chi-squared-distributionnon-central
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.
The $\chi^2_n$ distribution is defined as the distribution that results from summing the squares of $n$ independent random variables $\mathcal{N}(0,1)$, so: $$\text{If }X_1,\ldots,X_n\sim\mathcal{N}(0,1)\text{ and are independent, then }Y_1=\sum_{i=1}^nX_i^2\sim \chi^2_n,$$ where $X\sim Y$ denotes that the random variables $X$ and $Y$ have the same distribution (EDIT: $\chi_n^2$ will denote both a Chi squared distribution with $n$ degrees of freedom and a random variable with such distribution). Now, the pdf of the $\chi^2_n$ distribution is $$ f_{\chi^2}(x;n)=\frac{1}{2^\frac{n}{2}\Gamma\left(\frac{n}{2}\right)}x^{\frac{n}{2}-1}e^{-\frac{x}{2}},\quad \text{for } x\geq0\text{ (and $0$ otherwise).} $$ So, indeed the $\chi^2_n$ distribution is a particular case of the $\Gamma(p,a)$ distribution with pdf $$ f_\Gamma(x;a,p)=\frac{1}{a^p\Gamma(p)}x^{p-1}e^{-\frac{x}{a}},\quad \text{for } x\geq0\text{ (and $0$ otherwise).} $$ Now it is clear that $\chi_n^2\sim\Gamma\left(\frac{n}{2},2\right)$.
The difference in your case is that you have normal variables $X_i$ with common variances $\sigma^2\neq1$. But a similar distribution arises in that case: $$Y_2=\sum_{i=1}^nX_i^2=\sigma^2\sum_{i=1}^n\left(\frac{X_i}{\sigma}\right)^2\sim\sigma^2\chi_n^2,$$ so $Y$ follows the distribution resulting from multiplying a $\chi_n^2$ random variable with $\sigma^2$. This is easily obtained with a transformation of random variables ($Y_2=\sigma^2Y_1$): $$ f_{\sigma^2\chi^2}(x;n)=f_{\chi^2}\left(\frac{x}{\sigma^2};n\right)\frac{1}{\sigma^2}. $$ Note that this is the same as saying that $Y_2\sim\Gamma\left(\frac{n}{2},2\sigma^2\right)$ since $\sigma^2$ can be absorbed by the Gamma's $a$ parameter.
If you want to derive the pdf of the $\chi^2_n$ from scratch (which also applies to the situation with $\sigma^2\neq1$ under minor changes), you can follow the first step here for the $\chi_1^2$ using standard transformation for random variables. Then, you may either follow the next steps or shorten the proof relying in the convolution properties of the Gamma distribution and its relationship with the $\chi^2_n$ described above.
To prove the theorem we are going to need the following intermediate result.
Theorem. Let $\mathbf{y} \sim N_n \left(0, \sigma^2 \mathbf{I}_n \right)$ and let $ Q = \sigma^{-2} \mathbf{y}^{\prime} \mathbf{A} \mathbf{y}$ for a symmetric matrix $\mathbf{A}$ of rank $r$. Then if $\mathbf{A}$ is idempotent, Q has a $\chi^2 (r)$ distribution.
The theorem extends to the other direction as well but we only need the sufficiency so we will just prove this and we will do so using the eigenvalue-eigenvector decomposition of an idempotent matrix.
It is important to note that we first consider the case of mean zero. We will relax this assumption afterwards. But for now recall that for a square matrix of rank r, say $\mathbf{A}$
$$\mathbf{A} = \sum_{i=1}^{r} \lambda_i \mathbf{c}_i \mathbf{c}_i^{\prime}$$
where the lambdas are the eigenvalues and the $\mathbf{c}_i$s the corresponding eigenvectors. All pretty standard so far. Now if we additionally restrict $\mathbf{A}$ to be symmetric two things happen:
The eigenvalues are real-valued
Eigenvectors corresponding to different eigenvalues are orthogonal
These are consequences of the so-called Spectral Theorem of linear algebra and you can consult any good textbook for a proof. What does that do for us? You are about to see why symmetry is required. Let's write down the decomposition of our $\mathbf{A}$ in our quadratic form and see what happens.
