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4

For $\alpha_1, \alpha_2 \gt0$ We have $$\int_{0}^1 u^{\alpha_1-1} (1-u)^{\alpha_2-1} \, \mathrm du = \text{B}(\alpha_1,\alpha_2) \tag{1} $$ Where $\text{B}(a,b)$ is Beta Function $$ \begin{align} \Gamma(\alpha_1) & =\int_0^\infty\ e^{-x} x^{\alpha_1-1}\,\mathrm{d}x \tag{2}\\ \Gamma(\alpha_2) & =\int_0^\infty\ e^{-y} y^{\alpha_2-1}\,\mathrm{d}y ...


3

It is known as the chain rule. A justification can be seen, assuming of course $\Pr[C] > 0$ and $\Pr[A\cap C] > 0$, as $$\begin{align} \Pr[A\cap B \mid C ] &= \frac{\Pr[A\cap B \cap C ]}{\Pr[C]} = \frac{\Pr[A\cap B \cap C ]}{\Pr[A\cap C]}\cdot \frac{\Pr[A \cap C ]}{\Pr[C]} \\ &= \Pr[B\mid A\cap C ]\cdot \Pr[A \mid C ] \end{align}$$ where the ...


3

Surprisingly, the answer is no. Consider the case $n=2$ with probability space $\{0,1\}^4$ and $X_1, X_2$ the first two coordinate functions and $Y_1, Y_2$ the second two. The probabilities of the $16$ different configurations are $$\begin{array}[cccc]{} x_1 & x_2 & y_1 & y_2 & p(x_1,x_2,y_1,y_2)\cr 0 & 0 & 0 & 0 &1/16\cr 0 ...


2

You are right in saying that the teacher did not strictly define the experiment and thus it is hard to tell whether or not there is a bias. For example you could also ask was the selection process random. However I think the idea the teacher was trying to get across was that if I have $n$ objects and randomly partition the set into $A$ and $B$ with the same ...


2

It's much easier than you think! Since by hypothesis we know that the conditional mean is $ E[X_i \mid Y_i] = Y_i $ and $Y_i$ is uniform with mean zero, by using the tower property we find: $$ n\cdot E [ \bar{X}_n ] = E\left[\sum_{i=1}^n X_i \right] = \sum_{i=1}^n E[X_i] = \sum_{i=1}^n E[E[X_i \mid Y_i]] = \sum_{i=1}^n E[Y_i] = 0 $$


2

Let us evaluate the CDF of a Gaussian distribution: $$f(x) = \frac{1}{\sqrt{2\pi\sigma^2}} \int_\mathbb{R} \exp \left( - \frac{(x-\mu)^2}{2\sigma^2}\right)\ dx \label{orig}\tag{1}$$ Let us make the transformation in $\ref{orig}$ by $\mu \mapsto N \mu$ and $\sigma \mapsto \sigma \sqrt{N}$. $$\begin{align*} f(x) &= ...


2

Consider an election for state governor between a conservative candidate and a progressive candidate where 90% of the people in the state would vote conservative but the issues didn't interest them very much and only 5% of conservatives would actually go to the polls on election day. In other words, $90\%\cdot5\%=4.5\%$ of the total state population ...


2

A finite Markov chain always has at least one steady-state distribution. If the transition matrix is $A$, each column of $A-I$ sums to $0$, so $A-I$ doesn't have full rank, and there is at least one nontrivial solution to $Ax=x$. On the other hand a Markov chain with an infinite state space doesn't have to have a steady-state distribution. For example, ...


1

The error may be in stating $p=0.3$, since instead using $p=0.5$ leads to $$\Phi\left(\frac{101-100}{\sqrt{50}}\right)-\Phi\left(\frac{99-100}{\sqrt{50}}\right) \approx 0.112463.$$ Of course a continuity correction would be better as $\Phi\left(\frac{101.5-100}{\sqrt{50}}\right)-\Phi\left(\frac{98.5-100}{\sqrt{50}}\right) \approx 0.167996$ while ...


