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62 votes

Shouldn't the joint probability of 2 independent events be equal to zero?

There is a difference between independent events: $\mathbb P(A \cap B) =\mathbb P(A)\,\mathbb P(B)$, i.e. $\mathbb P(A \mid B)= \mathbb P(A)$ so knowing one happened gives no information about ...
Henry's user avatar
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20 votes

Why don't we see Copula Models as much as Regression Models?

The first and most important reason is that standard regression models had a one to two-hundred year headstart on copula models (depending on exactly where you count the genesis of regression models ...
Ben's user avatar
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19 votes
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Proof that joint probability density of independent random variables is equal to the product of marginal densities

By definition, the random variables $X_1,\dots,X_n$ are independent iff $$ \Pr(X_1\in B_1,\dots,X_n\in B_n) = \Pr(X_1\in B_1)\dots\Pr(X_n\in B_n) $$ for every choice of Borel sets $B_1,\dots,B_n$. ...
Zen's user avatar
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18 votes
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intuitive difference between joint probability and conditional probability in this example

You actually had your answer right there. $P(H=hit)$ is the marginal probability. It reads "The probability of getting hit.". It is the proportion of people that got hit crossing the street, ...
wiwh's user avatar
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14 votes

Why don't we see Copula Models as much as Regression Models?

A reason might be that regression and copulas do not answer the same question. Copulas are about the joint distribution while regression is about a conditional distribution or just the conditional ...
Richard Hardy's user avatar
13 votes

Shouldn't the joint probability of 2 independent events be equal to zero?

What I understood from your question, is that you might have confused independent events with disjoint events. disjoint events: Two events are called disjoint or mutually exclusive if they cannot ...
Umair Rafique's user avatar
12 votes
Accepted

Joint distribution in layman's terms

As a concrete example, suppose I toss a coin and roll a die one after the other. As you know, there is a probability distribution associated with the outcomes of both (discrete uniform distributions, ...
Will's user avatar
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11 votes
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Minimum CDF of random variables

Let $x$ by any number. Consider the event $\min(X,Y)\le x$. It can be expressed as the union of two events $$\min(X,Y)\le x = (X\le x) \cup (Y \le x),$$ shown by the overlapping yellow and green ...
whuber's user avatar
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11 votes
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What is the number of parameters needed for a joint probability distribution?

It takes $3\times 2 \times 2 \times 3 = 36$ numbers to write down a probability distribution on all possible values of these variables. They are redundant, because they must sum to $1$. Therefore ...
whuber's user avatar
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11 votes
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Joint probability measure

Joint Distributions and Expectation In general, the joint distribution of random variables $X$ and $Y$, defined on a common probability space $(\Omega, \mathcal{A}, \mathbb{P})$ and taking values in ...
Artem Mavrin's user avatar
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11 votes

Why don't we see Copula Models as much as Regression Models?

A short answer is that in practice for many applications we don't need the joint probability distributions. A cynic would say that it's also because the users don't event understand what is a joint ...
Aksakal's user avatar
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10 votes

Difference between joint density and density function of sum of two independent uniform random variables

Following up on Glen_b's answer, and in an attempt to dumb it down a bit more, the following illustrations shows how the bivariate or joint pdf of $X$ and $Y$, both independent and standard uniform ...
Antoni Parellada's user avatar
10 votes

Derivative of the Joint Distribution Interpretation

The first-order partial derivatives of a multivariate joint distribution function can be considered as giving the density of the differentiated variable, jointly with the cumulative probability of the ...
Ben's user avatar
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10 votes
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How to derive the joint distribution in these 3 models?

All of these DGPs generate a standard bivariate Gaussian with: $$ \begin{pmatrix} X \\ Y \end{pmatrix}\sim N\left(\begin{pmatrix} 0 \\ 0 \end{pmatrix},\begin{pmatrix} 1 & \sigma_{xy} \\ \sigma_{xy}...
Carlos Cinelli's user avatar
9 votes
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What does it mean to factor a joint distribution?

