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0

The definite discussion on congruential random number generation is in Knuth's "Seminumerical Algorithms" (volume 2 of his "The Art of Computer Programming", Addison-Wesley). Lots of other types of generators have been proposed since, but as far as I know none has undergone the same theoretical and practical scrutiny. The last edition of the book recommends ...


1

I think this is related to monotone likelihood ratios $-$ https://en.wikipedia.org/wiki/Monotone_likelihood_ratio $-$ which does imply FOSD.


1

Let $P_n(m)$ be the probability of reaching $m^{th}$ position at time $n$. The proper way to attack this sort problem is not to compute any sum explicitly. Instead, one should try to prove following sum over $m$ is independent of $n$. $$\mathcal{S}_n \stackrel{def}{=}\sum_{m=-\infty}^\infty P_n(m)$$ The main reason is with minimal modification, the ...


0

You can model a random walk using the binomial theorem. Model a single step as $px + \frac{1-p}{x}$, and $n$ steps as $\left(px + \frac{1-p}{x}\right)^n$. The coefficient of $x^i$ is the probability that, after $n$ steps, the person is at position $i$ (assuming $l=1$). From the binomial theorem, $$\begin{align} \left(px + \frac{1-p}{x}\right)^n & = ...


0

Going through the pdf is feasible, but it is easier to note that $$E(\ln X)=\int_0^\infty \ln x\cdot\left(\frac{1}{\theta} e^{-x/\theta}\right)\,dx.$$ Let $x=u\theta$. Then $\ln(x)=\ln\theta+\ln u$ and our integral becomes $$\int_0^\infty (\ln\theta) e^{-u}\,du +\int_0^\infty (\ln x)e^{-x}\,dx.$$ The first integral is $\ln\theta$ and the second, by the ...


0

Given $x\in(0,1)$, the wanted average length is given by: $$ \int_{0}^{x}(1-z)\,dz + \int_{x}^{1}z\,dz = \frac{1}{2}+x-x^2 $$ and since the derivative of the RHS is $1-2x$, $x=\frac{1}{2}$ is a stationary point.


1

Strictly speaking it appears you have been given $\mathsf P(X_f=1\mid S_f=S)=g(S)$. The marginal probability of $X_f$ should not be a function of firm size (by definition). $$\mathsf P(X_f=1) = \int_\Bbb R f(s)\cdot\mathsf P(X_f=1\mid S_f=s)\operatorname d s $$ You want to find $f(S\mid X_f=1) = \dfrac{f(S)\cdot\mathsf P(X_f=1\mid S_f=S)}{\int_{\Bbb ...


1

This is possible. Note first that the constants in front of the exponential functions in $f_A$ and $f_B$ don't need to bother us, because we have $r$. This means we can concentrate on what happens in the exponents. The exponent of $f_A$ can be written as: $$-\frac{1}{2} x^T \Sigma_A^{-1}x + \mu_A \Sigma_A^{-1} x - \frac{1}{2}\mu_A \Sigma_A^{-1} \mu_A$$ The ...


7

The vector $\vec X = [X_1,\dots,X_N]^T$ has a rotationally invariant distribution. That is, if $A$ is any orthogonal matrix, then the distribution of $\vec X$ and $A \vec X$ are the same. Hence by letting $A$ be an orthogonal matrix that takes $[1,\dots,1]^T$ to $[\sqrt N,0,\dots,0]^T$, your problem is the same as computing $$ \Pr(\sqrt N X_1 \mid \sum ...


1

The semicolon separates variables from parameters. Think of it as there being a set of pdfs, each of which is defined in terms of a variable $x$, but which differ from each other in their parameter $\theta$. By setting the parameter you are basically picking out one specific pdf of all these possible pdfs. A function does not necessarily need to use its ...


1

$$ P(p) = \sum_{N_1 \geq N_2+2} \binom{4}{N_1}\binom{0}{N_2} p^{N_1+N_2}q^{4 - N_1 -N_2 }$$ $\binom{0}{x}$ is 1 if $x=0$ and 0 otherwise so the only $N_2$ That can affect the sum is $N_2 = 0$. similarly the other binomial coefficient is zero if $4=n_1<N_1$ so we don't have an infinite sum, $$P(p) = \sum_{ k= 2}^{4} \binom{4}{k} p^k q^{4-k} $$ Typing on ...


0

We deal with your question about why we multiply by $2$. We are interested in finding the cumulative distribution function $F(y)$ of $Y$. The only interesting part is when $-1\lt y\lt 1$, because $F(y)=1$ if $y\ge 1$ and $F(y)=0$ when $y\le -1$. It is geometrically perhaps a little easier to find $G(y)=\Pr(Y\gt y)$. Then the probability that $Y\le y$ is ...


