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$X$ follows a normal distribution $X \text{~} N(\mu, \sigma^2) $.

And there are $n$ samples. Then what is the distribution of $$\frac{1}{n} \sum_i x_i^2$$

I do understand $\frac{\sum_i x_i}{n} \text{~} N(\mu, \frac{\sigma^2}{n})$, but don't know how to apply them here. Anybody who can help please?

Sycorax
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G K
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    Aside from a scaling factor, it will be [noncentral chi-squared](https://en.wikipedia.org/wiki/Noncentral_chi-squared_distribution). Is this for homework? – Glen_b Oct 17 '18 at 22:14
  • Glen_b, I've been studying on my own. I don't know what a noncentral chi-squared dist is.. would you mind elaborating it? – G K Oct 18 '18 at 00:43

2 Answers2

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If $X_1,...,X_n \sim \text{IID N}(\mu, \sigma^2)$ then this statistic has a non-central chi-squared distribution. To see this, we observe that $X_1/\sigma,...,X_n/\sigma \sim \text{IID N}(\mu/\sigma, 1)$ so that:

$$\frac{1}{n} \sum_{i=1}^n X_i^2 = \frac{\sigma^2}{n} \sum_{i=1}^n \Big( \frac{X_i}{\sigma} \Big)^2 \sim \frac{\sigma^2}{n} \cdot \text{N-C Chi-squared} \Big( DF = n, NCP = \frac{n \mu^2}{\sigma^2} \Big).$$

Ben
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  • Could you please elaborate how it becomes a non-central chisquare? Is there a way to understand it only by using basic mathematical stat? – G K Oct 18 '18 at 00:45
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    Suppose $X\sim N(\mu,1)$. Then, $X^2$ is Non-central Chi-squared with 1 degree of freedom and $\mu^2$ as the non-centrality parameter. This can be shown, for example, using the method of moment generating function or transformation (CDF), involving messy algebra but still basic math stat. If $X\sim N(\mu, \sigma^2)$, then $X^2$ is Non-central chi-square with 1 d.f. and $\mu^2/\sigma^2$ as the non-centrality parameter. This turns out to be a mixture of central chi-squares with mixing poisson probabilities. – saipk Oct 18 '18 at 06:00
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It is (nearly) a noncentral chi-squared distribution by definition.

The difference is in a scaling factor.

Definitions

Both the central and noncentral chi-squared distribution with $n$ degrees of freedom are the distribution of the sum of the squares of $n$ independent distributed normal distributed variables.

$$\sum_{i=1}^{n} X_i^2$$

  • For the (central) chi-squared distribution you have $$X_i \sim N(0,1)$$.
  • For the noncentral chi-squared distribution you have $$X_i \sim N(\mu_i,1)$$ and $\lambda = \sum_{i=1}^{n} \mu_i^2$ is the non-centrality parameter.

Transformation

For more general variables $X_i \sim N(\mu_i,\sigma_i^2)$ you have that

$$\sum_{i=1}^n \frac{X_i^2}{\sigma_i^2} \sim \chi^2 \left( n,\sum_i (\mu_i / \sigma_i)^2 \right)$$

is a noncentral chi-squared distributed variable. So your case is a transformed version of it:

$$\frac{1}{n} \sum_{i=1}^n X_i^2 = \frac{\sigma^2}{n} \sum_{i=1}^n \frac{X_i^2}{\sigma_i^2}$$

The actual distribution function

Getting the actual expression for the distribution of the noncentral chi-squared distribution is not as easy as for the central chi-squared distribution (which can be derived easily using symmetry arguments). It will eventually be represented in terms of analytical functions (e.g. the modified Bessel function of the first kind).

Sextus Empiricus
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