1

More generally, if $X_1, \ldots, X_n$ are independent normal random variables with means $\mu_i$ and variances $\sigma^2_i$, and $c_1, \ldots, c_n$ are constants, $Y = c_1 X_1 + \ldots + c_n X_n$ is normal with mean $\mu_Y = c_1 \mu_1 + \ldots + c_n \mu_n$ and variance $c_1^2 \sigma_1^2 + \ldots + c_n^2 \sigma_n^2$. Here you're doing the case $n=3$ with ...


1

When finding the standard deviation, use the prior estimate of proportion, $\hat p$, which is to say the population's proportion, $0.4$. This gives $$\sigma = \sqrt{\frac{0.4\times0.6}{300}}\approx0.0282$$ $$z = \frac{0.44 - 0.4}{\sigma}\approx1.414$$ Then the normal distribution gives $P\approx0.9214$. The answer key's value comes from $z=1.41$, but ...


1

Convergence in distribution means that $F_n(x) \to F(x)$ for all points $x$ except the points of discontinuity of $F$. Since the distribution $F$ for a PMF consists of a sequence of "jumps", or discontinuities at the points $x$ where $P(X=x)>0$, $F_n(x)$ need not converge to $F(x)$ at these points for convergence in distribution. but we know that "a ...


1

Suppose you pick $n$ points. Consider one of the chosen points, call it $x$. The probability that the distance to its nearest neighbour is less than $r$ is the probability that at least one other point lies within the disk of radius $r$ centered at $x$. Since the remaining $n-1$ points are drawn uniformly from the total area $A$, this probability is simply ...


1

Let $X_i$ have mean $\mu_i$ and standard deviation $\sigma_i$, and write $X_i = \mu_i + \sigma_i Z_i$ where $Z_i$ are independent standard normal random variables. The joint density of $Z_1, \ldots, Z_n$ is $$f(z_1, \ldots, z_n) = (2 \pi)^{-n/2} e^{-(z_1^2 + \ldots + z_n^2)/2} = (2 \pi)^{-n/2} e^{-\|{\bf z}\|^2/2}$$ which is rotationally invariant, i.e. ...


1

You need to decide what type of errors are important. If the various values come from measuring a length with a ruler, the base value doesn't matter, just the size of the error. An error from 0 to 6 is just as bad as from 249 to 255. In other situations a relative eror is more important and and error from 1 to 2 is just as bad as one from 127 to 254. In ...


1

The sample space for your process $\tilde{u}_k$ relies on three spaces: An abstract sample space $\Omega$, with sample points $\omega_i:=(^i\omega_j)_{j \in \mathbb{N}}^{i \in R^+}$ Note that $|\Omega|=\aleph^1$. $\gamma(\omega): \Omega \to \{0,1\}^{\mathbb{N}}$, which represents all possible infinite sequences of $1$ and $0$. $u(\omega):\Omega \to ...


1

This is a proof for $\mu = 0$. When $\mu \neq 0$, the proof can be done with some slight modification. You want to prove $\sqrt{n}(1-X_n^{-1}) - G_n = \sqrt{n}(2-X_n^{-1}-X_n)$ converges to $0$ in probability, which is equivalently that it converges to $0$ in distribution. To do this, we can apply the second order delta method(for reference see this) with ...


1

It looks to me like the probability of not getting a ticket is $\frac34$. (So the probability of getting a ticket is $\frac14$. Here's an intuitive way of seeing this. Half the time you park for less than two hours and definitely don't get a ticket. The other half of the time, you park your car for more than two hours, and then you get a ticket if the ...


1

Like you said, lets have $t$ be the time since the officer's last visit, so $t\sim \mathcal{U}(0,2)$ Your parking time is $h\sim \mathcal{U}(0,4)$. Let $T:= \{4-t<h\}$ be the event that you get a ticket, as you correctly pointed out.Often the easiest way to tackle these problems is via conditioning. ...