-2

I think it has to do with the fact that continuity is implied by differentiability and integrability, and since the Dirac-Delta function is differentiable and integrable, it is continuous.


2

Of course, no deterministic algorithm can generate truly random numbers. The goal is to find an algorithm that gives output values that 'behave as if' they are independent and identically distributed according to some distribution--usually $Unif(0, 1).$ By 'behave as if' one means that the pseudo-random numbers pass a large battery of benchmark tests for ...


0

I don't know how much rigor is required, but intuitively in order to get $Y$, the the negative values of $X$ are not changed and the positive ones all become 0. I have noted that $X$ can be simulated as twice a BETA(2,2) random variable minus 1. (Showing that might be worthwhile in its own right.) You say you have a plot of the CDF of $Y$, but something is ...


0

As $X_i$ follows $N(0,\sigma^2)$, $\Sigma_{i=0}^nX_i$ follows $N(0,n\sigma^2)$. So $\bar{X}=\dfrac{\Sigma_{i=0}^nX_i}{n}$ follows $N(0,\dfrac{\sigma^2}{\sqrt{n}}).$ Thus $\dfrac{\sqrt{n}\bar{X}}{\sigma^2}$ follows $N(0,1)$. and $\dfrac{(n-1)s}{\sigma^2}$ follows $\chi_{n-1}^2$ and is independent of $\bar{X}$. so ...


1

By symmetry, we have $\mathbb P(X>0) = \frac12$ (you can verify this by computing $\int_0^1 f(x)\ \mathsf dx$). Since $Y=0\iff X>0$, it follows that $\mathbb P(Y=0)=\frac12$. For $-1<y<0$, we have \begin{align} \mathbb P(Y\leqslant y) &= \mathbb P(X\leqslant y)\\ &= \int_y^{-1} f(x)\ \mathsf dx\\ &= \int_y^{-1} ...


0

Consider that integrating against $\delta(y-f(x))$ will pick out $P_X(x)$ for all $x$ such that $f(x)=y$ for the given (fixed) $y$. It follows that (in the countable case) $$\int P_X(x)\delta(y-f(x))\, dx=\sum_{f(x)=y} P_X(x)$$ i.e. $P_Y(y)$ is given by the sum of the contributions $P_X(x)$ for all the $x$ that map to $y$.


0

Note that $\bar{X}$ may not be a standard normal. Your $X$ r.v. representing $t$-distribution $$X = \frac{Z}{\sqrt{U/n}}$$ where U = variance. IF you have a copy of Mood Graybill and Boes (or download from Colorado University), refer to page 249 and 250. Let Z be a standard normal and U be a chi-square distribution with n degrees of freedom, then we can ...


2

For continuous random variables, the probability of a tie is immeasurably small.   So we can say: $$V_1<\max\{V_2, V_3\} \iff V_2=\max\{V_1,V_2, V_3\} \cup V_3=\max\{V_1, V_2, V_3\}\quad\text{a.s.}$$ Since the events in this union are disjoint (almost surely), then they partition the conditioned space: $$\begin{align}\mathsf E[\max\{V_2,V_3\}\mid ...


0

An explicit derivation of Robert Israel's claim is fairly straightforward by induction: $$\begin{align*} \operatorname{E}[X^{2m}] &= \int_{x=-\infty}^\infty x^{2m} \frac{1}{\sqrt{2\pi}} e^{-x^2/2} \, dx \end{align*}$$ and with the choice $$u = x^{2m-1}, \quad du = (2m-1) x^{2(m-1)} \, dx, \quad dv = x e^{-x^2/2} \, dx, \quad v = -e^{-x^2/2},$$ we ...


0

Extended comment. Here is a simulation experiment (in R), that suggests $Var(X^3) = 15.$ x = rnorm(10^6); y = x^3 var(y); mean(abs(y) < 40) ## 15.02577; 0.999341 summary(y) ## Min. 1st Qu. Median Mean 3rd Qu. Max. ## -1.020e+02 -3.077e-01 1.118e-08 -2.277e-04 3.101e-01 1.409e+02 hist(y[abs(y)< 40], ...


3

Let $A$ and $B$ denote the number of requests in $(0,T]$ from clients $a$ and $b$ relatively. Then, $A$ and $B$ are Poisson random variables with parameters $\lambda_aT$ and $\lambda_bT$ respectively, and you have declared them to be independent random variables. Thus, $C = A+B$, the total number of requests received during $(0,T]$ is a Poisson random ...