$$\sigma^{-2} \mathbf{y}^{\prime} \mathbf{A} \mathbf{y} = \sigma^{-2} \mathbf{y}^{\prime} \left( \sum_{i=1}^r \lambda_i \mathbf{c}_i \mathbf{c}_i^{\prime} \right) \mathbf{y} = \sum_{i=1}^r \lambda_i \left( \sigma^{-1} \mathbf{c}_i^{\prime} \mathbf{y} \right)^2 \tag{1}$$
We have written our quadratic form as weighted squared projections onto orthogonal axes. Let's now investigate the distribution of $\mathbf{c}_i ^{\prime} \mathbf{y}$ and $ \mathbf{c}_j ^{\prime} \mathbf{y}$, $i \neq j$. By basic rules
$$\sigma^{-1} \mathbf{c}_i ^{\prime} \mathbf{y} \sim N(0, \underbrace{\mathbf{c}_i^{\prime} \mathbf{c}_i}_{=1} ) $$
as the eigenvectors are not unique and therefore can be rescaled without loss of generality to have length one. Next,
$$Cov\left(\sigma^{-1} \mathbf{c}_i ^{\prime} \mathbf{x} , \sigma^{-1} \mathbf{c}_j ^{\prime} \mathbf{x} \right) = \sigma^{-2}\mathbf{c}_i ^{\prime} I_n \mathbf{c}_j = 0, \ \ i \neq j $$
by the second implication of the Spectral Theorem. Thus our summands are uncorrelated. By the normality they are also independent. It is easy to see then that the sum consists of weighted $\chi^2$ random variables (weighted by the eigenvalues). In this thread it was asked whether this follows the $\chi^2$ distribution regardless. The answer is no of course.
Enter the idempotence. It easily follows from the definition of an idempotent matrix and the eigenvalue/eigenvector problem that an idempotent matrix has eigenvalues equal to either one or zero. Since by assumption the matrix $\mathbf{A}$ has rank $r$, there are $r$ eigenvalues equal to one. Therefore
$$\sigma^{-2} \mathbf{y}^{\prime} \mathbf{A} \mathbf{y} = \sum_{i=1}^r \left( \sigma^{-1} \mathbf{c}_i^{\prime} \mathbf{y} \right)^2 \sim \chi^2 (r) $$
And this completes the proof.
What happens now if the vector $y$ has nonzero mean? No harm done, we will just use the definition of the non-central $\chi^2$ distribution to conclude that if
$$\mathbf{y} \sim N_n \left( \boldsymbol{\mu}, \sigma^2 \mathbf{I}_n \right)$$
then
$$\sigma^{-2} \mathbf{y}^{\prime} \mathbf{A} \mathbf{y} ~ \sim \chi^2 \left(r, \boldsymbol{\mu}^{\prime} \mathbf{A} \boldsymbol{\mu} \right) $$
where the second term indicates the non-centrality parameter. (Due to force of habit, I will skip the division by $2$ but you can just do it your way).
We are now ready to prove the required result.
Theorem. Suppose $\Sigma$ is a $n \times n$ positive definite and symmetric matrix, $A$ is a $n \times n$ symmetric matrix with rank $r$, and $(A\Sigma)^2 = A\Sigma$.
Then $\mathbf{y} \sim N(\boldsymbol{\mu}, \Sigma) \implies Q = \mathbf{y}^{\prime}A\mathbf{y} \sim \chi^2 \left(r, \boldsymbol{\mu}^{\prime} \mathbf{A}\boldsymbol{\mu} \right)\text{.}$
We are given that
$$\mathbf{A\Sigma A \Sigma} = \mathbf{A\Sigma}$$
from which it follows that
$$\mathbf{A\Sigma A} = \mathbf{A}$$
and hence we may rewrite our quadratic form as
$$Q = \left( \boldsymbol{\Sigma}^{-1/2} \mathbf{y} \right)^{\prime} \boldsymbol{\Sigma}^{1/2} \mathbf{A} \boldsymbol{\Sigma}^{1/2} \left( \boldsymbol{\Sigma}^{-1/2} \mathbf{y} \right) \tag{2} $$
Since by assumption $\boldsymbol{\Sigma}$ is a positive definite matrix, its square root is always well-defined and you can check that using the eigenvalue/eigenvector decomposition (You can also check that equation ($2$) is equivalent to equation ($1$) !)
Of course you would agree that $\boldsymbol{\Sigma}^{-1/2} \mathbf{y} \sim N_n \left( \boldsymbol{\Sigma}^{-1/2} \boldsymbol{\mu} , \mathbf{I}_n \right)$. If we temporarily assume that $\boldsymbol{\mu}=0$ we would be in the situation of the first theorem, right? Well, not exactly - we still have to show that the middle matrix is idempotent but that is easy enough under our assumptions.
$$\boldsymbol{\Sigma}^{1/2} \mathbf{A} \boldsymbol{\Sigma}^{1/2} \boldsymbol{\Sigma}^{1/2} \mathbf{A} \boldsymbol{\Sigma}^{1/2} = \boldsymbol{\Sigma}^{1/2} \mathbf{A} \boldsymbol{\Sigma}^{1/2}$$
Therefore if $\mathbf{y} \sim N_n \left(\mathbf{0}, \boldsymbol{\Sigma} \right)$, $Q \sim \chi^2 \left(r \right)$ which implies that if we now switch back to the situation of nonzero mean, we would have
$$ Q = \mathbf{y}^{\prime}\mathbf{A}\mathbf{y} \sim \chi^2 \left(r, \boldsymbol{\mu}^{\prime} \mathbf{A}\boldsymbol{\mu} \right)$$
as required. $\square$
Best Answer
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.