1

From the paper referenced, the cost function to minimise is $$ \ C=\sum_{u,i}c_{ui}(p_{ui}-x_u^Ty_i)^2+\lambda(\sum_u\|x_u\|^2+\sum_i\|y_i\|^2)$$ Differentiating with respect to vector $x_u$ results in (note: as the vectors $x_u$ and $y_i$ are real vectors, their scalar product is commutative, see red) $$\begin{align}\frac{\partial C}{\partial x_u} &= ...


1

Let $x =$the length of time in months before a major repair occurs, and let $x_0 =$ the guarantee period also in months. We want : $P(x < x_0) = 0.05$, and convert to $z$ variable: $P\left(z < \dfrac{x_0-10}{3}\right) = 0.05 \Rightarrow \dfrac{x_0-10}{3} = -1.645 \Rightarrow x_0 = 10 - 3\cdot 1.645 = 5.07 \text{ months}$.


1

Let $ X_1 \sim N(23.95,7.44), X_2 \sim N(16.29,7.79) $, then, as mentioned by Dilip Sarwate in the comments, $ X_1 - X_2 \sim N(7.66,{\sqrt{7.44^2 + 7.79^2}}) $ Thus the problem becomes $$\mathbb{P}(X_1<X_2) = \mathbb{P}(X<0),\ \ where \ \ X = X_1 - X_2 $$ This can be calculated directly or reverting to the standard $ N(0,1) $ distribution as ...


1

How is it defined in the discrete case? In the general case, a median is defined as any number $M$ such that $P(\xi\leqslant M)\geqslant\frac12$ and $P(\xi\geqslant M)\geqslant\frac12$, or, equivalently, such that $F_\xi(M^-)\leqslant\frac12\leqslant F_\xi(M)$. Medians are not unique, in general. Exercise: Determine the set of medians of the ...


1

Similar distribution means the type of distribution is the same. Identical distribution means the type of distribution is the same and their parameters have exactly the same value. If question stated that X and Y have same distribution then their parameters should have same values. But if question stated that X and Y have same type of distribution that's ...


1

No, there isn't. Counterexample: The list (10, 10, 10, ..., 10) has the standard deviation 0. The list (-50, -10, -50, -10, ..., -50) has a standard deviation of approximately 19.9555. The list (10, 50, 10, 50, ..., 10) has a standard deviation of approximately 19.9555. The list (-10, -10, -10, ..., -10) has the standard deviation 0. More generally, ...


1

When you integrate with respect to $x$, you must also take into account that $x$ cannot be allowed to exceed $y$. Sketch the region of integration and you will see what I mean. Therefore, the correct calculation is $$f_Y(y) = \int_{x=0}^{\min(1,y)} \frac{4x^3}{y^3} \, dx = \frac{\min(1,y^4)}{y^3} = \min(y^{-3},y) = \begin{cases} y^{-3}, & 0 < y \le ...


1

The best way to understand this answer is to draw out a graph showing the ranges of $x$ and $y$. Draw a vertical line at $x=1$ to show that boundary value and draw the line $y=x$. Now shade in the region that corresponds to $y>x$. You should be able to see that the range of $y$ can be divided into two parts: $y>1$ and $0<y\le1$. Hopefully the limits ...


1

Once again an illustration that one should consider PDFs as functions defined on the whole product space. Here $p:\mathbb R^2\to\mathbb R$ is defined, for every $(x,y)$ in $\mathbb R^2$, as $$p(x,y)=4x^3y^{-3}\,\mathbf 1_{0\lt x\lt1}\,\mathbf 1_{y\gt x},$$ hence, by definition, the density $p_Y$ of $Y$ is defined, for every $y$ in $\mathbb R$, by ...


1

A common question. Another way to say this: To compute the variance of the sample, you divide by $5$. But is that what you are interested in? Is this a random sample of some large population? Are you not interested in estimating the variance of that population? If so, divide by $4$. Beginners may be confused by this. An unbiased estimate for the ...


1

Maybe you are looking for a statement like this: The higher the probability of high values of $X$, the more likely it is that the firm defaults. That is, you are comparing probability distributions of $X$. Mathematically, you could use the notion of first order stochastic dominance (FOSD). Suppose you have two different distributions for $X$. Denote the ...



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