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I shall try to give you a short answer (in yellow boxes) as well as a commentary on on how to get it. The first step is just to find the probability of absorption given a starting point. We do this with recurrence relationships. Let $a_n$ be the probability of being absorbed given that the starting point is at $n$. The reason I did this is because this ...


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We have to consider the moment generating function. $$\begin{align} M_Z(t) & = \mathbb E(e^{t(X+Y)}) = \\[0ex] & = \mathbb E(e^{tX}e^{tY}) \iff X indep. Y= \\[0ex] & = M_X(t)M_Y(t)= \\[2ex] & = (1-p_0+pe^t)^n (1-p_1+pe^t)^m \\ \end{align}$$ Which is the same result we get via convolution. The MGF allows to notice what are the parameters of ...


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No. Sorry. We look at the probability generating functions. $$\begin{align} \Pi_X(s) & = \mathsf E(s^X) \\[0ex] & = (sp_0-(1-p_0))^n \\[2ex] \Pi_Y(s) & = (sp_1-(1-p_1))^m \\[2ex] \Pi_{X+Y}(s) & = \mathsf E(s^{X+Y}) \\[0ex] & = \Pi_X(s)\Pi_Y(s) \\[0ex] & = (sp_0-(1-p_0))^n(sp_1-(1-p_1))^m \end{align}$$ Now, if $p=p_0=p_1$ then we ...


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You are being confused by the use of lower-case variables which don't correspond to the capital random-variables in the way you're anticipating.   The use of $F_Y(x)$ and $f_X(y)$, while legitimate, is counterintuitive.   They are 'dummy variables' and as such any token can be used; but why make a choice that is liable to confuse?   It's ...


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Note that the density is undefined if $y \le 0$, so the support of $G_1, G_2$ is on $(0, \infty)$. The distribution of each of $G_1, G_2$ is Gamma with parameters $\alpha = 1/2$, $\beta = 1$; thus their sum is also Gamma but with parameters $\alpha^* = 2\alpha = 1$ and $\beta = 1$; i.e., exponential with rate $1$. This is a consequence of the fact that the ...


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This is (an attempt at) a partial answer. I think that you should be able to decouple the server selection aspect from the composite load aspect, since the servers are identical with exponentially distributed service time $\mu$. That is, one can analyze the number in system (irrespective of distribution across the servers) as an ordinary M/M/$c$ system ...


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y in the integral is a dummy variable. It could have been anything.


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Sort of. It looks like your math is correct, you're taking the total probability ($M_0$) and dividing each portion by that probability, normalizing it. I'm assuming you have to find the expectation and variance for other parts of your problem, and those look fine as well. The only thing I can see to look out for is that you say "accepting" and "rejecting" ...


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Here is the answer obtained with Maple. with(Statistics): V := RandomVariable(Exponential(lambda)): X := RandomVariable(Normal(a, sigma)): M := V*X/k: PDF(M, t); $$\left| k \right| \int_{0}^{\infty }\!1/2\,{\frac {\sqrt {2}}{\sqrt { \pi }\sigma\,\lambda\,{\it \_t}}{{\rm e}^{-1/2\,{\frac {1}{{\sigma}^{2 }} \left( {\frac {tk}{{\it \_t}}}-a \right) ...


2

I don't see anything wrong in your calulations. Integration yields $$\begin{align*} \mathbb{P}(W_1 \leq r M) &= \frac{1}{r} \int_{\mathbb{R}} \int_{y<x} \frac{2(2x/r-y)}{\sqrt{2\pi}} \exp \left(- \frac{1}{2} (2x/r-y)^2 \right) \, dy \, dx \\ &= \frac{1}{r} \frac{2}{\sqrt{2\pi}} \int_{\mathbb{R}} \exp \left(- \frac{(2x/r-x)^2}{2} \right) \, dx \\ ...


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I'm not sure if you can actually expect too much of an improvement here. In some sense the binomial distribution is actually a (near) worst case for Berry-Esseen. By Chebyshev's inequality, a positive proportion of the mass of the binomial distribution is located between $np-\sqrt{npq}$ and $np+\sqrt{npq}$. Together with the pigeonhole principle, this ...


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Hints Here is an outline of one approach. What if the substring contains exactly 34 1's, can you find it's likelihood? What would it be for 34 0's? For either 34 1's, or 34 0's? What if it were 35, how would this change your answer? Can you generalize 34 and 35 above to $n$? Sum from $n = 34$ to $n = 20000$?


