Sum of independent random variables and Normalization

[joshthekid]

Hi,

Lets say I have N independent, not necessarily identical, random variable. I define a new random variable as

$$Y=Σ^{N}_{i=0} X_{i}$$

does Y follow a normalized probability distribution?

 

[BWV]

The sum of normalized squared values of X_i follows a Chi-squared distribution:

https://en.wikipedia.org/wiki/Chi-squared_distribution

 

[EngWiPy]

What is the distribution of the Xs?

 

[member 587159]

The sum of normalized squared values of X_i follows a Chi-squared distribution:

https://en.wikipedia.org/wiki/Chi-squared_distribution

Note that there are no squares in the sum, so your statement is not true.

 

[member 587159]

Hi,

Lets say I have N independent, not necessarily identical, random variable. I define a new random variable as

$$Y=Σ^{N}_{i=0} X_{i}$$

does Y follow a normalized probability distribution?

If each of the $X_i's$ is normally distributed, then yes. It will have mean sum of the means and variance sum of the variances.

 

[joshthekid]

If each of the $X_i's$ is normally distributed, then yes. It will have mean sum of the means and variance sum of the variances.

I'm talking about the generalized case where Xi can literally come from any valid probability distribution and the Xi do not have to be the same.

 

[member 587159]

I'm talking about the generalized case where Xi can literally come from any valid probability distribution and the Xi do not have to be the same.

Then there is no general conclusion to be made.

 

[Dale]

I'm talking about the generalized case where Xi can literally come from any valid probability distribution and the Xi do not have to be the same.

I agree with @Math_QED, and in fact for any valid probability distribution there may not even be a sensible sum operation, e.g. what if one X is the outcome of a coin toss and another X is gender.

 

[BWV]

Note that there are no squares in the sum, so your statement is not true.

Well, note that my statement is true, just was referring to the similarity to chi-squared

Would have to assume that the underlying distributions have finite variances so can be normalized (seems some confusion in the OP between normalized / standardized and normal) Normalized means the variable is rescaled to 0 mean and variance of 1. The Central Limit Theorem states that if you then sum these normalized variables the distribution of the sum converges to the normal distribution.

 

[member 587159]

Well, note that my statement is true, just was referring to the similarity to chi-squared

Would have to assume that the underlying distributions have finite variances so can be normalized (seems some confusion in the OP between normalized / standardized and normal) Normalized means the variable is rescaled to 0 mean and variance of 1. The Central Limit Theorem states that if you then sum these normalized variables the distribution of the sum converges to the normal distribution.

Misread your post then. Sorry!

 

[Stephen Tashi]

does Y follow a normalized probability distribution?

I think what you mean to ask is "does Y have a normal distribution?" or "Is Y normally distributed?" The adjective "normalized" doesn't refer to a specific family of probability distributions.

I'm talking about the generalized case where Xi can literally come from any valid probability distribution and the Xi do not have to be the same.

Perhaps you also mean to say that the Xi are mutually independent?

As others indicated, given only those facts we cannot conclude that the distribution of Y is a normal distribution.

If you do a web search on "generalized central limit theorem" you may be able to find theorems that give conditions where the sum of independent non-identically distributed random variables can be shown to be approxmately a normal distribution. This is a very technical topic. I don't know the details.