New answers tagged

1

This is just an extended comment. You might want to try a symbolic computation program such as Mathematica, Maple, or MATLAB. Here are the results for the mean and variance using Mathematica. (And maybe there is some simplification available?) dist = TransformedDistribution[Abs[x], x \[Distributed] VarianceGammaDistribution[\[Lambda], \[Alpha], \[Beta],...


1

This is related to Cover's theorem. A nice summary by Emin Orhan is given here. Ps: I would post this in a comment but don't have enough reputation.


1

It’s correct except the second because it’s either $Y_1=b, Y_2=c$ or $Y_1=c, Y_2=b$. That makes $2a(1-a)$. Your intution is correct, you’ll write the cases that lead to a specific outcome and accumulate the probabilities. Here, $$P(Y_1=b, Y_2=c)=P(Y_1=b)P(Y_2=c)=a(1-a)$$ A quick check: If you sum all three probabilies, it makes 1.


2

The problem with your reasoning is that you don't take into account the probability of the second card to be a heart or not (p = 0.5). P(A|B)=P(A|red and heart on second) + P(A|red but not heart on second draw) = (12/51)x0.5 + (13/51)x0.5 = 25/102.


2

Using intuition, knowing that the first card was red doesn't change the probability of choosing a heart that much, so it seems wrong that it could be almost double. $P[A|B \text{ or } C]$ is not necessarily equal to $P[A|B]+P[A|C]$, even if $B$ and $C$ are disjoint. Suppose $B$ and $C$ are disjoint, like they are in the example where $B$ means the first card ...


0

In the future please be more careful when asking your question. The screenshot of the problem is missing so much context; how do you expect others to know how $N_t$ and $S_{N_t}$ are defined? Furthermore there are careless typos throughout your professor's solution (writing $T$ sometimes instead of $t$, missing equals signs, misplaced parentheses, ...


1

You can estimate $N$ by maximum likelihood. I will write $N=3 K$, and call the $K$ sets for colors. You are pulling objects at random without replacement from the jar, until the first time you draw an already seen color. In your experiment that happened at draw 6. Write $X+1=6$, so $X$ is the last draw we see a new color. Define $$ A_x=\{\text{Draws $1,2,\...


3

It is tempting to think so, but a simple counterexample with a discrete probability distribution shows why this is not generally possible. Let $(X,Y,Z)$ take on the eight possible values $(\pm1,\pm1,\pm1).$ Let $0\le p\le 1$ be a number and use it to define a probability distribution $\mathbb{P}_p$ as follows: $\mathbb{P}_p=1/8$ whenever $Y+Z\ne 0.$ $\...


1

Thank you to Arya. $P(A) = 0.15$ $P(B) = 0.6$ $P(B|A) = 0.8$ $P(A|B) = \frac{0.15*0.8}{0.6} $ $P(A|B) = 0.2$


2

As suggested by the comment of @whuber, since the package documentation does not tell precisely what is plotted, the answer is to be found in the code of the plot method for the S3 class "gpd" which is the class of fitmodel object in the code chunk. The answer is on the line #270 of the R code. The shown points are $[i, \, e_i]$ with $$ e_i := \...


3

Since the wrapped Cauchy distribution is made of the superposition of truncated scaled Cauchy distributions$^1$ translated by $2n\pi$: $$f_{WC}(\theta;\gamma)=\sum_{n=-\infty}^\infty \frac{\gamma}{\pi(\gamma^2+(\theta+2\pi n)^2)}\qquad -\pi<\theta<\pi$$ it can be simulated by Simulating a regular Cauchy variate $X$ with scale$^2$ $\gamma$ Translating ...


1

Try covariate balancing, i.e. learn weights $w_i, u_j$ such that $\sum_{i} w_i X_i = \sum_j u_j Y_j$ and $\sum_{i} w_i X_i^2 = \sum_j u_j Y_j^2$ for two of the datasets $\{X_i\}$ and $\{Y_j\}$, then sample using propabilities proportional to the weight. This should balance the mean and the variance at least.


