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0

I doubt $Z$ has a simple distribution. But it might be possible to describe its moments. For example if you have a random variable $X_i$ with a Poisson distribution with mean $\lambda_i$ then its variance is also $\lambda_i$ and its second moment is $\lambda_i+\lambda_i^2$. So if you have a random variable $Y_i$ which is $X_i$ with probability $p$ and $0$ ...


0

This seems to be an old question, but I find it interesting hence a treatment: We want to show that $\displaystyle X_n=\frac{\sum x_i}{s_n} \sim N(0,1)$ for an appropriate choice of $\displaystyle s_n \rightarrow \infty $. To do so recall the Lindeberg's condition that $\displaystyle lim_{n \rightarrow \infty} \sum \int_{|z|\geq s_n ...


0

The mgf of $X^2$ can most easily be computed using the density: $$ M_{X^2}(t) = E[\exp(tX^2)] = \dfrac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty e^{-x^2/2 + t x^2}\; dx$$ The mgf of $X^2 + Y^2$ is then the square of this, and the mgf of $(X^2 + Y^2)/2$ is $$ M_{(X^2 + Y^2)/2}(t) = E[\exp(t(X^2+Y^2)/2)] = M_{X^2 + Y^2}(t/2)$$ Hopefully you may recognize the ...


1

You have to go through each case: $ \begin{cases} 1 & a+b<C \\ \frac{C^2}{2 a b} & (a\geq b\lor a\geq C)\land (a<b\lor b\geq C)\land C>0 \\ \frac{C-b}{a} & a+b=C \\ -\frac{a-2 C}{2 b} & a<b\land a<C\land b\geq C \\ -\frac{b-2 C}{2 a} & a>b\land b<C\land a\geq C \\ -\frac{a^2-2 a C+(b-C)^2}{2 a b} & ...


4

I would recommend a series of books, specifications, libraries and CAS programs. Books A Course in Number Theory and Cryptography, Neal Koblitz (very dense, but an amazing book) An Introduction to Mathematical Cryptography, Jeffrey Hoffstein, Jill Pipher, J.H. Silverman (very readable and excellent book, which is more up-to-date) An Introduction to ...


1

See "Testing statistical hypotheses" by E.L. Lehmann (196), Chap. 6 regarding regularity conditions. If you don't have access to that, here is a link to them. See p. 6-12 of the pdf. To summarize, you need to demonstrate a bit more than just consistency of the mode. Specifically, you need to show: $\lim_{n \to\infty} E(X_n)=\lim_{n \to\infty} \int xf_n(x) ...


0

Well, the CLT used in practice is about getting approximations to distributions. But to get a limit result, we need simplifications. While we are in practice interested in the distribution of an average for large, but finite, $n$, without letting $n \rightarrow \infty$ we cannot say much about the distribution, it can really be anything. So the limit is ...


0

I found one proof: First a guide to notation: $AB=A\cap B$ $A+B=A\cup B$ $\overline{A}=A^c$ Now my proof: $d(A,C)=d(A\overline{C})+d(C\overline{A})$ $d(B,D)=d(B\overline{D})+d(D\overline{B})$ ...


2

Let there be $n$ independent Bernoulli trials $X_1, X_2, \ldots, X_n$, each of which has probability $p$ of being correct, and $1-p$ of being incorrect; i.e., $\Pr[X_i = 1] = p$ for each $i = 1, 2, \ldots, n$. Let $Y = \sum X_i$ be the corresponding binomial random variable that counts the number of correct trials. If the majority rule applies to the ...


1

$$ \begin{eqnarray*} P\left(X_2> x\right)&{}={}&P\left(W_3> x,\ W_4> x\right)\newline &{}={}&\mathbb{E}\left[{\bf{1}}_{\left\{W_3> x,\ W_4> x\right\}}\right]\newline &{}={}&\mathbb{E}\left[{\bf{1}}_{\left\{W_3> x\right\}}{\bf{1}}_{\left\{W_4> x\right\}}\right]\newline ...


0

Here is a general approach. Know about your population size, how many total people will fit your demographic? Confidence interval: This determines how much higher or lower than the population mean you are willing to let your sample mean fall. Confidence level : How confident do you want to be that the true population parameter falls within your confidence ...


