(6)$ in General $P( a \le x \le b) = P(a)-P(b+1)$ so for an n sided dice you need to include $P(n+1)=0$ ( I see you have done that for your first vector )


1

Next step is to evaluate $E[Z|\Lambda]$. By the hint, given that $\Lambda=\lambda$ we know that $Z$ has a Poisson($\lambda$) distribution and therefore $E(Z|\Lambda=\lambda)=\lambda$ (since the mean of a Poisson distribution equals its parameter). Conclude $E(Z|\Lambda)=\Lambda$.


1

To find $c$: \begin{eqnarray*} 1 &=& \int_{y=0}^{1} \int_{x=y}^{2-y} cx\;dx\;dy \\ &=& c\int_{y=0}^{1} \left[x^2/2 \right]_{x=y}^{2-y} \;dy \\ &=& c\int_{y=0}^{1} \left(2-2y \right) \;dy \\ &=& c \left[2y-y^2 \right]_{y=0}^{1} \\ &=& c \left(2-1 \right) \\ \therefore\quad c &=& 1. \end{eqnarray*} For ...


1

The Fisher information is essentially the negative of the expectation of the Hessian matrix, i.e. the matrix of second derivatives, of the log-likelihood. In particular, you have $$l(\alpha,k)=\log\alpha + \alpha \log k - (\alpha + 1) \log x$$ from which you compute the second-order derivatives to create a $2 \times 2$ matrix, which you take the expectation ...



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