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0

Using the the multinomial theorem $$\frac{400!}{k!\ l!\ (400-k-l)!}(0.5)^k(0.2)^l(0.3)^{400-k-l}$$ is the probability of any combination. You want to sum all $(k,l)$ pairs such that $k+2l\leq 390$ This will be difficult to actually evaluate, you might want to use approximations.


0

Take some $Z$ following a normal distribution, and $X$ any other. Then set $Y=X-Z$. $X-Y$ follows a normal distribution, while $X$ doesn't.


2

Let $W$ (for example) be a random variable which takes on values $0$ and $1$, each with probability $\frac{1}{2}$. let $Z$ be standard normal, and suppose $W$ and $Z$ are independent. Let $X=W+Z$ and let $Y=W$. Then $X-Y$ is normally distributed, but neither $X$ nor $Y$ is normal. Remark: Note that if $X$ and $Y$ are normal and not independent, then $X-Y$ ...


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 } & ...


0

If the modulus sign means a normal mathematical modulus, then I think the key has to be wrong. (I don't see the reason why you need it, as square will take care of the sign anyway, unless we are working with complex variables. ) If we assume that h,b,k,c are independent variables in very general terms. $$ E[y^2]= E[th^2 + b^2+k^2+c^2 +2bk+2bc+2bh \sqrt t ...


1

One can use product probability space to model a Poisson process. Let $$\Omega = \prod_{k = 1}^\infty \mathbb{R}^+, \mathcal{F} = \sigma\left(\prod_{k=1}^\infty\mathcal{B}(\mathbb{R}^+)\right)\text{ and }\mu = \prod_{k=1}^\infty \mathrm{Poisson}(\mathbb{R}^+),$$ where $\mathcal{B}(\mathbb{R}^+)$ is the Borel $\sigma$-algebra on $\mathbb{R}^+$ and ...


0

$p_{U|Y}(k|j)$ is a conditional probability: $$ p_{U|Y}(k|j) = Pr(U=k|Y=j) = \frac{Pr(U=k \cap Y=j)}{Pr(Y=j)} $$ $p_{X|Y,U}(i|j,k)$ is also a conditional probability: $$ \frac{Pr(X=i \cap Y=j \cap U=k)}{Pr(Y=j\cap U=k)} $$ The product then becomes: $$ \begin{align} p_{U|Y}(k|j)p_{X|Y,U}(i|j,k) &= \frac{Pr(U=k \cap Y=j)}{Pr(Y=j)}\frac{Pr(X=i \cap Y=j \cap ...


1

$\begin{align} P(XY<1) & = \int_0^3 P(X<1/y \mid Y=y)f_Y(y)\operatorname d y \\ & = \int_0^3 \int_0^{\min\{2, 1/y\}} f_{X,Y}(x,y)\operatorname dx \operatorname d y & \tag{$\star$} \\ & = \frac 1 6\left( \int_0^{1/2} \int_0^{2} 1\operatorname dx \operatorname d y + \int_{1/2}^3 \int_0^{1/y} 1\operatorname dx \operatorname d y \right) ...


0

Draw a picture of the rectangle on which the joint density function lives. Draw the curve $xy=1$, that is, $y=\frac{1}{x}$. We want to integrate $\frac{1}{6}$ over the part of the rectangle that is below $y=\frac{1}{x}$. Note that $xy=1$ meets the upper boundary $y=3$ of the rectangle at $x=1/3$. So our probability is ...


1

See the figure: You must calculate the blue area and divide it by the full area ($2 \times 3 = 6$). You should find: $P[x\ y < 1] = \frac{1}{6} (1+\log (6))$.


1

Both averages are calculated correctly. They differ because there are two sampling models being employed -- in one approach we sample a class at random and ask what is the average; in the other approach we sample a student at random and ask what is the average class size the student sees. As for which number you would choose to represent the 'average' class ...


1

Let $N \in\{1,2, \cdots ,n\}$ denote the number of attempts. The conditional probability we are looking for is $$P(N=k|N\le n)=\frac{P(N=k\cap N\le n)}{P(N\le n)}.$$ First, $$P(N=k\cap N\le n)=\begin{cases}0,& \text{ if }& k>N\\ P(N=k)=(1-p)^{k-1}p,& \text{ otherwise. } \end{cases}$$ Second, $$P(N\le n)=\sum_{i=1}^n(1-p)^{i-1}p.$$ So, for ...


