# Normal with mean unequal to zero squared

It is a well-known fact that if $x_i \sim N(0,1), i = 1, \dots, n$, that then for $\nu \in \{1=1,\dots, n\}$ it holds that $\sum_{i=1}^\nu x_i^2 \sim \chi^2(\nu)$

I was wondering what now would happen if $x_i$ would instead be $\sim N(\mu,1), \mu \neq 0; i = 1,\dots, n$.

Could anyone shed some light on this and tell me if the resulting distribution (a) differs a lot from a $\chi^2$ distribution [I think not] and (b) whether this has a name?

• Commented May 6, 2015 at 11:17
• @COOLSerdash Thanks, that's what I was looking for :)! In the special case that $\sigma = 1$ and $\mu_i=\mu \ \forall i$, how would this deviate from a standard chi squared?
– rbm
Commented May 6, 2015 at 11:20
• I don't know what you mean exactly by "deviate" but it will be a Noncentral chi-square distribution with parameters $\nu = n$ and $\lambda = n\mu^{2}$ where $\nu$ denote the degrees of freedom. The distribution would have a mean of $\nu+\lambda = n(1+\mu^{2})$ and variance $2(\nu + 2\lambda) = 2(n+2n\mu^{2})$. So the mean is shifted by $+ \lambda$ in comparison with a chi-squared distribution with degrees of freedom $\nu$. And the variance is increased by $4\lambda$. Commented May 6, 2015 at 18:34
• @COOLSerdash That was what I meant. Thank you :)! If you make your comment an answer I will accept it.
– rbm
Commented May 6, 2015 at 18:35

This will be a Noncentral chi-squared distribution.

If $X_{i}\sim N(\mu, 1); i=1, \ldots, n$, then it will be a Noncentral chi-squared distribution with parameters $\nu = n$ and $\lambda = n\mu^{2}$, where $\nu$ denotes the degrees of freedom and $\lambda$ is the non-centrality parameter.

The mean of this distribution would be $\nu + \lambda = n(1 + \mu^{2})$ and its variance $2(\nu + 2\lambda)=2(n + 2n\mu^{2})$. So the mean is shifted by $+\lambda$ and the variance increased by $4\lambda$ in comparison with a standard chi-squared distribution.

As an example, let $n = 10,\mu = \sqrt{\frac{3}{2}}$ and $\sigma = 1$. The corresponding distribution of $\sum_{i=1}^{10}X_{i}^{2}$ is a Noncentral chi-squared distribution with parameters $\nu = 10$ and $\lambda = 15$. The mean is $10+15 = 25$ and its variance $2(10 + 2\cdot 10\cdot \frac{3}{2})=80$. The PDF looks like this: