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1

No, you don't need to find the pdf of $1/Y$. Use the Law of the Unconscious Statistician: $$ \mathbb E[1/Y] = \int_0^\infty \dfrac{1}{y} f_Y(y)\; dy$$ where $f_Y$ is the density for $Y$.


1

$\newcommand{\var}{\operatorname{var}}$ \begin{align} \var(b) = \var(\ \underbrace{(x'x)^{-1}x'}_{\begin{smallmatrix} \text{This is constant,} \\ \text{i.e. not random.} \end{smallmatrix}}\ \ y) & = (x'x)^{-1} x' \Big( \var(y)\Big) x (x'x)^{-1} \\[8pt] & = (x'x)^{-1} x' \Big( \sigma^2 I_{2\times 2}\Big) x (x'x)^{-1} \\[8pt] & = \sigma^2 ...


0

The way to approach these types of problem is the following. We begin by noting that we can calculate the $z$-value as follows: $$z=\frac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\frac{61-50}{7}=1.5714$$ And therefore we have that (using tables or an appropriate online tool): $$\mathbb{P}[Z<z]=\Phi(z)\approx 0.9420$$ Thus the $p$-value of this two-tailed test is ...


0

Same and identical mean the same word in the english language, so they should mean the same in maths too. however, "similar" and "identical" do not mean the same in the english language, so naturally in math they don't mean the same either. The word "similar" suggests there are resemblances but the two similar things are not all the same ; here by similar ...


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 ...


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 ...


0

Conditional probabilities: for the probability measure $\mathbb{P}$ (appropriately defined now, we don't like zero divisions), $$ \mathbb{P}\left(B \,|\, A\right){}={}\frac{\mathbb{P}\left(B \,\cap \, A\right)}{\mathbb{P}\left(A\right)} $$ Choose the probability measure $\mathbb{P}\left( \dot\,\,|\,\, C\right)$ and apply it to this definition above, ...


0

In general $\mathbb{P}\left(A\mid B\right)$ is defined by the equality $$\mathbb{P}\left(B\right)\mathbb{P}\left(A\mid B\right)=\mathbb{P}\left(A\cap B\right)$$ If $\mathbb{P}\left(B\right)\neq0$ then this comes to the same as $\mathbb{P}\left(A\mid B\right)=\frac{\mathbb{P}\left(A\cap B\right)}{\mathbb{P}\left(B\right)}$. Multiply both sides in the ...


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 ...


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 ...


0

This issue is that the pdf of a sequence of random variables that converge to another random variable (in distribution) may not converge uniformly to the limiting density. In the example given, the PDF would converge to a dirac delta at $x=2$, which cannot be approached uniformly. That's why they say its generally not true. There are special cases where it ...


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 ...


0

I'm assuming that Y, although unit less, has a natural ordering to it. There's no specific rule on when you MUST use a specific type of model. Really, you could even use a standard OLS regression (I would even recommend starting off with this, and then progress to more complicated models). In general, If Y in continuous (i.e can take values 2.75 or 3.45), ...


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 ...


2

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

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 ...


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: $$ n\cdot E [ \bar{X}_n ] = E\left[\sum_{i=1}^n X_i \right] = \sum_{i=1}^n E[X_i] = \sum_{i=1}^n E[E[X_i \mid Y_i]] = \sum_{i=1}^n E[Y_i] = 0 $$


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. ...


0

You need to look at the Gamma distribution and its expected value The gamma distribution takes two parameters $\alpha$ and $\beta$ you can think of $\Gamma(\alpha,\beta)$ as the time required for $\alpha$ events to occur given that these events occur randomly in a Poisson process with the mean time between events as $\beta$ So (a) this is simply the ...


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 ...


0

The sample space $\Omega$ is $\{0,1\}^\mathbb N$, and $\gamma_k(\omega) = \omega_k$ (i.e. an outcome tells you $\gamma_k$ for all natural numbers $k$). The $\sigma$-field is the Borel $\sigma$-field for the product topology, which is the $\sigma$-field generated by the $\gamma_k$.


0

First, $$\binom{20}1(0.9)^1(1-0.9)^{19}$$ isn’t what you want: you got the exponents back to front. You want the probability that exactly one ball is flat, which is $$\binom{20}1(0.9)^{19}(1-0.9)^1\approx0.27017\;.$$ Secondly, your arithmetic went astray somewhere, because $$\binom{20}1(0.9)^1(1-0.9)^{19}=1.8\times10^{-18}$$ exactly.


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

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

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 ...


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 ...


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

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} &= ...