1

Answer is 1/9.for example you take a 5 digit number then there are 9*10*10*10*10 ways then for successive number to have same digit there are 1*10*10*10*10 ways.probability is 1/9 by dividing


1

Expected value of a quotient is not necessarily the quotient of the expected values.


0

It turned out that Georgii's notation is quite intuitive. He doesn't use $\Omega$ to underline the difference between the actual sample set and the observations that staticticians use. $\Omega$. This is the 'real world', the place where observable things happen (e.g. a dice is thrown and the upper side contains six nice, black dots). One easily sees that ...


0

HINT: Without loss of generality assume that $\mu_1\le\mu_2$. Then $$\mu=x_1\mu_1+x_2\mu_2=x_1\mu_1+x_2\big(\mu_1+(\mu_2-\mu_1)\big)=(x_1+x_2)\mu_1+\ldots$$ Finish this off, and it gives you half of what you want, and a similar idea applied to $\mu_2$ gets you the other hald. Then try to apply the basic idea to the general case.


1

Let's get you started. Suppose without loss of generality that $\mu_1 \leq \mu_2$. Then we have that $$ \mu = x_1\mu_1 + x_2\mu_2 \leq (x_1 + x_2)\mu_2 = \mu_2.$$ Each other subcomponent of the proof looks extremely similar to this line.


3

Here is a piecewise formula for $f_z$. $$ f_z(z) = \begin{cases} z+1 & -1\leq z\leq 0 \\ 1-z & 0\leq z\leq 1 \\ 0 & \text{else}. \end{cases} $$ Computing a formula for $F_z$ is just a matter of integrating. $$ F_z(z) = \int_{-1}^z f_z(z) \, dz = \begin{cases} 0 & z\leq -1 \\ \frac{z^2}{2}+z+\frac{1}{2} & -1<z\leq 0 \\ ...


1

Counterxample: Let $X$ be distributed as a mixture of two variables $Y,Z$ with some known densities and mixing paramenter $\theta \in [0,1]$. Then $p_X(x_i) = \theta \, p_Y(x_i)+(1-\theta)\, p_Z(x_i) = \theta \,g(x_i)+h(x_i)$ and $$L(\theta)=\prod_{i=1}^n (\theta \,g(x_i)+h(x_i))$$ This is a polynomial of degree $n$ in $\theta$, so it can have several ...


0

for polynomial regression you are trying to create a model for your data with a simple form of a sum of linear terms imagine you have $n$ data points indexed from ${1,..i,..n}$ $y_i = a_0 + a_1x_i + a_2x_i^2 + ...+ \epsilon_i$ it is easier to see these equations in matrix form $\vec{y} = X\vec{a} + \vec{\epsilon}$ here $\vec{y},\vec{a},\vec{\epsilon}$ ...


2

No, $P(A \cap B) \neq P(A)P(B)$ For example, if $A$ is the event "when rolling a dice, it comes out 6" and $B$ is the event "it comes out 5" you can see that if you know that B happened (ie a 5 has been rolled) then you know that A has not happened In this case $P(A|B) = 0 \neq P(A) P(B)$ If B does not give you any information about A, then we say that A ...


-3

if x denotes beta distribution of 1st kind with parameters @ and 1 then -logx will follow gamma distribution with parameters @ and 1


1

It is because you have $6$ numbers as $2^x$ in the range $[0,100]$. For the same range you have $50$ even numbers. Since the probability of every number to happen is equal, you have $1/6$ for powers and $1/50$ for the even number. But I think your set must be restricted to $[0,100]$, else these probabilities will change.


0

In high dimensions, $X^T$ would be rank deficient and have a non-trivial kernel, ensuring the existence of $\Delta W \in null(X^T)-\{0\}$. Now plug this $\Delta W$ in Michael's illustration, and we're done.


1

A direct method would rely on induction: Let $Y \sim \operatorname{NegBinomial}(r,p)$ with $$\Pr[Y = y] = \binom{r+y-1}{y} p^r (1-p)^y, \quad y = 0, 1, 2, \ldots,$$ so $Y$ counts the random number of "failures" before the $r^{\rm th}$ "success" is observed, where the probability of "success" is $p$. Let $X \sim \operatorname{Geometric}(p)$ with $$\Pr[X = ...


0

try computing the MGF of ∑Xi . the MGF of geometric Xi is p[(1-q(e^t))^(-1)] . where q=1=p. and t is a constant chosen such that the quantity converges. now MGF of ∑Xi will be the product such n quantities. (as Xi's are independent. ) and , consequently you will get the MGF to be p^n[(1-q(e^t))^(-n)]. which is MGF of negative binomial distribution. by ...


