New answers tagged

0

I'm guessing that what you have in mind is a Poisson process. Two essential facts about the process described in your question are the following. Let $X$ be the number of sites falling within a particular region whose area is $a$. Then $\operatorname{E}(X) = a.$ If two regions do not intersect each other, then the numbers of sites falling within them are ...


0

If you graph $f(x,y) = c(2-|x|-|y|)$ over the region $|x|+|y| \le 2$ (i.e. the square with vertices $(2,0), (0,2), (-2,0), (0,-2)$), then you get a pyramid shape. The base of the pyramid is the square with vertices $(2,0), (0,2), (-2,0), (0,-2)$. The height of the pyramid is $f(0,0)$. The formula for the volume of a pyramid is $$\text{Volume} = \dfrac{1}{3}...


0

To move forward with this (correct) approach, you need to see (better yet, construct): 1) The graph of the function $|x|+|y| = \mbox{const.}$ in the $xy$-plane. (E.g., see the illustration for the case $||x||_{1}$ in https://en.wikipedia.org/wiki/Unit_sphere 2) The graph of the function $f(x,y) = \mbox{const} - (|x|+|y|)$. This graph is the "top ...


0

Each random variable $X_i$ is either $1$ or $0$. It is $1$ if the $i$-th trial is a success and $0$ otherwise. Thus, $$X_1+X_2+\ldots+X_n$$ is simply the total number of trials that were successes, i.e., the total number of successes. Suppose, for example, that $n=5$, and we have successes on trials $1,2$, and $4$; then $X_1=X_2=X_4=1$, $X_3=X_5=0$, and $...


2

The Law of Total Probability, is: $~\mathsf P(A)=\mathsf P(A, B)+ \mathsf P(A, B^\complement)$ In this case $A\equiv \{2X<c\}, B\equiv \{4Y\leqslant c\}, B^\complement\equiv\{4Y>c\}$ so we have: $$\mathsf P(2X<c) ~=~ \mathsf P(2X<c, 4Y\leqslant c)~+~\mathsf P(2X<c, 4Y>c)$$ Then it is just a matter of algebraic rearrangement: $$\mathsf P(...


0

The win outcome is defined when there is a majority. So at least $N=3$ trials would be needed. In some situations there is no majority. The possible win situations can be seen in the following table \begin{array}{ll} \text{Number of trials ($N$)} & \text{Number of required outcomes to win} \\ 3 & \text{3 or 2} ...


2

Not in general. $$\Pr\left(X_{1}+X_{2}>x\right)=\int\Pr\left(X_{1}+X_{2}>x\mid X_{2}=y\right)dF\left(y\right)=\int\Pr\left(X_{1}>x-y\right)dF\left(y\right)$$ where independence was used. Since the rv's are nonnegative this equals: $$\int_{0}^{x}\Pr\left(X_{1}>x-y\right)dF\left(y\right)+\int_{x}^{\infty}\Pr\left(X_{1}>x-y\right)dF\left(y\...


1

This is going to be a "spherical cow shooting milk in all directions in a frictionless vacuum" approach. The basic idea is fine, but variations may make this model inaccurate. You can make every point where a person can choose to change his path a node in a directed graph. Each street at an intersection would be a node, an exit would be a node, an onramp ...


3

$X_i-\mu$ is a standard normal random variable, hence $$ \mathbb{E}[|X_i-\mu|]=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}|x|e^{-\frac{x^2}{2}}\;dx=\sqrt{\frac{2}{\pi}}\int_0^{\infty}xe^{-\frac{x^2}{2}}\;dx=\sqrt{\frac{2}{\pi}}$$ Therefore the sum is equal to $$\frac{\sqrt{\pi}}{2n}\cdot n\sqrt{\frac{2}{\pi}}=\frac{1}{\sqrt{2}}$$ for all $n$. Even without ...


1

Fifth problem: $2$ lines are not in use if and only if $4$ are in use. $3$ lines are not in use if and only if $3$ are in use. $4$ lines are not in use if and only if $2$ are in use. So we need to add the probabilities that $4$, $3$, $2$ are in use.


