22 votes

A and B are independent. Does P(A ∩ B|C) = P(A|C) · P(B|C) hold?

No this is not in general true, as you can see from a simple counter example: Toss two independent coins. Event $A$ is coin 1 head. $P(A)=0.5$ Event $B$ is coin 2 head. $P(B)=0.5$ Event $C$ is either ...
George Savva's user avatar
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13 votes
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Dropping condition from conditional probability

This is equivalent to $X$ and $Y$ being conditionally independent given $Z=z$. We have $$ f_{X|Y,Z}(x|y,z) = \frac{f_{X,Y,Z}(x,y,z)}{f_{Y,Z}(y,z)} =\frac{f_{X,Y|Z}(x,y|z)f_Z(z)}{f_{Y|Z}(y|z)f_Z(z)} =\...
Doctor Milt's user avatar
  • 2,672
6 votes

Can $X_1$ and $X_2$ be independent conditioning on $X_1+X_2$?

$X_1 ~|~ X_1+X_2$ and $X_2 ~|~ X_1+X_2$ are not independent. They are perfectly negatively correlated distributions.
krkeane's user avatar
  • 2,180
6 votes

For normally distributed random variables, if X is independent of Y and X is independent of Z, is X independent of max(Y,Z)?

If $X$ were not independent of $\max(Y,Z),$ that would seem to contradict a basic theorem that when two variables $U$ and $V$ are independent, then measurable functions of them $f(U),$ $g(V)$ are ...
whuber's user avatar
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6 votes

Prove or disprove : $P[A|B] = P[B]$, the A and B are independent? Is this right?

The error in the quoted reasoning is in $P[A|A] = P[A]$. Instead we should have: $$P[A|A] = 1 \neq P[A]$$ For example, let the event be '$A = \text{it rains}$'. Say you live in the desert where it ...
Sextus Empiricus's user avatar
5 votes
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Variance of the product of two conditional independent variables

We can follow a similar derivation, assuming conditional independence of $Y$ and $Z$ given $X$: $$\begin{align}\operatorname{var}(YZ|X)&=E[Y^2|X]E[Z^2|X]-E[Y|X]^2E[Z|X]^2\\&=(\operatorname{var}...
gunes's user avatar
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4 votes
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Expectation of potential outcomes formula

I assume that by "conditional independence assumption" you mean $$ Y_{1i}, Y_{0i} \perp D_i | X_i.$$ The trick is simply to condition on $X_i$. Conditional on $X_i$, the propoensity score $p(X_i)$ ...
baruuum's user avatar
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4 votes
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Independence and conditional distribution

You are considering a random sample of size $N$. Each draw $i$ performed independently of any other draw $j$. Each draw is a draw of a random vector $(Y_i,X_i)$. Here is an illustration of all the ...
Jesper for President's user avatar
4 votes
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How to reason about independence of combinations of events?

For inspiration, let's examine Venn diagrams of each set. Elements of $\mathcal{X}=A\cap B^c\cap D$ are (a) in $A;$ (b) not in $B;$ and (c) in $D$. This region is highlighted in yellow. Elements of ...
whuber's user avatar
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4 votes

Are two coin flips conditionally independent if we know that the coin is biased towards heads?

The quoted section is implicitly assuming that the event $C = \{ \theta > 0.5 \}$ is sufficient to fully describe the parameter, and so it attains conditional independence of the observable coin ...
Ben's user avatar
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4 votes
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Does mutual independence of X, Y, Z implies conditional independence of X and Y, given Z

Yes it does. If $X,Y,Z$ are mutually independent then you can say $p_{X,Y}(x,y) = p_X(x) \cdot p_Y(y)$ $p_{X,Y \mid Z}(x,y\mid z) = p_{X,Y}(x,y)$ $p_{X\mid Z}(x\mid z) = p_X(x)$ $p_{Y\mid Z}(y\mid ...
Henry's user avatar
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4 votes
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Intuition of conditional independence in DAGs

Flow of dependencies in DAGs are determined by the d-separation criteria, and two variables are d-separated if and only if every path (without considering the orientation of the arrows) is blocked. ...
DaSim's user avatar
  • 336
3 votes
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Given random variables $X,Y,Z$, under what conditions is $P(Y|X)=P(Y|X,Z)$?

No, it's not required. Being $Y$ and $Z$ conditionally independent on $X$ directly implies $$P(Y|X)=P(Y|X,Z)$$ because $$P(Y|X,Z)=\frac{P(Y,Z|X)}{P(Z|X)}=\frac{\overbrace{P(Y|X)P(Z|X)}^{\text{...
gunes's user avatar
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3 votes
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Order of Conditional Independence Tests

Jeremy's own answer is correct, but I would like to add some details, and why the order matters at all. The SGS Algorithm Suppose there is a set of variables $\mathcal{P}$ from which we wish to ...
Frans Rodenburg's user avatar
3 votes

Can $X_1$ and $X_2$ be independent conditioning on $X_1+X_2$?

