New answers tagged self-study
1
The generic approach is to find the distribution of the rv $X-Y$, denoted $Z$, from the joint distribution of $(X,Y)$, which is a convolution exercise. Since the change of variables from $(X,Y)$ to $(Z=X-Y,Y)$ has Jacobian one
$$\left|\dfrac{\text d(x,y)}{\text d(z,y)}\right|=\left|\dfrac{\text d(z+y,y)}{\text d(z,y)}\right|=\left|\begin{matrix}1 &1\\0 &...
2
You seem to substitute $P_1$ for $P(A|H_1)$, which is wrong. It's actually $P(A|H_1)\geq P_1$ because $A$ doesn't win only when the second toss is Heads, it can still win when the second toss is tails; an example case is tails, tails, heads, heads.
2
This kind of calculation is in general handled with using the joint distribution of the two variables $X$ and $Y$.
You state that $X$ and $Y$ are i.i.d. $U(0, 1)$'s, and this uniquely specifies their joint distribution. Since they are independent, the joint density function factors:
$$f_{X, Y}(x, y) = f_X(x) f_Y(y) = 1 \times 1 = 1$$
Given this, you can ...
0
Here is a general algorithm I have used in the past to draw ROC
Sort the data by score from lowest to highest
Choose lowest score as the cut-point
Calculate sensitivity and specificity using the cut-point from step 2. Save sensitivity and 1-specificity
Repeat steps 2 and 3 for each unique score from lowest to highest
Create a step plot where the x-axis is 1-...
10
TLDR: probabilities are not required to build a ROC curve, only a numerical scale supporting the decision.
I'm studying the ROC Curve, and I was wondering if there is any
classification algorithm that doesn't return the output class as a
result of a certain threshold from the probabilities of the algo?
I previously let this question slip because I ...
7
Support Vector Machines and $k$-Nearest Neighbors come to mind.
(See here for a motivation for short answers. Longer answers are always welcome.)
2
Your attempt is correct, but for this particular multiple choice question we can find the cut-off point $c$ directly given that the critical region is of the form $-\sum\limits_{i=1}^n \ln(1-X_i)^2 <c$.
As you might have already noticed, $$X_i\sim \mathsf{Beta}(1,\theta)\implies 1-X_i\sim \mathsf{Beta}(\theta,1)\implies-2\theta\ln (1-X_i)\sim \chi^2_2$$
...
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The second part:
By the Var definition:
$$Var(X) = E(X^2) - [E(X)]^2$$
$$Var(X_n-b) = E(|X_n-b|^2) - [E(X_n-b)]^2$$
$$\lim_{n \to \infty}Var(X_n-b) = \lim_{n \to \infty}E(|X_n-b|^2 ) - \lim_{n\to \infty}{[E(X_n-b)]^2}$$
the Quadratic mean tell us that:
$$ \lim_{n\to \infty} E(|X_n-b|^2 ) = 0$$
and the first part of this problem defined:
$$\lim_{...
0
Here is my simulation repeated 3000 times of how much rice consumption we can expect for the whole population for a given week. At week 0 we initialize the population, the simulation starts from week 1 to week 40 (10 months).
set.seed(420)
res=vector("list",3000)
for (i in 1:3000) {
pop=100
con=rep(5,100)
for (j in seq(1,10*4)) {
pop=round(pop-...
1
Indeed, because $X_2+X_3\sim N(0,2)\implies\left(\frac{X_2+X_3}{\sqrt 2}\right)^2\sim \chi^2_1$, independently of $X_1\sim N(0,1)$, $$\frac{\sqrt 2X_1}{|X_2+X_3|}=\frac{X_1}{\sqrt{\left(\frac{X_2+X_3}{\sqrt 2}\right)^2}}\sim t_1$$
Alternatively, since $\frac{X_2+X_3}{\sqrt 2}\sim N(0,1)$, independently of $X_1\sim N(0,1)$,
$$\frac{X_1}{\left|\frac{X_2+X_3}{...
0
To see if model A is nested in model B, it is not enough to compare the symbolic model structure, but see What is a "symbolically nested" model?. What matters is that, for every set of values of the parameters in A, we can find parameters for B that gives the same predicted values. And that is clearly the case for your first example, so A is ...
2
Begin with the hint given. Take $x=0.3$. Can you find the value of $\theta\in[-1,1]$ that maximizes $\frac{1+0.3\theta}{2}$?
