Non-zero mean and finite-variance Gaussian Squared R.V has Non-central Chi squared distribution, but how?

If $$X \sim \mathcal{N}(\mu,\sigma)$$

then $$X^2 \sim \frac{e^{-\frac{\left(\mu +\sqrt{x}\right)^2}{2 \sigma ^2}} \left(e^{\frac{2 \mu \sqrt{x}}{\sigma ^2}}+1\right)}{2 \sqrt{2 \pi } \sigma \sqrt{x}} \hspace{3 mm}, \hspace{3 mm} x>0$$

If $X^2$ has been known as non-central chi square distribution ($\mathcal{X^2(1,\lambda)}$) then how to calculate the non-centrality parameter in context of above distribution of $X^2$, so that both the distributions become equal? Any help please.

• You are mistaken in your premises: $X^2$ does not have a $\chi^2(1,\lambda)$ distribution. It is $\sigma^2$ times such a distribution. By definition, the noncentrality parameter of that distribution is $\lambda=(\mu/\sigma)^2$.
– whuber
Oct 13 '14 at 4:31
• I mean a non-central chi square distribution, which has two parameters: degree of freedom and non-central parameter.
– kaka
Oct 13 '14 at 7:50
• Yes, it is clear that's what you mean--but you are wrong to suppose that the $X^2$ you exhibit here has such a non-central $\chi^2$ distribution. It does not. After you rescale it by $1/\sigma$, it will have that distribution, whose PDF is given by a modified Bessel function $I_{-1/2}$. This fact is readily checked after you recognize that $I_{-1/2}(z) = \sqrt{1/(2\pi z)}\left(\exp(z)+\exp(-z)\right)$.
– whuber
Oct 13 '14 at 16:23
• @whuber Nice description. So $\mathcal{N}(\mu,\sigma)$ doesn't have a non central Chi-Square distribution. so does the distribution of $X^2$ where $X \sim \mathcal{N}(\mu,\sigma)$ has any particular name?
– kaka
Oct 13 '14 at 21:28
• A multiple of a distribution occurs so often and is so simply related to the distribution itself that it need not have any special name. For that reason, for example, the square of a Normal$(0,\sigma)$ distribution has no special name, either: it's just $\sigma^2$ times a $\chi^2(1)$ distribution.
– whuber
Oct 13 '14 at 21:33

If $$X \sim \mathcal{N}(\mu,\sigma),$$ and $$Y \sim \chi^2 \Big(k=1,\lambda=\Big(\frac{\mu}{\sigma}\Big)^2 \Big),$$
then $$X^2\stackrel{d}{=}\sigma^2 Y.$$
In words, if X is normal random variable with non-zero mean and variance and Y is non-central chi squared random variable with one degree of freedom and non-central parameter $\lambda=(\frac{\mu}{\sigma})^2$, then $X^2$ is equal in distribution to $\sigma^2$Y.
It should be $X^2$ in the conclusion.