1

I'll refresh future readers on the variables you defined. $A$ is whether the callee becomes a new customer. $B$ is whether the callee has used a rival company in the previous year. The information you've been given is: $P(A) = 0.15$ $P(B \mid A) = 0.8$ $P(B) = 0.6%$ You tried to compute $P(B \mid A)$ by multiplying $P(B \mid A) \times P(A)$. That's not ...


0

OK, I'll bite (though this is not strictly a programming question, it is also not really a question about statistics, but more of a combinatorial puzzle, and a nice one at that) I'll re-formulate the question a bit: suppose each kid in a class of n pupils draws a real number from a continuous (not necessarily normal) distribution, once every day on d ...


0

A table like this is typically visualized in a calibration plot, also called a reliability diagram. (If you search for this term, you may want to exclude the term "chemistry", because a "calibration curve" is also used in analytic chemistry, and it's a different thing.) You start out by binning the interval $[0,1]$, as in the "range&...


2

You are correct: for a transformation $T$ we have $$ P(T(X) = a) = P(X \in T^{-1}(\{a\})) = \sum_{x \in T^{-1}(a)} f_X(x). $$ If $T$ is invertible then $$ P(T(X)=a) = f_X(T^{-1}(a)) = (f_X\circ T^{-1})(a) $$ so we are indeed just plugging $T^{-1}(a)$ into the density of $X$. As to why the continuous and discrete cases are different, in one sentence I'd say ...


0

I think you've mostly interpreted the table correctly. Horses are binned by their betting odds of winning, and the "exp" column seems to show the mean probability of winning among the horses in each bin (according to the betting odds). So, for horses that are unlikely to win according to the betting odds (0<p<0.01), they have, on average, ...


1

One of the most popular ways to learn a (deterministic) function is Gaussian Process (GP) regression. It is commonly phrased in the Bayesian framework so that our 'prior beliefs' are $$ f(\cdot) \sim GP(m(\cdot), C(\cdot, \cdot))$$ where $m(\cdot)$ represents our prior expected value of $f(x)$ for any $x$. It's usually a 'rough and ready' approximation. E.g. ...


2

For the first part of your question we have that: We have that $Z\geq 0$ thus the domain can be decomposed as $Z\geq 0= (Z\leq \theta \mathbb{E}[Z]) \cup (Z> \theta \mathbb{E}[Z])$ $$\mathbb{E}[Z] = \int_{Z\geq 0} z p(z)dz = \int_{Z\leq \theta \mathbb{E}[Z]} zp(z)dz + \int_{Z> \theta \mathbb{E}[Z]}zp(z) = \\ \mathbb{E}[Z\times 1_{Z\leq \theta \mathbb{E}...


1

Prediction usually refers to the outcome, in this example whether team A wins or loses a particular game against team B. The 95% chance that team A will win is referred to as the estimated probability. Let's consider the simple example where there is no possibility of tie (e.g., sudden death). There are only two outcomes: A - Team A wins, and B - Team B wins ...


0

In my mind the most intuitive role of marginal likelihood is indeed as a normalization factor. I'll elaborate more on this. The bayes theorem is: $$ P(\phi|X) = \frac{P(X|\phi)P(\phi)} {P(X)} $$ Now let's explore the numerator components: $$P(X|\phi)P(\phi)$$ Prior multiply by the likelihood using a specific parameter - How well can we explain the data using ...


4

In Bayesian statistics, the marginal likelihood $$m(x) = \int_\Theta f(x|\theta)\pi(\theta)\,\text d\theta$$ where $x$ is the sample $f(x|\theta)$ is the sampling density, which is proportional to the model likelihood $\pi(\theta)$ is the prior density is a misnomer in that it is not a likelihood function [as a function of the parameter], since the ...


1

Non-probabilistic machine learning models do not handle uncertainty about the parameters. They simply return point estimates for the parameters. You may use additional techniques (e.g. bootstrap) to learn something about the uncertainty. Many of the available solutions (e.g. using dropout also at the prediction time) are thought of as approximations of the ...


1

The covariance here will not be zero because the sample mean includes the variable $X_1$ as part of it; a higher values of the latter variable is going to give you a higher sample mean and a lower value is going to give you a lower sample mean. Consequently, we would expect these two things to be positively correlated. Now, using the fact that the ...