1

If there are $k$ lines not in use, then it is the same as $6-k$ lines being in use. If between $2$ and $4$ lines, inclusive, are not in use, then that is the same as between $2$ and $4$ lines, inclusive, being in use. ($2$ lines not being in use is the same as $4$ lines being in use, etc.) If at least $4$ lines are not in use, then this is the same as at ...


3

Your last line should have been $$m_{X+Y}(t) = \int_{-\infty}^\infty e^{ts} f_{X+Y}(s) \mathop{ds}= \int_{-\infty}^\infty e^{ts} \int_{-\infty}^\infty f_X(s-y) f_Y(y) \mathop{dy} \mathop{ds}.$$ Making the change of variables $s=x+y$ gives you the answer.


-1

Exact Probability: $\sum_{x > 120}^{900} {900 \choose x} \cdot \left(\frac{1}{9}\right)^x \cdot \left(\frac{8}{9}\right)^{900-x}$. Approximate Value: 0.0166779, using the Mathematica code: Sum[Binomial[900, x]*(1/9)^x *(8/9)^(900 - x), {x, 121, 900}] // N If on the other hand, as some of the comments appear to suggest, you desire an approximate method ...


0

By the symmetry $(u,x) \mapsto (-u,-x)$, the integral over the region $$ |x| < 2 \qquad \text{and} \qquad u < x $$ is the same as the integral over the region $$ |x| < 2 \qquad \text{and} \qquad u > x $$ and adding them together gives $$ 2 \int_{-2}^2 \int_{-\infty}^{x} f(u) \, \mathrm{d}u\, \mathrm{d}x = \int_{-2}^2 \int_{-\infty}^{+\infty} ...


1

This is ugly, but may show how, even having no clue what one is looking for, it's still possible to try something (and possibly arrive at the answer, albeit not in the most elegant way). The important thing being: if you have no clever idea, do the non-clever systematic cumbersome thing. It might work. I don't like numbers, so instead of $2$ I used $a$. The ...


0

Switching the order of integration, we have $$\int_{-2}^2\int_{-\infty}^xf(u)dudx=\int_{-2}^2\int_u^2 f(u)dxdu+\int_{-\infty}^{-2}\int_{-2}^2f(u)dxdu$$ $$=\int_{-2}^2(2-u)f(u)du+4\int_{-\infty}^{-2}f(u)du$$ $$=2\int_{-2}^2f(u)du+4\int_{-\infty}^{-2}f(u)du$$ $$=4\int_{-2}^0f(u)du+4\int_{-\infty}^{-2}f(u)du$$ $$=4\int_{-\infty}^0f(u)du$$ $$=4\times\frac{1}{2}$...


3

tl;dr: look at the characteristic function, invoke independence of the summands to get a product of characteristic functions, compute the closed-form expression, and finally squint hard at the resulting expression to recognize a known characteristic function. (This is a good method whenever you have a r.v. defined as the sum of independent (possibly ...


4

If $n$ is the number of tosses, then $T=n-H$, so $H-T=2H-n$. Since $n$ is large, $H$ has a close to normal distribution, mean $n/2$, and variance $(n)(1/2)(1/2)$. So $2H-n$ has close to normal distribution, mean $0$ and variance $n$. Let $W$ be a normal with mean $0$ and variance $n$. Let us find the mean of $|W|$. This is $$\int_{-\infty}^\infty \frac{|w|}...


1

$VarX_n=EX_n^2=n^{2s},VarS_n=\sum_{j=1}^n j^{2s}$ so chebychev's inequality gives $P(|S_n/n|>\epsilon)\le \sum_{j=1}^nj^{2s}/(n^2\epsilon^2)\sim \int_1^nx^{2s}dx/(n^2\epsilon^2) \sim n^{2s+1}/(n^2\epsilon^2)\to 0$ iff $2s-1<0, s<1/2$. I assumed $s\neq-1/2$ for the integral approximation, but that case is analogous.


