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awarded  Curious
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Jun
18
comment 3 random numbers to describe point on a sphere
The link here: math.stackexchange.com/questions/44689/… , may be informative for your application.
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awarded  Notable Question
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awarded  Popular Question
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awarded  Yearling
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awarded  Notable Question
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awarded  Popular Question
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comment How Do I Get This Joint Density Function?
I think there is a deterministic relationship between $X$ and $Y$. Therefore, the joint distribution of $X$ and $Y$ will take non-zero values along the indices which map to and fro each other, and will take zero values otherwise.
Jul
28
comment Conditional Probability, Lack of Dependence on a Parameter
The way I see it is: If the conditional distribution of $Z$ given $X$ and $Y$ is the same as the conditional distribution of $Z$ which depends only on $Y$, then $X$, $Y$, $Z$ form a Markov chain in that order. If $p(Z\mid X,Y) = p(Z\mid Y)$, then the Markov chain is formed. If $Z = f(Y)$ and $Y$ is known, then our characterization of $f(Y)$ knowing the additional information $X$ is the same as the characterization not knowing $X$ since $Z$ is a deterministic function of only the random variable $Y$.
Jul
28
accepted Conditional Probability, Lack of Dependence on a Parameter
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awarded  Revival
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awarded  Popular Question
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comment $(A\cup B)\cap C = A\cup(B\cap C)$ if and only if $A\subset C$
@SjoerdSmaal : Hi, to solve the counterpart, it is also possible to use table method in this question : math.stackexchange.com/questions/435433/… . Whenever the two sides of the counterpart match, it is desired to show $x \in A \Rightarrow x \in C$ is a true statement.
Jun
18
answered $\mathbb E[\frac{\partial}{\partial\theta}\log f(X;\theta)]^2$ and $\mathbb E[\frac{\partial^2}{\partial\theta^2}\log f(X;\theta)]$
Jun
16
comment $\mathbb E[\frac{\partial}{\partial\theta}\log f(X;\theta)]^2$ and $\mathbb E[\frac{\partial^2}{\partial\theta^2}\log f(X;\theta)]$
If it helps (and without giving the answer outright), the probability distribution function has a well-known name and is oft-analyzed, and the specific expectations you are trying to compute are also well-known and oft-analyzed.
Jun
12
answered Naive bayes: Log odds derivation
Jun
10
answered Difficulty in proving a function is convex
May
29
comment Naive bayes: Log odds derivation
The reasoning from step 17 to step 18 is the same as the reasoning that leads equation 3 to equation 6, in the referenced document.
May
19
comment Conditional Probability, Lack of Dependence on a Parameter
@snarski: There is no requirement that $X$ and $Y$ are independent.