1

I found that there was a correlation of -0.5109730443013045 between the number drawn and its frequency, and a p-value of 0.090274 for that correlation. For a binomial test for a number being odd or even, I got a p-value of 0.00621804354. When I adjusted the odd numbers for the difference in means (that is, added (even mean-odd mean) to the frequencies of odd ...


2

Looking at the dodecahedron, if we roll it and discard everything higher than six, then this "truncated dodecahedron" functions precisely as a standard six-sided die: every roll between 1 and 6 has equal probability $\frac{1}{6}$. We thus concatenate a number of rolls from the cube (where every roll between 1 and 6 has equal probability $\frac{1}{6}...


1

From Wikipedia: A Bayesian network (also known as a Bayes network, belief network, or decision network) is a probabilistic graphical model that represents a set of variables and their conditional dependencies via a directed acyclic graph (DAG). Bayes' rule is used for inference in Bayesian networks, as will be shown below. A better name for a Bayesian ...


0

In response to I did not fully understand why we are multiplying the intermediate conditionals i.e.: $...Pr(Burglar|Storm=T)xPr(Cat|Storm=T)...$ and Is there a logical explanation for the multiplication of the intermediate probabilities - or is it just a trick? Denoting each random variable with their first letter, the model represented using the ...


0

In Bayesian Networks, there are three types of junctions, https://en.wikipedia.org/wiki/Bayesian_network. In your case, you have a Fork junction, $Storm \rightarrow Burglar \ \& \ Storm \rightarrow Cat.$ Each junction case has some conditional or unconditional independence properties. For your case, the Fork junction has $$Burglar \perp Cat \ | \ Storm $...


4

An obligatory Dilbert comic: If you have a random number generator that at random generates "4" with probability $p$, the probability of observing it $n$ times in a row is $p^n$, assuming that the draws are independent. Notice that the more times you observe "4", the smaller the probability gets, but it will never go down to zero. There ...


0

Checking for the randomness can be viewed as "discover patterns". In your random generator example, we can calculate the probability of certain events (for example consecutive 4 for 10 times) and conducting experiment to verify our assumption. For example, we know certain thing is very less likely to happen and it is happening all the time (say ...


10

You mentioned that odd-even pattern, so let's investigate that. Category Observed Expected # Expected odd 92 110 50% even 128 110 50% And test with just these two categories.... Chi squared equals 5.891 with 1 degrees of freedom. The two-tailed P value equals 0.0152 That would generally be considered significant This kind of result only happens fifteen ...


5

As an addition to the other answers, let me offer you a visual way to inspect the differences between the expected and observed frequencies: A (hanging) rootogram, invented by John Tukey (see also Kleiber & Zeileis (2016)). In the figure below, the square roots of the expected counts are displayed as red dots. The square roots of the observed frequencies ...


14

Following @doubled's (+1) chi-squared test, a remaining question is whether 220 draws from the machine are enough to detect an actual small bias. Maybe the odd numbered balls are a bit heavier, lighter, or less round in such a way that they are slightly more likely to be be drawn. Maybe the true probability distribution is $(6,4,6,4, 6,4,6,4, 6,4,6,4)/60.$ ...


23

To determine whether the results seem to indicate some shenanigans were afoot, we can test it! To begin, we need to specify what our null hypothesis is. I'll take the moment here to stress (as subsequent answers have pointed out in more detail) the importance of forming hypotheses before seeing the data we will use for testing this hypothesis. For example ...


0

Here's what I ended up doing. My original question may have been a little ambiguous, but this solution gave me what I needed. Rather than using the probabilities that Player A would beat Player B and so on, I assumed each player would have a "score" $S_i$ sampled from some distribution $D_i$. So $S_A \sim D_A$, $S_B \sim D_B$, and $S_C \sim D_C$, ...


0

Agree with the top answer that cross-entropy, as a distribution dissimilarity measure, may be more narrowly applicable to situations when we are comparing an estimated probability distribution (q) against the true probability distribution (p). If we only consider the true probability distribution p, its entropy is the expected value of the corresponding log-...