Take $x=0.8$. Can you find the value of $\theta\in[-1,1]$ that maximizes $\frac{1+0.8\theta}{2}$?
Take $x=-0.5$. Can you find the value of $\theta\in[-1,1]$ that maximizes $\frac{1-0.5\theta}{2}$?
Take $x=-0.9$. Can you find the ...
1
The first step of finding the MLE is to write out the likelihood function (or log-likelihood function). In the present problem, with only a single observed data point, the latter is given by:
$$\ell_x(\theta) = \ln(1 + x \theta) + \text{const.} \quad \quad \quad \text{for } -1 \leqslant \theta \leqslant 1.$$
This is the function you need to maximise to ...
3
The question seems to be asking you to formally state a hypothesis.
The "hint" makes a critical fallacy of conflating a "test statistic" with the "effect of interest". In our case we're interested in average price of a phone, it's "expectation". So we either define the average price of either type of phone, or the difference between them. If I were grading ...
0
In order for it to be a proper probability distribution, $\pi$ should itself be a probability distribution, such that
$$\forall k \quad \pi_k \geqslant 0$$
and
$$\sum_{k=1}^K \pi_k= 1$$
Hence
$$\int f(x) dx = \sum_{k=1}^{K}\pi_k \int N(x|\mu_k, \sigma_{k}^2) dx = \sum_{k=1}^{K}\pi_k \cdot1 = 1$$
and non-negative everywhere due to non-negativity of $\...
2
Here, it means the normal PDF:
$$\mathcal{N}(x|\mu,\sigma^2)=\frac{1}{\sqrt{2\pi\sigma^2}}e^{-(x-\mu)^2/2\sigma^2}$$
The $\mu,\sigma^2$ in given side means that you can treat them as known quantities.
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$N(x|\mu, \sigma)$ combines the two notations: $x \sim N(\mu, \sigma)$ and $p(x| \mu, \sigma)$. So it reads: $x$ is normally distributed with parameters $\mu, \sigma$.
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The $X_1$ coefficient will only remain unchanged in the two models if $X_2$ has 0 covariance with either $Y$ or with $X_1$.
0
This may be easier to see by expanding the matrix notation (which, through its inconsistent use of transposes, might be a little confusing):
$$f(\mathbf x) - y = x^\prime\alpha - (\beta^\prime \mathbf{x} + \varepsilon) = \sum_{i=1}^p (\alpha_i-\beta_i)x_i - \varepsilon.$$
Its square expands into three terms according to the power of $\varepsilon,$
$$(f(\...
2
This is a nice and neat simulation that I can demonstrate with R code. I would start off by defining the initial values:
population <- 100
consumption_rate <- 5
Next, we could define a few functions to vary the population parameters, as specified in the 3 steps:
## Take a uniform value in 0.01-0.02 as a percentage of our current population
...
4
There are two steps to solving this problem. The first is based on the realization that what you actually observe are observations $y_i = 1_{(x_i < 0)}$ which are drawn from a Binomial distribution with probability parameter $p(\mu) = \text{P}(x<0 |\, \mu, 1)$, with $\text{P}$ of course being the cumulative Normal distribution.
We form the maximum ...
0
For $M_n^{(1)}$ we must first check that the process is integrable. Note that $\theta\mapsto |e^\theta+e^{-\theta}|$ is an even function which is increasing on $(-\infty,0)$ and decreasing on $(0,\infty)$ and hence attains its maximum $2^{-n}$ at $\theta=0$. Now,
\begin{align}
\mathbb E[|M_n^{(1)}|] &= \mathbb E\left[\left|\frac{2e^{\theta S_n}}{(e^\...
0
The answer is a log-transformation, since biomass is strictly positive and the data were apparently right-skewed.
1
Indeed, by the Law of Total Variance: for any $Z, W$
$$V(Z) = E[V(Z\mid W)] + V[E(Z \mid W)]$$
Let $Z = Y - E(Y\mid X)$, and $W=X$.
Then
$$E(Z \mid W) = E\Big[Y - E(Y\mid X) \mid X\Big] = E(Y\mid X) - E(Y\mid X) = 0$$
So
$$V[E(Z \mid W) = V[E(Y - E(Y\mid X) \mid X) = V(0) = 0 $$.
So we are left only with the term
$$E[V(Z\mid W)]$$
Remember that $E(...