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I'm guessing you're just starting probability. You are correct --- and some of the comments here might be confusing. So here is the full story in what I hope is the notation of your text. Sorry if this is redundant. The probability assignment to the $n+1$ points in the sample space is as follows: $P(\{a\}) = .3,$ and $P(\{b_i\}) = .7/n,$ for $i = 1,\dots, ...


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One way to look at it is through conditional probabilities. If you have $P(A)=0.3$, $P(B)=0.7$ and $P(b_i \mid B)=1/n$ then by law of total probability \begin{align} P(b_i) &= P(b_i \mid B)P(B) + P(b_i \mid A)P(A)\\ &= \frac{1}{n}0.7 + 0 \end{align} (using that $A$ and $B$ are disjoint) and \begin{align} P(a) &= P(a \mid B)P(B) + P(a \mid ...


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What you have done is correct. You are finding the probability that out of 7 weeks, there are exactly 3 weeks in which she wins exactly 2 prizes. Isn't that what you want? This does of course include the possibility that in some of these weeks she does win a different number of prizes, or indeed none at all.


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$$\left(p+q\right)^{n}=\sum_{k=0}^{n}\binom{n}{k}p^{k}q^{n-k}$$ Taking the derivative w.r.t. $p$ on both sides we find: $$n\left(p+q\right)^{n-1}=\sum_{k=0}^{n}\binom{n}{k}kp^{k-1}q^{n-k}$$ Multiplying with $p$ on both sides we find: $$np\left(p+q\right)^{n-1}=\sum_{k=0}^{n}\binom{n}{k}kp^{k}q^{n-k}$$ If $p+q=1$ then this equality can be written as: ...


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Let $X$ be a random variable with a CFD $F(x)$ having a PDF $f(x)$. Consider the transformation: $$ u=\Phi^{-1}(F(x)),$$ where $\Phi^{-1}$ is the inverse of the standard normal CDF. For the PDF $f_U(u)$ of this function $u=\Phi^{-1}(F(x))$ one has (with $\varphi(u)$ the PDF of the standard normal distribution) under some regularity conditions on $f(x)$: $$ ...


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I had the same question with Jalaj, and I found your answer is very right. But could you please give me one or some references about this uniform distribution of the vectors? I always appreciate your help causse I need it for writing a paper. Thanks in advance, Viet Hoa


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The antiderivative is easy. Note that ,since you want the integral to equal $1$ : $$\begin{align*} \int_{0}^{\infty}xe^{-cx^2}\, {\rm d}x =1 &\Leftrightarrow \frac{1}{2c}\int_{0}^{\infty} 2c xe^{-cx^2}\, {\rm d}x =1 \\ &\Leftrightarrow \frac{1}{2c}\left [ -e^{-cx^2} \right ]_0^{\infty}=1\\ &\Leftrightarrow 2c = 1\\ &\Leftrightarrow ...


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for $f(x)=e^{-cx^2}$ you have $f'(x)=-2cxf(x)$, so integration by parts gives $\int xf(x)dx = \frac{1}{-2c}\int f'(x)dx=\frac{1}{-2c}f(x)+C$. now the definite integral. to have a probability-mass function, $c$ has to be positive (otherwise the limit in infinity is intinite) and then $1=\int xf(x)dx = \frac{1}{-2c}f(x)|_0^{\infty}=\frac{1}{-2c}(0-1)$, so ...


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Use "fitdist" as given here. You should provide a hypothesis.


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You have to find the Taylor series expansion of your function $$G(z) = \frac{1}{2} \frac{3+z}{3-z}=\frac{1}{2} \frac{6+z-3}{3-z}= \frac{3}{3-z}-\frac12=\frac{1}{1-z/3}-\frac12$$ Now, $$\frac{1}{1-t}=\sum_{n=0}^{\infty} t^n$$ Hence $$G(z)=-\frac12+\sum_{n=0}^\infty \frac{1}{3^n}z^n$$ $$G(z)=\frac12+\sum_{n=1}^\infty \frac{1}{3^n}z^n$$ The probability ...


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It seems that the departure process is indeed a Poisson process. See for example the first line of the paper: Newell, G. F. "The $M/G/\infty$ Queue." SIAM Journal on Applied Mathematics 14.1 (1966): 86-88.


