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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.

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    $\begingroup$ 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 $\lambda=(\mu/\sigma)^2$. $\endgroup$
    – whuber ♦
    Commented Oct 13, 2014 at 4:31
  • $\begingroup$ I mean a non-central chi square distribution, which has two parameters: degree of freedom and non-central parameter. $\endgroup$
    – kaka
    Commented Oct 13, 2014 at 7:50
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    $\begingroup$ 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 $I_{-1/2}(z) = \sqrt{1/(2\pi z)}\left(\exp(z)+\exp(-z)\right)$. $\endgroup$
    – whuber ♦
    Commented Oct 13, 2014 at 16:23
  • $\begingroup$ @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? $\endgroup$
    – kaka
    Commented Oct 13, 2014 at 21:28
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    $\begingroup$ 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. $\endgroup$
    – whuber ♦
    Commented Oct 13, 2014 at 21:33

1 Answer 1

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As already answered by whuber in above comments:

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.

Moreover, a scaled non-central chi square variable doesn't have the non-central chi squared distribution.

It should be $X^2$ in the conclusion.

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