1
Yes, it is enough. If two statistics are in 1-1 correspondence $T_1(\mathbf X)=h(T(\mathbf X))$ with 1-1 function $h$, then
$$
\mathbb E_\sigma[g(T_1)]\equiv 0 \iff \mathbb E_\sigma[g(h(T))]\equiv 0 \iff g(h(T)) = 0 \text{ a.s. } \iff g(T_1)=0 \text{ a.s.}
$$
Step-by-step:
You can say that $T_1=\log(\prod^n_{i=1}(x_i-x_i^2))$ is complete. It means that ...
0
So you have $y = X\beta + U$ where $X = \begin{bmatrix}1 & 0\\ \vdots & \vdots \\ 1 & 0 \\ 1 & 1 \\ \vdots & \vdots \\ 1 & 1 \end{bmatrix}$ with $n_1$ 0's and $n_2$ 1's in the dummy variable vector.
Then, find $\begin{bmatrix} \hat{\beta}_{0} \\ \hat{\beta}_{1}\end{bmatrix}=\hat{\beta} = (X'X)^{-1}X'y$.
Note that $X'X = \begin{...
0
Here is a classical counterexample:
Let $\Pr(z_n=0)=(n-1)/n$ and $\Pr(z_n=n)=1/n$. Then, for any $\epsilon$,
$$\Pr[|z_{n}-0|<\epsilon]=\Pr(z_n=0)=(n-1)/n\rightarrow 1,$$
i.e. $z_{n}\to_p0$. However, $$E(z_n-0)^2=n^2\cdot 1/n\rightarrow\infty,$$
so that
$z_{n}\nrightarrow_{\mathrm{m.s.}}0$.
The reason you cannot just apply the expected value operator ...
1
You can start by expressing $\hat{\beta_0}$ as a linear combination of $y_i$, similar to $\hat{\beta_1} = \sum{k_iy_i}$:
$\hat{\beta_0} = \bar{y} - \hat{\beta_1}\bar{x} = \frac{1}{N}\sum{y_i}-\sum{k_iy_i}\bar{x}=\sum{(\frac{1}{N}-k_i\bar{x})y_i}=\sum{l_iy_i}$
1
We can look at a more general case that I believe will help solve this problem.
\begin{eqnarray*}
P(X|X+Y=n) &=& \frac{P(X=x, Y=n-x)}{P(Z=n)}\\
&=& \frac{P(X=x)P(Y=n-x)}{P(Z=n)} \\
&=& {n \choose x} \left( \frac{\lambda_1}{\lambda_1+\lambda_2} \right)^x \left( \frac{\lambda_2}{\lambda_1+\lambda_2} \right)^{n-x}
\end{eqnarray*}
Which ...
0
Yule Walker equations uses autocorrelation function to estimate the parameters in the linear time series model. It's not used for solving ACF. But, you also need to calculate ACF values, in which you use method of moments method to estimate it from data you have.
2
The argument is incorrect: it is not because
$$\int_{-\infty}^\infty g(T(x))\exp(\frac{-x^2}{2\sigma^2} + \frac{x\alpha}{\sigma})\text{d}x=0$$that $g\circ T$ is necessarily zero. (The argument does not even use the specific functional form of $T$.) Furthermore, as pointed out by @whuber, the integral in your approach should be on $\mathbb R^n$ rather than $\...
2
It looks like you are trying to figure out how to prove that one thing is not a function of another thing. This is generally quite simple using a proof by counter-example. For any data vector $\mathbf{x}$, let $S(\mathbf{x}) \equiv \sum x_i$ denote its sum. If $T$ can be written as a composite function of $S$ then this would mean that $S(\mathbf{x}) = S(\...
2
The information provided by the statement "Suppose the probability that the student will finish the exam in less than $x$ hours is $x/2, ...$ refers to the cumulative distribution function of the amount of time required to finish the exam, not the probability distribution function. With a one-hour time limit, $x$ cannot be greater than $1$, at which time we ...
0
W >= X^tX imply that W^-1 <= (X^tX)^-1
0
The key is to use : W >= X^tX imply that W^-1 <= X^tX and you can multiply by
X^tX (S.D.P) no problem the inequality still work ....
Note that (A >= B imply that A - B is S.D.P)
0
Since this is a self study question I will give you some hints, instead of the complete answer.
I don't know calculus, so we have to do it logically. First, having uniformly distributed predictions on the range [0, 2/3] and [1/3, 1] is for the sake of AUC the same as having 2 predictions for each class, let's say
P_A = 1, 2 and
P_B = 2, 3.
Method 1, ...
