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We can have that $A,B$ are independently distributed but dependent on $C$. For instance, consider $A,B \sim \mathcal{N}(\mu =C,\sigma^2=1)$ where $A$ and $B$ are independent and identical normal distr …

To test two probabilities for independence one way is to check whether $P(A, B) = P(A) \cdot P(B)$. ... … Is there a standard way to assess for independence by taking into account the confidence intervals? Using confidence intervals is not a standard way to test for independence. …

Imagine a plot of the joint distribution of $X,Y$. The condition $X+Y <a$ can be depicted by a diagonal line $y=a-x$ as a boundary. The points below it satisfy the condition. Second image Next, cons …

You can compute an exact distribution by using a repeated convolution. Here is an example with r code: n = 10 ### a vector of probabilities for the number of rupees p = rep(0,n*5+5+1) p[1+c(0,1,3,5)] …

If the integrand $f(x,y)$ can be partitioned into a product of functions of the individual differentials $$f(x,y) = g(x) h(y)$$ then we can write a double integral as a product of single integrals. $ …

This sounds a bit like the idea behind collider bias, the reverse of confounding bias. Confounding bias: x and y are found to be correlated because both are caused by z. Collider bias: x and y are fo …

Fair independent coin The probability of event a (5th case is heads $H_5$) given event b (already 4 heads $H_4H_3H_2H_1$) $$\underbrace{P(H_5|H_4H_3H_2H_1)}_{\text {P(a given b)}} = \frac{\overbrace{P …

joint probability $Z=z$ and $J=j$ $$\begin{align} \\P(Z=z , J=j) &= f_j(Z) \prod_{i \neq j} 1-F_i(z)\\& = \lambda_je^{-\lambda_j z} \prod_{i \neq j} 1 - (1-e^{-\lambda_i z}) \\&= \lambda_j \prod_{i} …

The following is a counterexample Say the following events (for binary values of X, Y and Z) have equal probabilities. X Y Z probability 0 1 1 1/4 1 0 1 1/4 0 0 0 1/4 1 1 0 1/4 Then $P(X=1 …

The answer is no. An example is for a vector $X$ of size 1 and $Y$ of size 2. Let $X$ be Bernoulli variable with equal probability. Let $Y$ be distributed, with equal probability, among $(1,1)$ or $(0 …

Intuition by using more experts The case of 2 experts is a bit difficult (we explain later why). It becomes more intuitive when you consider a choice based on a majority in multiple expert advice who …

The answer below gives is first an interpretation of how 'unpaired' relates to independence. … An unpaired way of dependency Independence between samples occurs if the outcomes of the two variables are unrelated. …

If you are after visualizing the relationships, then you might use some network visualization algorithms. I did this once with the tags of cross validated (Disclaimer: I am not an expert at this). I m …

Covariance is zero for $X$ and $Y$ (and also for $A$ and $B$), and in the first case there is independence but in the second case there is no independence. …

But that doesn't imply independence. This makes your case an example of: Why zero correlation does not necessarily imply independence …