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Conditional Independence involving four events

When $A, B, C, H$ are events, this property seems wrong. For example, when $H = \Omega$, then it amounts to say "if $A$ and $C$ are independent, $B$ and $C$ are independent, then $A \cap B$ and $...
Zhanxiong's user avatar
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3 votes

Independent random variables with $X^2 + Y^2 =1$

If $X$ and $Y$ are independent and $X^2+Y^2=1$, then $X^2$ and $Y^2$ are almost surely constant. Indeed, let $U=X^2$ and $V=Y^2$. Then $U=1-V$ and the random variables $U$ and $V$ are independent ...
Davide Giraudo's user avatar
6 votes
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Independent random variables with $X^2 + Y^2 =1$

Example ... $X$ and $Y$ are independent, and both have the scaled Radermacher distribution $$ P(X=1/\sqrt2) = P(X=-1/\sqrt2) = 1/2. $$ Then $X$ and $Y$ are not constant, they are independent, but ...
GEdgar's user avatar
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Independent random variables with $X^2 + Y^2 =1$

Consider two random variables $X$ and $Y$. Assume for the sake of contradiction that $X$ and $Y$ are independent and satisfy the equation $X^2 + Y^2 = 1$ almost surely. This implies that the pair $(X, ...
Snowball's user avatar
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0 votes
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Prove the existence of $g$ such that $\mathbb E (e^{aX_t} ) = e^{g(a) t}$ for a Compound Poisson Process $X_t$

By the Tower Property: $$E[e^{aX_t}] = E[E[e^{a(Y_1 + \cdots + Y_n)}] | N_t = n]$$ And since $(Y_i)$ are iid, we simplify the above to: $$E[E[(e^{aY_1})]^n | N_t = n]$$ As $Y_i$ is independent of $N$, ...
FD_bfa's user avatar
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1 vote

Independence assumption for interarrival time

You have to keep in mind the separation between a physical process and a model. We observe some process in the real world (customers arriving at a server) and want to construct a model that helps us ...
msantama's user avatar
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1 vote

Independence of Sample Mean and Sample Variance

Yes it is a characterization for the normal distribution shown first by Geary (1936) that the sample mean and sample variance are independent if and only if the population distribution is normal. ...
Amir's user avatar
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1 vote

Does Independence hold for $\sigma$-Algebras Generated by Disjoint Subsets of an independent Sequence

I think I understand how to prove it now. Please let me know, if my reasoning is correct. Big thanks to @IzzakvanDongen. Let $\mathcal{A}$ and $\mathcal{B}$ denote the sets of all possible finite ...
MathMaestro's user avatar
4 votes

Does Independence hold for $\sigma$-Algebras Generated by Disjoint Subsets of an independent Sequence

I will show the argument when the subsets of $\mathbb{N}$ are finite. Let $A=\{a_1,...,a_n\}\subset \mathbb{N}$. Let $(X_n)_{n \in \mathbb{N}}$ be random variables. Then, $(X_{a_1},...,X_{a_n}):\Omega\...
Snoop's user avatar
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4 votes
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Prove that $Z_n$ is independent of $(Y_{i,n})_i$

First, note that $Z_n$ is $\sigma(Z_{n-1}, Y_{1,n-1}, Y_{2,n-1}, \ldots)$-measurable. By induction on $n$, this shows that $Z_n$ is $\sigma(Y_{i,j} : i\geq 1, 1\leq j < n)$-measurable. It now ...
Brian Moehring's user avatar
0 votes

Independence vs. Measurability

Assuming that X is non-trivial, we can find that $E[X|\mathcal{G}]=EX$, if X is $\mathcal{G}$-measurable, and $E[X|\mathcal{G}]=X$, if X is independent to $\mathcal{G}$. Thus, we can conclude that if ...
zs Tan's user avatar
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2 votes
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$X$ $Y$ are independent so are $f(X)$ and $f(Y)$ if $f$ is continuous

It's easier than that. If $X_1$ and $X_2$ are independent, if we set $Y_i = f(X_i)$ then for any two measurable sets $A,B$ we have $$P(Y_1 \in A, Y_2 \in B) = P(X_1 \in f^{-1} (A), X_2 \in f^{-1} (B)) ...
Jose Avilez's user avatar
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0 votes
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Equivalent definition of independent increments of a stochastic process.

The problem becomes easy if you assume $X_0=0$ surely, as you indicated in your comments. The main idea is to consider the vectors $(X_0, X_{t_1}, ..., X_{t_n})$ and $(X_0, X_{t_1}-X_0, ..., X_{t_n}-...
Michael's user avatar
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2 votes
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How to get this formula for expectation of continuous-time urn process

As a toy example, consider a Yule process $(Y_t, t\geq 0)$: It's law can be described as follows: $Y_0=1$, a.s. If $Y_t=n$, then the transition to $n+1$ happens at rate $n$. This is very similar to ...
David's user avatar
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