2

We start with the Pearson differential equation: $$f'(x)=-\frac{\left(a_1 x+a_0\right) }{b_2 x^2+b_1 x+b_0}f(x),$$ Define $g(x)=b_2 x^2+b_1 x+b_0$ (using a trick in Diaconis et al.(1991)). Consider (f g)'(x), also written $(f(x) g(x))'=f'(x) g(x) +f(x) g'(x)$. We have $$(f g)'(x)=(-a_0 + b_1 - a_1 x + 2 b_2 x) f(x)$$ We assume that the distribution has ...


0

If we normalize the pdf to unit area after the fact, $$E(x)=\int_{-\infty}^\infty xp(x)dx=\frac{\int_0^\theta x^2dx}{\int_0^\theta xdx}=\frac{2\theta^3}{3\theta^2}=\frac23\theta.$$ This is compatible with the location of the centroid of the triangle. Then $$E(x^2)=\int_{-\infty}^\infty x^2p(x)dx=\frac{\int_0^\theta x^3dx}{\int_0^\theta xdx}=\frac{2\theta^...


0

How precise the sketch is? Because, generally the algebraic form of a straight line passing through the origin is $f_X(x)=ax$. But lets assume that $a=1$ then, $$ \int_{0}^{\theta}x = 1 \to\theta=\sqrt{2}. $$ Now you can proceed with the calculation. Othewise, if $a\neq0$, then $a = 2/\theta^2$ using the same argument of $\int axdx=1$. In this case the ...


0

The pdf cannot be $f(x)=x$ on $[0,\theta]$ (and zero elsewhere) because then it would not integrate to one (unless $\theta=2$). The pdf is shown as a straight line, so let $f(x)=kx$. We need to determine $k$. The area under the pdf is the area of a triangle with height $k\theta$ and base $\theta$, i.e the area is $$\frac{k\theta^2}{2}.$$ We need to choose $k$...


0

For the first inequality, the concavity is not needed. Also the condition $\Psi(a)>0$ is irrelevant, since the inequality reads $0\le 0$ otherwise. Yet let us prove it for $\Psi(a)>0$. Define the probability density on $[0,a]$: $f(x) = x\psi(x)/K$, where $K = \int_0^a x\psi(x)dx$. For a random variable $\xi$ with this density, $\mathsf E[\xi^2]\ge \...


2

You could use a summation, as noted in the comments. Or we can notice that $P(X> x) = P(X\geq x+1)$. In words, this means we have fail the first $x$ trails; success has to happen on the $x+1$ trial, or the $x+2$ trial, etc. The probability of failing on any particular trial is $$1-P(\text{double ones}) = 1-\frac 16\frac 16 =\frac{35}{36}.$$ Hence $$P(X\...


0

If the two functions are probability density functions, and the measurements are independent, then by definition: PDF_GC(measurement_g, measurement_c) = PDF_G(measurement_g)*PDF_C(measurement_c) You state that you have not performed any normalisation, so it is plausible that your functions are actually not probability density functions.   Normalise! ...


0

$f(\lambda|x) \propto \lambda^x \cdot \exp(-2\lambda)= \lambda^{(x+1)-1} \cdot \exp(-2\lambda)$. This is a Gamma density with parameters $x+1,-2$, ie Gamma($\lambda|x+1,-2$)


0

$P_n(A)$ is a random variable and it does have its own probability distribution, if you know (or impose) a distribution on the sample. That is precisely what you did when you set $\Pr[X \in A] = p$. But the reason why $P_n(A)$ is empirical is because, a priori, there is no assumption on the sampling distribution: the calculation of the statistic $P_n(A)$ ...


1

Since $\bar{F}(x) = 1- \mathbb{P}(X \leq x) = \mathbb{P}(X>x)$, we have $$x \bar{F}(x) = \int_{\{X>x\}} x \, d\mathbb{P} \leq \int_{\{X>x\}} X \, d\mathbb{P}$$ for any $x>0$. If $\mathbb{E}(X)<\infty$, then we can let $x \to \infty$ using the dominated convergence theorem to conclude $$\lim_{x \to \infty} x \bar{F}(x)=0.$$ If $\mathbb{E}(X)...