0

Here are some suggestions: First calculate the histogram of your data. Then discretize your density function. After that use a discrete version of some distance. It can even be the Euclidean distance, KL- divergence, or Hellinger distance. There are many of them. Another way is to consider kernel density estimation (KDE). After than you can use the ...


0

How about taking the Kullback-Leibler divergence between the empirical distribution (of the sample), and the target distribution?


0

I never heart about it. Expectation of something is expectation of something. Even for a very common example, expectation of a Gaussian random variable, there is no specific name. Most common notation is howeverr $\mu$. In case we talk about $\mu$, without the definition, it doesnt mean automatically the mean of Gaussian distributed random value. But for the ...


1

Following and improving on Julia's line of thought, here's the function to calculate the number of lists of length $n$ at Kendall-Tau distance $d$ from some permutation: public static int getCount(int n, int d) { if (d < 0) return 0; if (n < 1) return 0; if (d == 0) return 1; if (d == 1) return n-1; if (d > n * (n-1) / 2) return 0; ...


0

An "engineering" explanation: say you have a sampling grid in 2D using $2$ samples per dimension, you will have a $2\times 2$ grid with $4$ samples. $4^{1/2}=2$ Now you do the "equivalent" for 3D, you will have $2\times 2\times 2$ grid with $8$ samples with the same sampling density $8^{1/3}=2$. A 3D grid of $3\times 3\times3=27$ samples would have a ...


0

Presumably the sample mean would not be an efficient statistic in the case of a Cauchy distribution, which has infinite mean. See http://en.wikipedia.org/wiki/Cauchy_distribution


0

With educational research, unless you have a very well-calibrated test, the quantitative differences (i.e., magnitudes), are not overly informative. However, a useful measure is the fraction of students that "didn't fail". Define your "success variable" ($S$) as follows: For each student $i$, let their pre-test score be $I_i$, and postest score be $P_i$. ...


1

Notation: $\langle x,y \rangle$ is the inner product between $x$ and $y$. It is sometimes called the dot product, denoted by $x \cdot y$. In any case we have $\langle x,y \rangle = \sum_{i=1}^n x_i y_i$. You want an approximate solution to $Ax=b$. The trick is to choose it such that $Ax-b$ is orthogonal to the span of $A$. In other words you want $\langle ...


0

Can you clarify exactly what you do not understand? $X^T$ is the transpose of $X$. $X^{-1}$ is the inverse matrix of $X$. Does this help?


0

$$F(x) = \int_{-\infty}^x f(y) dy.$$ Note that $f$ needs to be appropriately defined on the whole line; for example, for an exponential variable, $f$ is zero for $x<0$.


0

We have that $(X,Y)$ is uniformly distributed over $S$, where $$S=\{(x,y)\in\mathbb R^2 : 0<y<1, 0<x<5y\}. $$ Let $R\subset S$ be the region $$\{(x,y)\in S: x + y > 1\}.$$ Then to find the probability $\mathbb P\{(X,Y)\in R\}$, we would integrate the density function over this region. But since the density function is constant, this boils down ...


1

If $P(X=Y)=1$ then for any measurable $A$ we have:$$P(X\in A\wedge X\neq Y)\leq P(X\neq Y)=0$$ so that $P(X\in A\wedge X\neq Y)=0$ and consequently: $$P(X\in A)=P(X\in A\wedge X=Y)+P(X\in A\wedge X\neq Y)=P(X\in A\wedge X=Y)$$ Likewise we also find:$$P(Y\in A)=P(Y\in A\wedge X=Y)$$ This can be applied to find: $$P(X\in A)=P(X\in A\wedge X=Y)=P(Y\in ...


0

You're almost there. You know that $X=Y$ on a set $A$ with $P(A)=1$. Now you just need to show that for any $B$ in your sigma algebra, $P(X\in B)=P(Y\in B)$. Since $B$ is somewhat arbitrary, it could contain elements not in $A$. So first prove $P(X\in B)=P(X\in B\cap A)$. Now conclude the result.