1

Goal: show $P(Y_n \le t) \to P(X \le t)$ [for $t$ at which $F_X(t):=P(X \le t)$ is continuous]. Hints: $P(Y_n \le t) = P(Y_n \le t, |X_n - Y_n| > \epsilon) + P(Y_n \le t, |X_n - Y_n| \le \epsilon)$. $P(Y_n \le t, |X_n - Y_n| > \epsilon) \le P(|X_n - Y_n| > \epsilon)$. What does the right-hand side converge to? $P(Y_n \le t, |X_n - Y_n| \le \epsilon)...


1

The first thing to try is always Chebyshev's inequality -- especially with well-behaved distributions where moments are likely to be enough to decide the question. $$\mathrm{var}[S_p]=\frac{1}{p^2}\sum_{i,j=1}^p \mathrm{cov}[a_iU_i^2,a_jU_j^2]$$ Now, correlations are bounded by 1,so: $$\left|\mathrm{cov}[a_iU_i^2,a_jU_j^2]\right|\leq \sqrt{\mathrm{var}[...


1

Turning my comments into an answer: These Venn diagrams are a good starting point for understanding how the probabilities of two events interrelate. If two events are indeed independent, then you can compute $P(A \cap B) = P(A) \times P(B)$. In fact, independence is defined in terms of these joint and marginal probabilities. In the example you've shown, they'...


2

how would you know with 100% certainty it wasn't random? You wouldn't. This gets into why there are many different probability interpretations.


12

No, it is not. In order for that to be true, $A$ and $B$ should be conditionally independent given $\theta$.


0

I read that if $X_1, X_2$ are 2 random variables with different excess kurtosis, their joint distribution cant be elliptical. Is there an intuition or proof of that? It is not very clear to me. In my opinion the best intuition is the follow. Elliptical distributions deal with some stylized facts in finance returns, like: fat tails and tail dependence. ...


1

I feel your question concerns more the interpretation than anything else, because there seems to be a hidden assumption that Your telephone number is randomly selected -> they call Under this assumption I would rephrase the question as A: they call. B: you're not at home at dinnertime when they call. Now your being home at dinnertime is independent of ...


1

Time-to-ruin in the discrete-time random walk: From your question, I take it that you are referring to a discrete-time version of the gambler's ruin problem. Without loss of generality we can consider the wealth of the gambler denominated in units equivalent to the betting amount. We will assume that the wealth of the gambler is a positive whole number of ...


2

I'll give a fairly simple example, but this should give you an idea for how the idea is useful in other scenarios. Suppose you fit a linear regression to some data which is strictly positive. I.e. $Y > 0$ (e.g. $Y=\text{house prices in your area}$ $x=\text{square footage of house}$). Fitting a linear regression might give us predictions which are ...


3

I had the same issue once upon a time. There is a very nice, clearly spelled out example in Manning & Schütze's Foundations of Statistical Natural Language Processing 178-181. In particular, the example on p. 179 considers an example where the corpus contains 14307668 tokens. It shows that for two words w1 and w2, if w1 appears 42 times in the corpus, w2 ...


0

By "$f_{X_1+\ldots, + X_n}(x) \sim f_{X_1}(x) \quad \text{for }x\to \infty$", do you mean that the pdf of $\sum X_i$ minus the pdf of $X_1$ converges to 0? I don't think that's always true. I'm imagining a counterexample where $f_{X_1}$ has an infinite number of increasingly narrow peaks where the pdf takes value 1, separated by troughs. Suppose ...


2

The antiderivative of a Gaussian function has no closed form, but the integral over $\mathbb{R}$ can be solved for in closed form: \begin{align} \int_{-\infty}^{\infty} \exp(-x^2) dx = \sqrt{\pi} . \end{align} Since $\exp(-x^2)$ is an even function (graph is symmetric about the $y$-axis), we can split this into two equal parts $$ \int_{0}^{\infty} \exp(-x^2) ...


2

It doesn’t really matter which notation, though there’s a subtle reason you might prefer Murphy’s. We could create an arbitrary variable $u$ of the correct type and measure the loss $\ell(y, u)$. Now Murphy’s version of the loss is $- \log p(y \mid u)$. Your proposed formula requires you to ‘remember’ how $u$ was created. That doesn’t make it wrong, but ...


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