1

There are a couple of slightly different ways you can think about this. Firstly, it's helpful to rewrite the $f_X(x)$ term in the denominator of your conditional density as $\int_{\mathbb{R}} f_{X,Y}(x, y) dy$ - that is, the integral over the joint density of the event you're conditioning on. By a similar token, the conditional density for $X$ conditional ...


1

$Y=X^2+Z$ means $Z=Y-X^2$, so we let $z(x,y)=y-x^2$ After affirming that there is a bijection between $(X,Z)\leftrightarrow(X,Y)$ , (because why?), we then can directly apply the Jacobian change of variables transformation: $$\begin{align}f_{X,Y}(x, y) =&~ f_{X,Z}(x, z(x,y))~\big/\Big\lvert\dfrac{\partial\big(x,z(x,y)\big)}{\partial\big(x,y\big)}\Big\...


2

We can calculate the expected number of 1's using the linearity of expectation. Let $X_1$ denote the random variable that is $1$ if the 4-sided die rolls a 1, and $0$ otherwise. Similarly, let $X_2$ denote the indicator variable for the 6-sided die, and $X_3,\ldots, X_7$ for the other dice. We have $X_1+X_2+\cdots +X_7$ indicating the total number of 1's ...


0

The true Fabius function is no-where analytic. This implies a lack of an analytical function describing it. The linked page a has a discussion concerning a function that approaches the Fabius function when taken to infinity. (though it gets reasonably close at around n=20): Recursive Integration over Piecewise Polynomials: Closed form?


2

Based on your assumption, $X$ and $Y$ are jointly normal. Let $\rho$ be the correlation, that is, \begin{align*} \rho = \frac{E\big((X-E(X))(Y-E(Y)) \big)}{\sqrt{E\left((X-E(X))^2 \right)}\sqrt{E\left((Y-E(Y))^2 \right)}}. \end{align*} Then, $Y-E(Y)$ and \begin{align*} Z:=\frac{X-E(X)}{\sqrt{E\left((X-E(X))^2 \right)}} -\rho \frac{Y-E(Y)}{\sqrt{E\left((Y-E(Y)...


2

These questions make little sense from the classical (Kolmogorov's axiomatic) probability theory point of view. For example, let $\xi$ be $U[0,1]$, and $\eta = \xi$ if $\xi$ is irrational and $1/\pi$ otherwise. Then these variables have the same distribution, yet $\eta$ never takes rational values. So the answers to both questions would be $0$. However, ...


0

it is correct, in your example, $P(X=4)=0$, and your calculations are correct, moreover since for any $x\geq 3, F(x)=1$, you can state $P(X>3)=1-P(X\leq 3)=1-F(3)=0$, so $X$ does not take values $>3$


0

The time of each event occurring is uniformly distributed from 0 to 10 seconds. What is the probability that the events will occur within 2seconds of each other? The probability will be the integral of the pdf over where the condition occurs within the support. That is you want $0\leq x\leq 10$ and $0\leq y\leq 10$ and also $(x-2)\leq y\leq (x+2)$ . $$\...


1

Outline: Draw the $10\times 10$ square with corners $(0,0)$, $(10,0)$, $(10,10)$, $(0,10)$. Draw the two lines $y=x+2$ and $y=x-2$. We want the probability of falling in the part $K$ of the square that is between these two lines. This is the area of $K$ divided by $100$. It is easier to find first the area of the part of the square that is not in $K$. ...


0

For simplicity, let $T_1$ be the shelflife of molecule 1 after it enters the system, at time $0$, and $T_2$ be that of molecule 2 after it enters the system, at time $\Delta t$.   Then the event of molecule 1 decaying before molecule 2 is represented by: $\{T_1< T_2+\Delta t\}$ Noting that the supports of the exponential distribution asserts that $...


1

$$(h-1)h'=2 $$ is a separable differential equation: it can be simply re-written as $$ \frac{d}{dt}\left(\frac{h^2}{2}-h\right) = 2 $$ from which: $$ (h-1)^2=(4t+C) $$ and $$ h = 1\pm\sqrt{C+4t} $$ readily follow. Have also a look at the generating function for Catalan numbers.