0

Use the Taylor series approximation $$ h(Y) - h(\mu) \approx h'(\mu)(Y - \mu), $$ where $\mu = E(Y)$. Then, we have $$ [h(Y) - h(\mu)]^{2} \approx [h'(\mu)]^{2}(Y - \mu)^{2}. $$ Hence, $$ Var(h(Y)) = E\{ [h(Y) - h(\mu)]^{2} \} \approx [h'(\mu)]^{2}E\{ (Y - \mu)^{2} \} = [h'(\mu)]^{2}Var(Y). $$ Now, since $h'(\mu) = \frac{2 - v}{2}\mu^{-v/2}$ and $Var(Y) = ...


0

A rotation matrix R is the same as an orthonormal basis that preserves orientation ($\det(R)=1$). Hence, to create a uniform distributed random rotation matrix, we need to pick three orthogonal random unit vectors, make sure that the orientation is correct and concatenate them into a matrix. First pick two random unit vectors $u$ and $v$. For numerical ...


0

To show the convergence to $0$ is not almost sure, note $$\{X_n = n+1\} \subset \{|S_n| \geq n/2\} \cup \{|S_{n-1}| \ \geq n/2\}.$$ The probabilities on the left have infinite sum. By Borel-Cantelli, it therefore occurs with probability $1$ that $|S_n|/n \geq 1/2$ infinitely often. For the convergence in probability I quote the following result from ...


1

Let $W=\min(|X-1|, 5-X)$. We find the cdf of $W$. The cdf of $Y$ can then be obtained easily by pushing everything forward by $1$. Apologies for the length. I wanted to put in lots of detail, and do not have time to tighten things up. Draw carefully the curves $y=|x-1|$ and $y=5-x$. Everything depends on that. Note that $y=|x-1$ is $y=1-x$ to the left of ...


1

Yes.   That the probabilities of those four data points sum to one indicates that they are most certainly all of the possible outcomes.   Then: $$\mathsf P\{X\leq 3\} = \mathsf P_{X,Y}(1,1)+\mathsf P_{X,Y}(2,2)+\mathsf P_{X,Y}(3,3)\qquad\color{red}{\checkmark}$$


0

Note that $P(\tau\leq n)=P(\max_{1\leq k\leq n} S_k \geq C)$. So you might want to have a look at a paper by Janssen and van Leeuwaarden: "Cumulants of the maximum of the Gaussian random walk".


1

Building on Michael Hardy's answer. I use the notation $\bar h = E(h)$ and $\bar k = E(k)$ for readability. Since $h,k$ are independent, $E(hk) = \bar h \bar k$. By definition, $E(h^2) = \sigma_h + \bar h^2$ and $E(k^2) = \sigma_k + \bar k^2$. We have, $$\begin{align*}E((\sqrt{t}h+k)^2)& =E(th^2+2\sqrt{t}hk + k^2)\\ &= ...


2

$\newcommand{\E}{\operatorname{E}}\newcommand{\var}{\operatorname{var}}$ It follows from independence that $\E(hk)=\E(h)\E(k)$. If $\E(h)=\E(k) = 0$ then $\E(h^2)= \var(h)$ and $\E(k^2)=\var(k)$ and we have \begin{align} \E\left((\sqrt{t}\, h + k)^2\right) = \E(th^2 + 2\sqrt t\,hk + k^2) & = t\E(h^2) + 2\sqrt t\E(h)\E(k) + \E(k^2) \\[10pt] & = ...


2

The probability that two independent continuous random variables are exactly equal is 0. And, of course, normal random variables are continuous. Continuous random variables put 0 probability on any one point. A more interesting problem might be the probability that the two variables are within one unit of each other. For discrete random variables the ...


1

Notice that the pdf of $X$ is $$f_X(x)=\begin{cases} \frac{1}{18},&\text{ if } -9\le x \le 9\\ \ 0,& \text{ otherwise. }\end{cases}$$ We can treat $F_Y(y)=P(Y<y)$ as the sum of two probabilities: $$F_Y(y)=P(Y<y\cap |X|>3)+P(Y<y\cap |X|\le 3).$$ First, $$P(Y<y\cap |X|>3)=P(0<y \cap |X|>3)=\begin{cases} 0,&\text{ if } ...


0

You are correct that $Y$ takes values in $(-12,12)$ and "mainly" uniformly. However, this is a 'mixed' distribution with positive probability at the point $0$. So you are not going to find a PDF for $Y$. Think it through in terms of intervals of $X$ in order to get the CDF of $Y.$ If you have ready access to R, you might run the following code for a ...


1

For reference, I am repeating the PDF (or PMF) table of $X_2.$ Then $E(X_2) = \sum_{i=0}^6 v_i p_i,$ the terms and total of which are in the third row. (My notation may different from that of your book, so make sure you connect the table with whatever formula your book has.) value 0 0 1 1 0 0 1 1 Total prob. .04 .08 .08 ...