0

As non-statician I dislike percentages and my $p$ stand for your p%. Maybe something like: $$X:=US$$ where $U$ and $S$ are independent random variables. This with $P(U=1)=p$ and $P(U=0)=1-p$ and with $S$ having PDF $f$. $U$ somehow states whether there is a shock or not. $S$ somehow "measures" a shock (and is probably meant to be positive). $X$ is ...


1

Here's one way of viewing it. We want to write $$ \begin{bmatrix} y_1 \\ \vdots \\ y_n \end{bmatrix} = \hat\alpha\begin{bmatrix} 1 \\ \vdots \\ 1 \end{bmatrix} + \hat\beta \begin{bmatrix} x_1 \\ \vdots \\ x_n \end{bmatrix} + \begin{bmatrix} \hat\varepsilon_1 \\ \vdots \\ \varepsilon_n \end{bmatrix} $$ and choose the values of $\hat\alpha$ and ...


1

The random variables $\bar{X}-\bar{Y}\,,\,\, X_i-\bar{X}$ and $Y_i-\bar{Y}$ are jointly (and marginally) normal, since they are linear combinations of independent, Normally distributed, random variables. Consequently, a well known property of Multivariate Gaussians guarantees that $$ \begin{eqnarray*} \mathbb{C}ov\left(\bar{X}-\bar{Y}\,,\, ...


1

Since Y is a positive random variable, only outcomes with w=x+y>x=z are possible. Because all positive values are possibilities for Y and X, the pdf will be posive for all z=x>0 and all w=x+y>x=z. It follows that pdf is positive at {(z, w): z>0, w>z}. No need to compute the pdf to determine where pdf is positive in above reasoning. In fact, the above ...


1

For the geometric version, you drew a picture, which will serve us well for the integral. I would rather call the random variables $X$ and $Y$, but draw the $y_1$-axis where the $x$-axis usually is, and draw the $y_2$ axis where the $y$-axis usually is. We are integrating over a certain region, which is actually a trapezoid. Let us integrate first with ...


0

As stated in the comments, since the CF of $\frac{X_n}{n}$ is $\frac{\lambda}{\lambda-nit}$, the CF of $U_n$ is given by: $$\frac{\lambda^n}{\prod_{k=1}^{n}(\lambda-kit)},\tag{1}$$ so its MGF is given by: $$\frac{\lambda^n}{\prod_{k=1}^{n}\left(\lambda+kt\right)},\tag{2}$$ and its distribution is given by the inverse Fourier transform of $(1)$ that can be ...


1

You can certainly say that someone in group A is $\frac {77}{52}$ times more likely to like singing than someone in group B.


1

First, I am assuming the project is about or heavily involves mathematics. To come up with a good project you should find a problem that needs to be addressed (problem statement), decide on how your project will address that problem (purpose statement), and be able to explain why what you are doing matters. Here are some slides that go over this process. ...


1

Lets look at it like you need a total of $60$ percentage points. Right now you have $$.05(60)+.05(53)+.3(47)+.1(66)=26.35\text{ percentage points}$$ So, you need to now achieve $$.05(\text{A3})+.05(\text{A4})+.3(\text{T2})+.10(\text{P2})$$ such that the sum of that expression is $33.65$. If you are looking for what single grade could you score on them ...


1

You don't need any calculus: it's a matter of simple algebra. If $x>0$ and $y>0$, then $0<x<x+y$, and since $x+y>0$, we can divide by $x+y$ to get $0<\frac{x}{x+y}<1$. (And you don't mean the minimum and maximum of a bivariate "equation." You mean bounds on the values of a bivariate function.)


0

$$\newcommand{\cov}{\text{Cov}} \begin{align*}\cov(X,Y)&=E\bigg[(X-E(X))(Y-E(Y))\bigg]\\ &=E\bigg[XY\bigg]-E\bigg[X\bigg]E\bigg[Y\bigg]\\ \cov(a-bX+Y,X)&=E\bigg[\bigg(a-bX+cY-E(a-bX+cY)\bigg)\bigg(X-E(X)\bigg)\bigg]\\ &=E\bigg[\bigg(-bX+cY-E(-bX+cY)\bigg)\bigg(X-E(X)\bigg)\bigg]\\ ...


0

I like this answer taken from http://mathforum.org/library/drmath/view/74065.html : " It may be clearer to you if you think of probability as the fraction of the time that something will happen. If event A happens 1/2 of the time, and event B happens 1/3 of the time, and events A and B are independent, then event B will happen 1/3 of the times that event A ...



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