1

For the case of elliptically contoured distributions (of which the Gaussian is a special case), the distribution of the norm of Mv is available in the literature (See for example 1 and 2) M. Rangaswamy, D.D. Weiner, and A. Ozturk, "Non Gaussian Random Vector Identification Using Spherically Invariant Random Processes," Aerospace and Electronic Systems, ...


1

Hint: the probability that $X+Y\equiv n\pmod{10}$ is $$ \sum_{k=0}^9\Pr[X=k]\Pr[Y\equiv n-k\pmod{10}] $$ Now consider the sum $$ \sum_{k=0}^9\Pr[Y\equiv n-k\pmod{10}] $$


0

Let me just comment on your approach, and note the difference between what you do and the $\mod 10$ problem. Remember what the ranges of your random variables $X$ and $Y$ are: The $X$ can attain the values in the set $\{0,1,\ldots,9\}$, and $Y$ can attain the values $0,1,\ldots$ In your sum, $\text{Pr}(X=k)$ only has a non-zero value for $k\geq 0$, and ...


0

Your attempt almost solves the question, but indeed there is a reason for the mod $10$. Your solution does not take into account that $z$ can only be a number between $0$ and $9$. Here is my solution: Let $z \in \lbrace 0,..., 9 \rbrace$. Then we have due to independence: $$ Pr(Z= z ) = Pr(X+Y = z) = \sum_{i=0}^{9}{Pr(X = i) Pr(Y = z- i \textrm{ mod } 10)} ...


1

The distribution of $1/F_{n,m}$ is the same as the distribution of $F_{m,n}$ (with the roles of $n$ and $m$ interchanged). The reason why that happens isn't hard to see. In this case $m$ is the same as $n$, so the distribution of $F_{7,7}$ is the same as the distribution of its reciprocal. Thus if you find $$ \Pr(F_{7,7}\le x_2) = 0.05 $$ then you can ...


0

Is this what you are looking for? $$\displaystyle \frac{\bar{X}\,-\,\mu}{S} \sqrt{n}= \displaystyle \frac{\frac{\bar{X}\,-\,\mu}{\sigma}}{\frac{S}{\sigma}} \sqrt{n}= \displaystyle \frac{\frac{\bar{X}\,-\,\mu}{\sigma}\sqrt{n}}{\sqrt{\frac{S^2}{\sigma^2}}} = \displaystyle ...


1

I will assume that $X$, $Y$ are the numbers of trials until the first success, where the probability of success on any trial is $p$. Then $\Pr(X=j)=\Pr(Y=j)=(1-p)^{j-1}p$. Let $A$ be the event $X=i$ and let $B$ be the event $X+Y=n$. Then by the definition of conditional probability, we have $$\Pr(A|B)=\frac{\Pr(A\cap B)}{\Pr(B)}.$$ The probability of ...


2

Hint: Write $F_X(x) = E[I\{X \le x\}]$ and use Fubini's theorem.


0

The asymptotic value is $\frac{1}{2}$. Let $X_i$ be an i.i.d. sequence of standard Gaussian random variables. Then $Y_k = \sum \limits_{i = 1}^k X_i^2 \sim \chi^2_k$. Now note that $$P(Y_k < k) = P(Y_k \le k) = P\left(k^{-1/2}(Y_k - k) \le 0\right) \xrightarrow{k \to \infty} \frac{1}{2}$$ because $k^{-1/2}(Y_k - k) \xrightarrow{d} \mathcal{N}(0, 2)$ by ...


1

Well, let's review a few helpful definitions that will clarify how to calculate the Kullback-Leibler divergence here. By definition the summation of the parameters of the mutlinomial distribution is 1; i.e., $$\sum_{m=1}^k\theta_m=1$$, where $\theta_m$ is the probability of the $m^{th}$ outcome occuring. The probability mass function (PMF) of the ...


0

If there are N raffle tickets sold to N different people, then each ticket holder has the same probability of winning the prize. The distribution is uniform.


0

Here's a way to go : We know that the distribution is a gaussian vector and since we know the covariance matrix, we can compute the density function $f(x_1,x_2,..,x_k)$. Using this we can compute the cumulative distribution function for $ \max_{i=1,2,..,k} |X_i|$. $$ P(\max_{i=1,2,..,k} |X_i| < x )= \int_{]-x,x[^{k}} f(x_1,x_2,..,x_k)dx_1dx_2...dx_k ...


1

If $0\le u\le 1$ then $$ \Pr(U\le u) = \Pr\left( \frac X \theta \le u\right) = \Pr(X\le \theta u) = \int_0^{\theta u} \frac{2(\theta - x)}{\theta^2} \, dx = 1 - (1-u)^2. $$ Hence $$ f_U(u) = 2(1-u) \text{ if }0\le u\le 1 $$ and $=0$ if $u>1$ or $u<0$.