It's possible if one of them is constant - for example if $X_1$ has a Bernoulli distribution and $X_2$ is always equal to zero.
fblundun's user avatar
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3 votes
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is it possible that $X_{j}$ and $X_{k}$ are independent of each other conditioning on $Z = f(X_1,\cdots, X_N)$?

Yes, it's possible. Suppose $X_i$ are Rademacher variables ($\pm 1$ with equal probability) and $Z=\sum_i X_i^2$. Then $Z$ is constant, so conditioning on it has no effect. Less trivially, suppose $Z=\...
Thomas Lumley's user avatar
3 votes

If $X$ and $Y$ are uncorrelated random variables, then under what condition is $E[X \mid Y] \approx E[X]?$

One situation where this is interestingly almost true is when $X$ and $Y$ are both projections from the same high-dimensional distribution, ie, $X=a^TZ$, $Y=b^TZ$ for high-dimensional $Z$. Hall & ...
Thomas Lumley's user avatar
3 votes
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Joint distribution where random variables always exist in the same orthant

It depends on what you mean by "$P.$" Let's interpret this as asking for the joint probability density function. Let the PDF of $X$ be $f.$ For instance, $f(\mathbf x) = (2\pi)^{-n/2}\exp(-...
whuber's user avatar
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3 votes
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Informative Censoring vs. Random Censoring vs. Conditionally Independent Censoring

Lagakos provided a formal definition of "noninformative censoring" in Biometrics 35: 139-156, 1979. It's the most general case in which the standard formulation of (partial) likelihood for ...
EdM's user avatar
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3 votes
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Notational confusion about conditional independence in Pearl 2009

I think $X, Y, Z, W$ should all be interpreted as random variables, and "$YW$" is not the product of $Y$ and $W$, but the random vector $(Y, W)$. Therefore, the expression means: \begin{...
Zhanxiong's user avatar
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2 votes
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Is the example distribution conditionally independent?

You already computed: $P(y_1=1|x)=0.6$ $P(y_1=0|x)=0.4$ $P(y_2=1|x)=0.4$ $P(y_2=0|x)=0.6$ $P(y_3=1|x)=0.3$ $P(y_3=0|x)=0.7$ The values of $P(Y|x)$ that you are asking about (i.e. the joint ...
suckrates's user avatar
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2 votes
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Conditional independence and joint distributions in graphical models

You're right about $𝑝(𝑎,𝑐|𝑏)=𝑝(𝑎|𝑏)𝑝(𝑐|𝑏)$, it is the definition of conditional independence. For the formula, without any information about the variables/events, we normally have the ...
gunes's user avatar
  • 56.9k
2 votes

show that A and B are independent

$p(D|A,B,C) = p(D|C)$, otherwise we couldn't decompose $p(A,B,C,D)$ like this in the first place. This reduces the fraction to : $p(A,B) = p(A)p(B)*\frac{p(D|C)}{p(D|C)} = p(A)p(B)$ Which is ...
jkm's user avatar
  • 2,155
2 votes
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How would I find $P(X \ne Y)$ given independent conditional probability mass functions?

Let's begin by solving the general problem of determining the chance that $X\ne Y$ when $X$ and $Y$ are independent random variables. It's almost always easier to subtract the chance that $X=Y$ from ...
whuber's user avatar
  • 321k
2 votes
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Bishop PRML Question 8.10: d-separation

Reading the conditional independences depicted on the graph, we have $$ p(a,b,c,d)=p(a)\,p(b)\,p(c\mid a,b)\,p(d\mid c). $$ First, using Fubini, we get \begin{align} p(a,b) &= \int \int p(a,b,...
Zen's user avatar
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2 votes

Bishop PRML Question 8.10: d-separation

You can show the desired result by showing that a and b are not d-seperated by the note d by using the traditional d-seperation criteria. a and b are d-seperated by d if every path from a to b is ...
Abm's user avatar
  • 352
2 votes
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Conditional Independence Example

However, these examples feel unnatural because the X′s are typically covariates and Y is typically a response (at least with the naive Bayes classifier) (...) it's not clear to me which data ...
Carlos Cinelli's user avatar
2 votes

Conditional independence: conditioning on an empty set of random variables

Yes, I would say, $X \perp\!\!\!\perp Y$ can be thought of as conditional independence, but conditioning on the full sample space $\Omega$. The probability of the empty set impossible event is zero, ...
kjetil b halvorsen's user avatar
2 votes
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Causality: Models, Reasoning, and Inference: Notation Question Concerning Graphoids

$\newcommand{\ci}{\!\perp\!\!\!\perp\!}$The notation $YW$ here stands for the set of variables $\{Y, W\}$. Thus, for instance, $(X\ci YW|Z)$ means$P(X, Y, W |Z) = P(X|Z)P(Y, W|Z)$, for all ...
Carlos Cinelli's user avatar
2 votes
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Is Pearson's chi-squared test of independence conditional on marginal distributions?

The "exactness" of Fisher's test depends on the assumption that each of the row totals and column totals (the "margins") is given as part of the experimental design. That's what's ...
EdM's user avatar
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