1

The easy way for this problem, as is the case for many pdf problems, is to work with CDF's instead. Here, since $f(x) = \frac{x-1}{2}$ on $(1,3)$, $$ F(x) = \left\{ \array{0 & x\leq1\\ \frac{(x-1)^2}{4} & 1< x < 3 \\ 1 & x\geq 3 }\right. $$ And this needs to match the CDF of the uniform distribution ojn $(0,1)$ $$F(y) = y$$ So $$y= \frac{...


1

A variant of this problem comes up a lot when you're trying to simulate something using a Monte Carlo code. Here's how I would obtain $u(x)$, and it doesn't require the solution of any differential equations: The CDF $F(x)$ is given by $$ F(x) = \int_1^x f(w) dw = \frac{(x-1)^2}{4} $$ $F(x)$ is uniformly distributed between 0 and 1 (my wording here may ...


1

The two word answer is "polar coordinates". In more detail, let $f:S^{n-1}\to\Bbb R$ be a continuous function. Then $$ \eqalign{ \Bbb E[f(X)] &=\int_{\Bbb R^n}f(x_1/z,\ldots,x_n/z)(2\pi)^{-n/2}e^{-z^2/2}\,dx_1\cdots dx_n\cr &=(2\pi)^{-n/2}\int_0^\infty\left[\int_{S^{n-1}} f(u)\,\sigma_{n-1}(du)\right]e^{-r^2/2}r^{n-1}\,dr\cr &=c_n\int_{S^{n-1}} ...


0

The question was answered on MathOverflow by Iosif Pinelis: The paper you refer to says (in line 1 of Sect. 2.2) that $\mathcal{F}$ is the unit ball of a reproducing kernel Hilbert space (not the entire RKHS). So, $|f(a)|=|\langle f, \phi(a)\rangle|\le\|f\| \|\phi(a)\|\le1\times\sqrt{\langle\phi(a),\phi(a)\rangle}=\sqrt{k(a,a)}\le\sqrt K$ for all $f\in\...


1

HINT If $X$ is a random variable with pdf $f(x)$ then $$ \mathbb{E}[g(X)] = \int_{-\infty}^\infty g(x) f(x) dx. $$ Can you apply this to your problem? What is $g(x)$? Can you integrate?


4

If $F$ is a CDF and it must be shown that $F$ is the CDF corresponding with the uniform distribution on $[0,1]$ then it is enough to prove that $F(y)=y$ for every $y\in(0,1)$. This because the values of $F$ on elements not in $(0,1)$ are determined by the following rules for a CDF: If $y\geq1$ then $1\geq F(y)\geq F(z)=z$ for every $z\in(0,1)$ and ...


3

According to Earliest Known Uses of Some of the Words of Mathematics, the term "characteristic function" in this sense (actually its French equivalent "fonction caractéristique") was first used by Poincaré in 1912 (except that with his notation, that function was what we now call the moment generating function). To me as an analyst, the terminology never ...


2

I will try to start from the simplest case possible and then build up to your situation, in order to hopefully develop some intuition for the notion of convolution. Convolution essentially generalizes the process of calculating the coefficients of the product of two polynomials. See for example here: Multiplying polynomial coefficients. This also comes up ...


1

Your criterion for positive definiteness is incorrect. A positive determinant is necessary but not sufficient for positive definiteness (Sylvester's criterion for positive definiteness of a symmetric matrix requires all leading principal minors to be positive.) At the point $(0,0)$ the Hessian matrix for the bivariate standard normal is $ H:=\left(\begin{...


0

Lemma 1 (Levy's Equivalence Theorem): Let $X_1,X_2,\cdots,X_n,\cdots$ be independent random variables and $S_n=\sum_{i=1}^{n}{X_i}$. Then $$S_n\overset{a.s.}{\rightarrow}S\Leftrightarrow S_n\overset{\mathbb{P}}{\rightarrow}S\Leftrightarrow S_n\overset{d}{\rightarrow}S \ .$$ Lemma 2 : Let $\left( \Omega,\mathcal{F},\mu \right)$ be a measure space and ...



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