0

No. It's much simpler than that. If $X$ has PDF $f_X(x;\theta)$, then $U = X/\theta$ is just a scale transformation of $X$, so its PDF is directly given by $$f_U(u;\theta) = \theta f_X(\theta u;\theta).$$ The idea is to calculate this and simplify it accordingly.


1

You are essentially asking for the distribution of the maximum order statistic of $n$ iid exponential random variables. That is to say, what is $$\Pr\left[\max_{1\le k \le n} X_k \le x\right]?$$ Recall that $$\Pr[X_k \le x] = 1 - e^{-x}, \quad k = 1, 2, \ldots.$$ Since the $X_k$s are iid, it easily follows that $$\Pr\left[\max_{1\le k \le n} X_k \le ...


2

It is absolutely not correct that at least one should be infinite. It is true that for every real number $a$, infinitely many of the random variables will be more than $a$, but every one of them will be finite. But the fact that for every $a>0$, infinitely many of them are $>a$ is enough to imply that the limit is infinite. I'm going to follow up on ...


1

In analog-to-digital conversion a quantization error occurs. This error is either due to rounding or truncation. When the original signal is much larger than one least significant bit (LSB), the quantization error is not significantly correlated with the signal, and has an approximately uniform distribution. The RMS error therefore follows from the variance ...


2

Consider a "spinner": an object like an unmagnetized compass needle that can pivots freely around an axis, and is stable pointing in any direction. You give it a spin and see where it comes to rest, measuring the resulting angle (divided by $2\pi$) as a number from $0$ to $1$.


1

We need to make some assumption about the process of distributing points. I will assume that points are distributed one at a time, that any person is just as likely to get the point at stage $k$, and that the decisions for the various points are independent. Under these assumptions, if random variable $X$ is the number of points Alicia gets, then $X$ has ...


0

That depends on the probability you are considering (I know, typical useless mathematician-answer). Usually one has $f: V \to [0, +\infty)$ or at least $f: V \to [0,+\infty]$. However, for this to work we would need our probability measure to be continuous. If you are more on the physical side of life, then you probably want to also consider the delta ...


1

Imagine the selected elements as black beads and the others as white. Add one more black bead (so there are $b+1$ black beads and $a+1$ beads in all) and close the row of beads into a circle. The circle is divided into $b+1$ segments, each consisting of some (possibly zero) white beads followed by one black bead. By symmetry, each of these segments has ...


1

Start out with the $b$ selected elements in a row, and then insert the remaining $a-b$ elements one by one, with equal probability for every possible insertion point (including the two ends). This gives you all possible arrangements of the selected elements with equal probability (since each arrangement is obtained in $(a-b)!$ different ways). In the ...


1

The expected value is a sort of average value over a large # of trials. Suppose an elementary event has a Pr = $\dfrac{1}{100}$, i.e. 1 in 100 On an average, how many trials would be needed to get the event once ? 100, isn't it, as the term itself suggests ? If an elementary event has a probability p, $E[x] = \dfrac{1}{p}$ More complex computations are ...


1

Another approach is to use the Law of Iterated Expectation, and partitioning on the event of succeeding on the first (next) try. Let $\bar X$ be the expectation of the count of tries until you encounter a six.   If you succeed on the first try, the (conditional) expectation is $1$; this condition has probability $\tfrac 1 6$ of occurring.   If you ...


1

This is because you're computing a probability, not an expected value. There is no need to introduce two variables. Let $X$ the number of times you roll the die before you get a $6$. You have $P(X=k)=p(1-p)^k$, for $k\geq 0$, with $p=\frac{1}{6}$ (by the way this is a geometric distribution). If you want the expected value, then you have to compute: ...


1

The Moment Generating Function is rather nice. $$ \mathbb E[e^tX] = \mathbb E\left[ \mathbb E[e^tX \mid K]\right] = \mathbb E \left[ ((1+e^t)/2)^K \right] = \dfrac{1}{n} \sum_{k=1}^n \left(\dfrac{1+e^t}{2}\right)^k = \dfrac{((1+e^t)/2)^{n+1} - (1+e^t)/2}{n(e^t-1)/2}$$


1

ETA: Did's comment reminds me that the lower limit on the sum should be no lower than $1$. Let $K$ be the number of coins flipped: \begin{align} P(X = x) & = \sum_{k=x}^n P(K=k) P(X=x \mid K=k) \\ & = \sum_{k=x}^n \frac{1}{n} \binom{k}{x} \frac{1}{2^k} \\ & = \frac{1}{n} \sum_{k=x}^n \binom{k}{x} \frac{1}{2^k} \end{align} ...



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