I Sum independent normal variables

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A finite sum of independent normal random variables is indeed normally distributed, and independence is a necessary condition for this property. Counterexamples involving dependent normal variables, such as those correlated but not jointly normal, demonstrate that their sums may not retain a normal distribution. The discussion also touches on the possibility of transforming correlated variables into independent ones through linear transformations, provided they are jointly normally distributed. Additionally, the Central Limit Theorem indicates that sums of random variables from any distribution with finite variance converge to a normal distribution. Overall, the conversation emphasizes the critical role of independence in maintaining normality in sums of random variables.
  • #31
BWV said:
but you have to define normal by some generating function.

A normal random variable ##\eta## can , in many different ways, be represented as the sum of two other independent normal random variables.

I don't understand how you want to restrict the types of representations of ##\eta## that are permitted.
 
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  • #32
  • #33
mathman said:
Added note: These random variables are uncorrelated.
Is that correct? The intension of the η=σξ example is to randomly flip the sign on half of ξ, which does not change the distribution, but half of η is perfectly negatively correlated to ξ, and this information is recovered in P(ξ+η=0)=P(σ=−1)=1/2.

ISTM there are problems with the construction on this, but they are above my pay grade. ξ is a function on ℝ, while σ is discrete. How would you actually flip the sign randomly on half of ℝ? any countably finite number in σ would not change the outcome of the distribution, and if σ in uncountably infinite, then how is P(ξ+η=0) not 0?
 
  • #34
BWV said:
How would you actually flip the sign randomly on half of ℝ?
Why would that be necessary? If we grant that you can take a random sample from ##\xi##, in order to realize a sample of ##\xi \sigma## you only need one to decide whether to flip the value of that particular random sample by using one realization of ##\sigma##.

The probability of any particular realization of a sample from a normally distributed random variable is zero, so if zero probability events are going to be a conceptual problem, they are already present in the idea of taking a sample from ##\xi##.
 
  • #35
BWV said:
Is that correct? The intension of the η=σξ example is to randomly flip the sign on half of ξ, which does not change the distribution, but half of η is perfectly negatively correlated to ξ, and this information is recovered in P(ξ+η=0)=P(σ=−1)=1/2.

ISTM there are problems with the construction on this, but they are above my pay grade. ξ is a function on ℝ, while σ is discrete. How would you actually flip the sign randomly on half of ℝ? any countably finite number in σ would not change the outcome of the distribution, and if σ in uncountably infinite, then how is P(ξ+η=0) not 0?
Covariance ##E(\xi\eta)=E(\sigma\xi^2)=E(\sigma)E(\xi^2)=0##, since ##E(\sigma)=0## while ##\sigma## is independent of ##\xi## and both means ##=0##.
 
  • #36
mathman said:
Covariance ##E(\xi\eta)=E(\sigma\xi^2)=E(\sigma)E(\xi^2)=0##, since ##E(\sigma)=0## while ##\sigma## is independent of ##\xi## and both means ##=0##.
So this is a good example to show that even uncorrelated is not enough to guarantee that the sum is normal. It requires independence.
 
  • #37
FactChecker said:
So this is a good example to show that even uncorrelated is not enough to guarantee that the sum is normal. It requires independence.
Independence is not necessary If the variables are jointly normal, even if correlated, the sum will be normal.
 
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  • #38
mathman said:
Independence is not necessary If the variables are jointly normal, even if correlated, the sum will be normal.
I stand corrected. I should have just said that uncorrelated is not enough.
 
  • #39
Stephen Tashi said:
Are their theorems that deal with the general question? - Given a set of random variables ##\{X_1,X_2,..,X_M\}##, when does their exist some subspace ##S## of the vector space of random variables such that ##S## contains each ##X_i## and ##S## has a basis of mutually independent random variables?
"Linearly independent" presumably implies we are considering a set of random variables to be a vector space under the operations of multiplication by scalars and addition of the random variables.
Suppose we take "transformed" to mean transformed by a linear transformation in a vector space. If the vector space containing the random variables ##{X_1,X_2,...X_M}## has a finite basis ##B_1,B_2,...B_n## consisting of mutually independent random variables then (trivially) for each ##X_k## there exists a possibly non-invertible linear transformation ##T## that transforms some linear combination of the ##B_i## into ##X_k##.

If the smallest vector space containing the ##X_i## is infinite dimensional (e.g the vector space of all measurable functions on the real number line) , I don't know what happens.

I don't recall any texts that focus on vector spaces of random variables. Since the product of random variables is also a random variable, the topic for textbooks seems to be the algebra of random variables. But that approach downplays the concept of probability distributions.
Would be interesting to see if there is something similar to "Decouple" a basis for such space, so that any two ##X_i, X_j## are independent but the span is preserved.
 
  • #40
WWGD said:
Would be interesting to see if there is something similar to "Decouple" a basis for such space, so that any two ##X_i, X_j## are independent but the span is preserved.
That would require a change of the independent variables to a different set.
 

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