# Chi-squared dist. converges to normal as df goes to infinity, but

1. Feb 22, 2013

chi-squared dist. converges to normal as df goes to infinity, but....

This is surely going to sound naive, but at least this will make it easy to answer.

For a chi-squared distribution, if k = the degrees of freedom, then
[a] k = μ = (1/2) σ2
as k goes to infinity, the distribution approaches a normal distribution.

But when I put these two together, I get
[c] as k goes to infinity, the mean and the variance become infinite
which would seem odd for a normal curve.
What am I getting wrong here? Thanks in advance.

2. Feb 22, 2013

### Staff: Mentor

Re: chi-squared dist. converges to normal as df goes to infinity, but.

The curve for every k gets closer and closer to a normal distribution with the same mean and variance with increasing k.
If you scale the distribution in an appropriate way, you get something approaching a normal distribution with mean 0 and variance 1.

3. Feb 22, 2013

Re: chi-squared dist. converges to normal as df goes to infinity, but.

mfb, thanks very much. That makes sense.

4. Feb 23, 2013

### ssd

Re: chi-squared dist. converges to normal as df goes to infinity, but.

Putting k=μ (mean of the normals, I presume) appears weired, k is positive integer ( being the number of normals summed here), and -< μ< ∞ is real. Also that, if all means of the initial normal distributions are not 0, the then the resulting chi sq is non central.

Last edited: Feb 23, 2013
5. Feb 23, 2013

### Staff: Mentor

Re: chi-squared dist. converges to normal as df goes to infinity, but.

Where is the problem in different gaussian distributions which all have an integer as expectation value?
The chi-squared distribution is positive for positive values only, but for large k, the gaussian distribution is a reasonable approximation (its part <0 is negligible).

6. Feb 23, 2013

Last edited: Feb 23, 2013
7. Feb 24, 2013

### ssd

Re: chi-squared dist. converges to normal as df goes to infinity, but.

Please check again. I am talking of μ as normal mean.... you are mistaking μ as chi sq mean. "μ =k" CAN NOT be consequence of any literature definition, where ever written....lodge a request for correction there. And of course, I stand correct about non centrality.... please go through the derivation of n.c. chi sq.

About integer and real part: I did not say that a particular value of normal mean cannot be integer. But I say, taking normal mean as integer is weired. The first loophole arises in context of the present problem as the fact that μ is differentiable but k is not.

Last edited: Feb 24, 2013
8. Feb 24, 2013

Re: chi-squared dist. converges to normal as df goes to infinity, but.

In that case, I am not sure of your question, because you referred to the original μ=k, and in the original context, μ is the mean of the chi squared distribution.
I'm also not sure whether this is a suggestion for me to go through it myself, or to write down the derivation here in this post. In the latter case, probably another contributor would do a better job of it than I would.

9. Feb 24, 2013

### ssd

Re: chi-squared dist. converges to normal as df goes to infinity, but.

Well, if μ is assumed as chi sq mean, no issues (is it not obvious from my posts). The original post is some what misleading with (unnecessary) involvement of μ as the chi sq mean... where k clearly stands for that. Without clarification, μ has been naturally presumed as the originating normal mean. I understood your problem in a completely wrong way altogether.
Hope it clarifies my statements.
PS. "going through" in common jargon probably does not mean writing down. :)

Last edited: Feb 24, 2013
10. Feb 24, 2013