1

You can first simplify the problem: Let 1 denote an odd number and 0 and even number. Case 1: P(X=0) $11111$ $00000$ The even number can be arranged in 5! ways. The odd numbers can be arranged in 5! ways, too. Thus you have the factor $(5!)^2$ in all cases. In each coloumn 0 can be in the first row or in the second row. This gives an additional factor of ...


1

Let us grind it out. There are five odd and five even. So we might as well assume that we use five $0$'s and five $1$'s. Let $X$ be the number of columns with two $0$'s. We want the probability distribution of $X$. The random variable $X$ can only take on the values $0$, $1$, and $2$. So we really have only two probabilities to compute, since if we know ...


0

Yes. Your answer is correct. Ákos Somogyi showed one method to obtain it; by integration, substitution, then derivation. Another method is to use the change of variables transformation. $y=1-\lvert x\rvert$ means $x =\pm(1-y)$ The standard formula is modified because the support of $X$ is subdivided into two intervals which both map to the support ...


0

$X(\omega)$ takes on the value $k$ if and only if we reach $r$ successes for the first time after $k$ tries. $G_i(\omega)$ takes on the value $k_i$ if and only if we reach $1$ successes for the first time after $k_i$ tries You want to show that $X(\omega) = k \iff G_1(\omega) + G_2(\omega) + \dots + G_r(\omega) = k_1 + k_2 + \dots + k_r = k$ That is, for ...


0

We use it because Expectation is Linear.   That is the property that means the expectation of a sum of random variables is the sum of the expectations of the variables. So if we can express the random variable $X$ as a series of random variables with easier to find expectations, the we can find the expectation of $X$ easier. $$\begin{align} \mathsf ...


0

I don't know MatLab, but looking at your code, it looks like you are making a list of values $X_1, X_2, ..., X_n$ and taking the mean of that list. However, what you want is to take the mean of many $X_n$s for one fixed $n$ (the number of iterations).


1

Since the the map $1-|x|$ is not invertable on $[-1,1]$, you cannot apply the general formula, you have to do some calculations instead: a) Integrate over $[-1,y]$ to get the CDF: $$\int_{-1}^y \frac{1-x}{2}\mathrm{d}x=C+\frac{2y-y^2}{4}$$ As it is a CDF, $C=\frac{3}{4}$, so $$F_X(x)=\frac{3+2x-x^2}{4}$$ b) $|X|\leq 1$, so: ...


1

It looks like you're asking why the variance and expected value of a boolean variable isn't itself boolean. The problem goes away if you consider variance to be a measure of the variable rather than the variable itself, and consider 'expected value' to be a form of scaled probability rather than the domain value to 'expect'. For example, given instances ...


1

It's no different than an integer variable with values $\{0,1\}$. Naturally, both EV and variance will be in the real interval $[0,1]$. What you have to do is simply find the distribution of $X_2$ from the table - this will be parameterized by $p\in[0,1]$ with $P(X_2=1) = p$ and $P(X_2=0)=1-p$. The expected value will then be $\mathbb E(X_2) = p$ and the ...


0

A geometric random variable $G_i$ is the number of Bernoulli trials needed to get one success. Your variable $X$ is the number of trials needed to get $r$ successes, and so if these trials are carried out one after another, you must first have success $1$, success $2$, $\ldots$, success $r$; the number of trials to get success $1$ is $G_1$, the number of ...


0

We compute $$ \begin{align*} \lim_{T\to\infty}\frac1{2T}\int_{-T}^T\mathbb E[X^2(t)]\mathsf dt &=\lim_{T\to\infty}\frac1{2T}\int_{-T}^T\mathbb E\left[\frac{A_0^2}2(1 + \cos(2(\omega_0t+\theta)\right]\mathsf dt\\ &=\frac{A_0^2}2 \lim_{T\to\infty}\frac1{2T} \int_{-T}^T (1 + \mathbb E[\cos(2(\omega_0t + \theta))])\mathsf dt\\ &= \frac{A_0^2}2 ...


0

They are in general correlated. Say $Z$ has a pdf $p_Z(z)=\delta(z)$, i.e. $Z$ is always $0$ (it is not random). This is a special case of a random variable, so $X+Z=X$, therefore $X+Z$ and $Y$ are correlated.


1

$X_n$ converges in probability to $0$ since $$\mathbb{P}(|X_n|>\epsilon) = \mathbb{P}(|X_n|=1) = \frac{1}{n} \stackrel{n \to \infty}{\to} 0$$ for any $\epsilon>0$. However, $X_n$ does (in general) not converge almost surely: Since the sets $\{|X_n|>\frac{1}{2}\}$, $n \in \mathbb{N}$, are independent, we have by the Borel Cantelli lemma that ...