2

Obviously $W=\frac{X}{X+Y+Z}$ belongs to $[0,1]$. Moreover, for any $\alpha\in[0,1]$ we have: $$ \mathbb{P}[W\leq\alpha] = \mathbb{P}[X\leq \alpha(X+Y+Z)]=\mathbb{P}[X\leq\frac{\alpha}{1-\alpha}(Y+Z)]. $$ Can you compute the last probability? It is useful to recall that the ratio distribution between two uniformly distributed random variables over $[0,1]$ ...


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Answer to this question can be found in this post in the CrossValidated stackexchange forum.


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Assuming $\epsilon$ is a vector and that "$\ge$" must be considered component-wise. $\mathbb{P}(X^n\ge\epsilon)=\mathbb{P}\left(\bigcap_{i=1}^m\{X^n_i\ge \epsilon_i\}\right)\le\min_{i=1}^m\mathbb{P}(X^n_i\ge \epsilon_i)$, so the if part is true. It actually holds the stronger $$\exists j\ \liminf_n \mathbb{P}(X^n_j\ge\epsilon_j)=0\Longrightarrow ...


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


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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, ...


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It always helps to sketch the support of the joint distribution; i.e., the region of $(X,Y)$ such that the density is positive. In your case, you can see that this region must be in the upper half plane ($y \ge 0$), and for a given value of $y$, $x$ must be between $-y$ and $y$; so this region is triangular and is bounded to the right by $y = x$ and to the ...


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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 ...


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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 ...


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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.


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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

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 ...


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 ...


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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 ...


0

Okay, we need the true bounds of integration, which are a little different from what you mentioned. Here is a Wolfram Alpha plot of the bounds: If you integrate this with $x$ as the free variable, you end up with a two-stage integration, since the $y(x)$ function changes at $x=1$. As a result, I would recommend integrating with $y$ as the free variable. ...


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$$ \hat k >x \text{ if and only if }[X_1>x\ \&\ \cdots\ \&\ X_n>x] $$ 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) = ...


0

This is not a mixture of Gaussians, whose cumulative distribution function is a convex combination of cumulative distribution functions of Gaussian random variables having different parameters. And similarly in terms of pdfs. The distance squared is a Chi-Squared random variable with degrees of freedom equal to the dimension. Look it up. The distance in ...


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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 )


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: $$\lim_{n\to\infty}\mathbb P(X_n\leqslant t)=\mathbb P(X\leqslant t). $$


0

If $Z$ is independent of $X$ and independent of $Y$, then $(X,Z )\sim (Y,Z)$. Indeed, if $s,t\in\mathbf R$, then \begin{align} \mathbb P(X\leqslant s,Z\leqslant t)&=\mathbb P(X\leqslant s)\mathbb P(Z\leqslant t)\quad\mbox{ since $X$ and $Y$ are independent}\\ &= \mathbb P(Y\leqslant s)\mathbb P(Z\leqslant t)\quad \mbox{ becauseĀ }X\sim Y \\ ...


0

Have a look at the following question and its answer, from which I've restated the answer. gauss circle problem on a hexagonal lattice Basically the answer is known for any regular crystallographic group, it seems. For your case, the number for a circle of radius $r$ and sector angle $\theta$ would seem to be $$ N(r;\theta,x,x_0)= \frac{\theta ...


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

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 ...


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 ...


0

Since the random variable $X$ has not been precisely defined, we will supply a (speculative) definition. The game consists of rolling the two dice repeatedly. We stop when we roll a sum divisible by $4$ or when the total number of rolls reaches $6$. Let $X$ be the number of rolls. It is indeed true that the probability of getting a sum divisible by $4$ on ...


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 ...


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Hint: Find equation of line that connects $(2,0)$ and $(4,0.3)$ Find equation of line that connects $(4,0.3)$ and $(6,0.7)$ Find equation of line that connects $(6,0.7)$ and $(12,1)$


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No, It is incorrect, you can test by inserting values of x, for example F(4)= 1. Just integrate the function piecewise, your formula does only work for a uniform distribution (not piecewise), you would have to adapt it as: $F(X)=c\,(x-a)$ where c is the partial probability e.g. 3/20. $$ \int \frac{3}{20}\,dx = \frac{3\,x}{20} $$ and shift it accordingly: $$ ...



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