0

In part (c) the answer will be 0 . This is because 9.3 lies outside the control limits i.e. is 8.725 and 9.25 so there will be 0 probability of its occurrence.


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

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 ...


0

Assuming you have a standard normal distribution table handy... (a) You can use the table once you have converted your random variable (let's call it $X$) to the standard normal distribution $Z$. The conversion is $$Z = \frac{X - \mu}{\sigma}$$ where in this case $\mu = 250$ and $\sigma = 15$. So $$P(X > 270) = P\left(Z = \frac{X - \mu}{\sigma} > ...


0

As the comments suggest, $1$ test is equivalent to $x=0$, since none of the plugs would succeed. Likewise, $4$ tests means $x=3$ and $6$ tests means $x=5$. Since you know the probability distribution function, you just want to sum this over values from $x=3$ to $x=5$ (inclusive). This gives the answer that the answer key has.


0

I found this explanation in a website of the University of Texas: There are some types of Random Sample, one is Simple Random Sample. There are other types called Probability Samples. One kind of Probability Sample is Stratified Random Sample. Here is the information of the University of Texas. "A good example about how to select a SIMPLE RANDOM SAMPLE ...


0

Since multiple choices are possible for a single variable, this is equivalent to having a binary variable for each possible value for each possible variable. So then you'll get a $2 \times 2 \times N$ table, where one dimension is gender and one dimension is chosen/ not chosen and the last dimension is all the possible values for the variables. You want to ...


0

The problem term is $$ E\left[Y_{1}Y_{2}\right]=E\left[\left(X_{1}+2X_{3}\right)\left(X_{1}-2X_{3}\right)\right]=E\left[X_{1}^{2}\right]-4E\left[X_{3}^{2}\right]. $$ So, what is $E\left[X_{k}^{2}\right]$? Recall $$ \sigma^{2}=\text{Var}\left(X_{k}\right)=E\left[X_{k}^{2}\right]-E\left[X_{k}\right]^{2}=E\left[X_{k}^{2}\right]-\mu^{2} $$ so that $$ ...


0

You did this right. The short way to look at it is that $B+C-A$ is normally distributed with mean being $\mu = \mu_B+\mu_C-\mu_A$ and $\sigma^2 = \sigma^2_B + \sigma^2_C + \sigma^2_A$. The key point you need to know is that a variate made of the sum of two independent normal variates is itself normally distributed, even if the means of those two variates ...


0

great to see you are interested in GPs. Someone asked a similar question about how to make predictions on a GP and I think I gave a fairly comprehensive work through with sample data that they gave. I think seeing it in action will answer your question. Here is the link.


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 ...


0

If $X$ is a random variable with cdf $F_X(x)$, then any function of $X$, say $g(X)$, is also a random variable. Since $Y = g(X)$ is a function of $X$, we can describe the probabilistic behavior of $Y$ in terms of that of $X$. That is, for any set $A$, $$\Bbb P(Y \in A) = \Bbb P(g(X) \in A),$$ showing that the distribution of $Y$ depends on the function $FX$ ...


0

I think you're confusing the interpretation of the normal qq plot. A qq-plot generally is used to compare 2 probability distributions. In this case, to compare the distribution of your data to that of a normal distribution. The S shape you see in the graph means that your distribution is probably highly skewed or has heavier tails. Because of this, you ...


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 ...


1

$A_n$ is increasing in $n$ because the event $$ \bigg\{ \delta \sum_{i=1}^{n-1} X_i > b \bigg\} $$ is a subset of the event $$ \bigg\{ \delta \sum_{i=1}^n X_i > b \bigg\}, $$ simply because $X_n$ is nonnegative.


0

The notation $\mu'_r$ refers to the $r^{\rm th}$ raw moment of the random variable $X$; i.e., $$\mu'_r = \operatorname{E}[X^r]$$ whenever this expectation exists. We also have, for some function $g$ of a random variable $X$, $$\operatorname{E}[g(X)] = \int_{x\in \mathcal X} g(x) f_X(x) \, dx,$$ where $\mathcal X$ is the support of $X$, and $f_X$ is the ...


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 ...


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 ...


0

Replace the population moment by sample analog to obtain $\frac{1}{n}\sum_{i=1}^nY_i = \frac{\hat \theta}{2}$ and so $$\hat \theta = \frac{2}{n}\sum_{i=1}^nY_i$$


0

$\left(\dfrac{37}{38}\right)^{34}$ is the probability of losing $34$ bets in a row. If this does not happen then you have gained $35$ at least once and so are ahead after $34$ bets. So this is not a $Z$ score but, after subtraction from $1$, an exact probability answer to (a). You do not need to consider expectations or approximations.


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) &= ...



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