2
The result you are trying to prove is called the Gauss-Markov theorem, and there are a number of available proofs you can find with a quick internet search. It can be proved using some simple matrix algebra. Since your goal is to prove the theorem yourself, I will not give you the full proof, but hopefully I can get you started, and give some general tips ...
1
Your new density is: $$h(x)=pf(x)+(1-p)g(x)$$
So, by just substituting, you can calculate each moment and find the variance:
$$\begin{align}E[X]&=\int_{\mathcal{X}} x (pf(x)+(1-p)g(x))dx=p\int_{\mathcal{X}}x f(x)dx+(1-p)\int_{\mathcal{X}}x g(x)dx\\&=p\mu_1+(1-p)\mu_2\end{align}$$
Using similar approach for the second moment:
$$\begin{align}E[X^2]&...
1
Hint: You have specified that density $f$ holds with probability $p$. This should tell you something important about the distribution of the indicator $I$.
0
Theoretically speaking the outcomes of an experiment (experiment = a random procedure) can be numerical (i.e rolling a die) or can be mapped to numbers by the designer (i.e flipping a coin, with outcomes 1=head and 0=tail).
This numerical representation of the outcomes defines the random variables.
In these examples, we can tell that there is some chance ...
0
It always take $7$ min to succeed, which happens once.
It takes an average of $10$ min to fail.
The probability of failing is $2/3$ on each try, or $(2/3)^n$ for n tries.
So the total time is ($7 + 10\cdot \sum_n (2/3)^n$) min.
Which is $7$ min + $2 \cdot 10$ min $ = 27$.
2
Using the second equation that you wrote, you want to prove that
$$\frac{P(A \cap B)}{P(A \cup B)} \le \frac{P(A \cap B)}{P(A)}$$
If $P(A \cap B)=0$, then we are done. Suppose not, then this is equivalent to
$$P(A) \le P(A \cup B).$$
Try to prove this last inequality by showing that a set is another set's subset.
1
Guide:
We are not told if the rounds are independent of each other.
Suppose that they are independent and there is no draws.
Then the probability that $A$ wins against $B$ given $B$ won against $C$ is equal to the probability that $A$ wins against $B$, $0.7$.
For the third case, assuming it means a user win exactly $1$ match. Suppose the number of wins ...
0
Depending on the number of missing items and the number of auxiliary variables you could do multiple imputation with for example the package mice
5
Typical way to solve this:
The ant will either take path a and finish or take path b or c and be back in it's starting position.
Let $k$ be the number of times that the ant already has taken path b or c. Let $T_k$ be the expectation value for the time to finish for an ant that took already $k$ times the path. Then the expectation value for an ant with $k$ ...
2
Commenting on @Igor F.'s post (not enough reputation in this subforum to simply comment):
Both terms are geometric series:
$\sum_{i=0}^\infty i\cdot (\frac{2}{3})^i \overset{q=2/3}{=}\sum_{i=0}^\infty i\cdot q^i=q \frac{d}{dq}\sum_{i=0}^\infty q^i=\frac{q}{(1-q)^2}, \text{ for } |q|<1$, so for $q=\frac{2}{3}$ this equals $6$.
$\sum_{i=0}^\infty (\frac{...
24
$$T=7/3+(8+T)/3+(12+T)/3=9+2T/3$$
$$T/3=9$$
$$T=27$$
From start point, each of 3 paths are equally possible. Two paths lead you back to start point.
5
Let me join with an explanation which, to me at least, seems even easier:
First, observe that the paths $B$ and $C$ can be joined into a single path $X$, with the passage time equal to the mean times of $B$ and $C$, i.e. $10$ minutes. This is due to the fact that these paths have the same probability. If they didn't, we'd need to take a weighted average. ...
8
Hello donkordr and welcome to CV.
No, the mean does not answer the question unfortunately.
You can read your problem as a Markov Chain with two states: woods, and ant house.
Now, your transition probabilities are:
From the woods: (does not really matter as it's the goal) $p=1$ to stay in the woods
From the ant house: $p_1=\frac{2}{3}$ to stay at the ant ...
10
To answer this question, you have to sum over all possible paths the ant can take, and get the duration of that path, multiplied by the probability of taking that path. That is,
$$
E[T] = \sum_{\text{path} \in \text{possible paths}} p(\text{path}) T(\text{path})
$$
Every possible path takes Passage $A$ only once, but can take passages $B$ and $C$ any number ...
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