0

Are they also independent Gaussians? (If, say, they are all actually the same Gaussian, then you have a very different situation.) Assuming they are independent, there is such a statement. The key fact is that all norms on finite dimensional spaces are equivalent. That means that there exist constants $c,C>0$ such that $$c \| A \|_F \leq \| A \|_2 \leq ...


1

The probability that both $x$ and $y$ are less than $1/2$ is $1/4$. The probability that both $x$ and $y$ are less than $1/3$ is $1/9$. We want the former to be true, but not the latter, so the answer is $1/4 - 1/9 = 5/36$.


0

The formula for the same I think is $$P(X = x, Y=y) = \left(\dfrac{\left(\sum_{i,j=1}^{Max(x,y)}{4\choose i}{13\choose j}\right)}{{52\choose4}}\right)$$ Let us say you have 2 of the same suit and 3 of the same values, Combination of 2 suits(a,b) and three values (A,B,C), we get the following, and let us want the prbability that x = 2 and y =2 ...


0

I have an approach that seems to work. Write $a$ and $b$ as products of random unitary matrices $U_a \in U(M)$ $U_b \in U(N)$ with uniform (Haar) measure on the unitary group and $e_1=(1,0,0,\dots)$. $$ a = U_a e_1 $$ $$ b = U_b e_1 $$ Now the product you want is: $$ a^H Hb = e_1^TU_a^HH U_b e_1 $$ $H$ is a random unitary matrix with Gaussian measure ...


1

As an alternative to this routine computation, the result follows from the Lack of Memory property of the exponential distribution. Moreover, this doesn't depend on the exponentials having the same parameter. Think of $X$ and $Y$ as waiting times for events A and B respectively. The first of the two events occurs at time $\min(X,Y)$, and its probability of ...


0

This is a straighforward hypergeometric with $m=10$, $r=1$ and $n=X$. So $$P(Y=1)=\frac{{1\choose1}{9\choose{X-1}}}{10\choose X}$$ $$=\frac{9!}{(9-X+1)!(X-1)!}\cdot\frac{X!(10-X)!}{10!}$$ $$=\frac{X}{10}$$


1

It's a straightforward question if you put aside worrying about $X$ being a random variable for the moment and simply treat it as an unknown number. You had a pond of ten fish, and one of them was a piranha. You pulled out some number of them, call it $X$.   What is the probability that the piranha was among these? Let's assume that each particular ...


0

Since $X$ of the $10$ fishes are 'elected' the piranha has a chance of $\frac{X}{10}$ to belong to them. (Preassuming that each fish has the same chance of being elected)


0

It's a uniform distribution so we don't need to mess with integration, just measure the proportionate area of the event against the support (the unit square). $\{\langle X, Y\rangle\in (0;1)^2 : X−\sqrt{z^2-1}<Y<X+\sqrt{z^2−1}\}$ is the section of the unit square between the lines $y=x-\sqrt{z^2-1}$ and $y=x+\sqrt{z^2-1}$. That is also section of ...


1

Your terminology and notation are a bit strange. First, $p_2,$ which I'll call just $p,$ is a parameter, not an estimator. The number of successes in $n = 86$ Bernoulli trials is $S = \sum_{i=1}^{n} Y_i,$ which has the distribution $Binom(86,p),$ as you say. The estimator of $p$ is $\hat p = S/n = \bar Y.$ (a) If $p = .1,$ what is $P(\bar Y = S/86 \le ...


2

Consider the more general probability $$\Pr\left[ \mu - z \frac{\sigma}{\sqrt{n}} < \bar X < \mu + z \frac{\sigma}{\sqrt{n}} \right],$$ for some $z > 0$, where $$\bar X = \frac{1}{n} \sum_{i=1}^n X_i$$ is the sample mean of $n$ independent and identically distributed observations from a normal distribution with mean $\mu$ and standard deviation ...


-1

As I understand all random variables are asymptotically independent only. I do not think sample to sample independent exist. Your issue is always true for sample size.


0

Use trig identities such as $\cos (A +B) = \cos A \cos B-\sin A\sin B$ Then, as they are independent: $\mathsf E(\cos (A +B)) = \mathsf E(\cos A)\mathsf E(\cos B)-\mathsf E(\sin A)\mathsf E(\sin B)$ Thus if $B$ is zero mean and symmetrically distributed, you have $\mathsf E(\sin B)= 0$ (as Sine is an odd function). You must also show that for your ...



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