If I am understanding the passage and table correctly, there are essentially two answers: mathematical and qualitative. Qualitatively, the "different aspects of forecast quality" is essentially the ...
GeoMatt22's user avatar
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9 votes
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Maximum of a probability vector distributed as a Dirichlet variate

I am not sure there is a closed-form solution for the distribution of $p_{(k)}$ when $(p_1,\ldots,p_k)\sim\text{Dir}(\alpha_1,\ldots,\alpha_k)$ and the $\alpha_i$'s are different. At least, it ...
Xi'an's user avatar
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9 votes
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Distribution given sum

It can be instructional and satisfying to work this out using basic statistical knowledge, rather than just doing the integrals. It turns out that no calculations are needed! Here's the circle of ...
whuber's user avatar
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9 votes

How to make random draws from an unspecified distribution?

Presumably $w_1+w_2=1$ and $w_1,w_2 \geq 0$, so $f$ is a convex combination of $f_1,f_2$ and therefore a valid distribution (a mixture of $f_1,f_2$). Generate a Bernoulli($w_1$) random variable (i.e. ...
Batman's user avatar
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9 votes
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Uncorrelatedness + Joint Normality = Independence. Why? Intuition and mechanics

The the joint probability density function (pdf) of bivariate normal distribution is: $$f(x_1,x_2)=\frac 1{2\pi\sigma_1\sigma_2\sqrt{1-\rho^2}} \exp\left[-\frac z{2(1-\rho^2)}\right], $$ where $$z=\...
user158565's user avatar
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9 votes

Product of 2 Uniform random variables is greater than a constant with convolution

Some hints: Geometrical approaches are much easier for uniform RVs, but the general approach is to integrate the joint PDF in the region that satisfy $XY>\alpha$. The integral will basically look ...
gunes's user avatar
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9 votes

Product of 2 Uniform random variables is greater than a constant with convolution

Multiple answers and partial answers here, some for the more general problem of multiplying $n$ independent standard uniform random variables. For $n = 2,$ the PDF of the product $Z = XY$ is $f(z) = -...
BruceET's user avatar
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9 votes
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Is there a difference between marginal likelihood and likelihood of a marginal distribution?

The problem here is that although the observations $x_1,...,x_n$ are independent conditional on $\theta$, they are not independent conditional on $\alpha$ instead, so as a general rule: $$f(\mathbf{x}...
Ben's user avatar
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9 votes
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Expectation of a multivariate random variable

It looks like you might be getting tripped up with the indexing and summations. Here's how to handle the 2d discrete example using the same approach you were trying to take. The same logic applies to ...
user20160's user avatar
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9 votes
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Finding the conditional expectation given the joint density function

If you know the equation \begin{align} E[Y|X \leq 1/2] = \frac{E[YI_{[X \leq 1/2]}]}{P[X \leq 1/2]}, \tag{1} \end{align} you can skip the step of finding the conditional density $f_{Y|X \leq 1/2}$ by ...
Zhanxiong's user avatar
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9 votes

For what kind of distributions could the joint distribution be determined uniquely by marginal distribution and correlation?

This occurs for distributional families determined by mean and variance I am assuming here that the joint distribution is assumed to be within some stipulated distributional family (e.g., the normal ...
Ben's user avatar
  • 129k
8 votes

Minimum CDF of random variables

Since it says so in the title (though not repeated in the body of the question), I'm going to assume that $X$ and $Y$ are independent; otherwise, we can't say much. One of the key properties of ...
Danica's user avatar
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8 votes
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Do conditionals determine joint distribution?

[The following is an excerpt from our book Monte Carlo Statistical Methods on the reason why the joint distribution is determined by its conditional distributions, a result known as the Hammersley-...
Xi'an's user avatar
  • 107k
8 votes

Prove 2 identical uniform's are independent by computing the joint distribution

The joint distribution of $(A^-,C^ -)=(A-B,C-B)$ is given by its density \begin{align} f(a^-,c^-)&=\int f_A(a^-+b)f_C(c^-+b)f_B(b)\,\text d b\\ &=\int_0^1 \mathbb I_{(0,1)}(a^-+b)\mathbb I_{(0,...
Xi'an's user avatar
  • 107k
8 votes
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KL divergence for joint probability distributions?

KL divergence is defined between two distributions, period. If this is marginal or joint distributions is immaterial. You want them to have the same support. So you do the same as in single dimension. ...
kjetil b halvorsen's user avatar
8 votes

Finding the conditional expectation given the joint density function

I would refer to the marginal distribution of (capital) $X$ rather than to the marginal distribution of (lower-case) $x.$ The former is the random variable, and those are the things that have ...
Michael Hardy's user avatar

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