I got the problem with hints but I don't get the hints. Hope someone can shed light on my doubts.
Let $\mathbf{X}$ be an $n$-dimension isotropically distributed vector in complex space $\mathbb{C}^n$ with norm $\sqrt{\rho n}$.
$H$ is circularly complex Gaussian scalar random variable $\mathcal{CN}(\mu_H,\sigma_H^2)$.
$\mathbf{W}$ is circularly complex Gaussian vector $\mathcal{CN}(0,\sigma_w^2 \mathbf{I}_n)$.
$$\mathbf{Y}=H\mathbf{X}+\mathbf{W}$$
Prove that the conditional PDF of $\mathbf{Y}$ given $H=h$ is $$P_{\mathbf{Y}|H}(\mathbf{y}|h)= \frac{\Gamma(n)}{\pi^n\sigma_w^2(\left\Vert\mathbf{y}\right\Vert |h| \sqrt{\rho n})^{n-1}} \exp\left(-\frac{\left\Vert\mathbf{y}\right\Vert^2 + |h|^2\rho n}{\sigma^2_w}\right) I_{n-1}\left(\frac{2\left\Vert\mathbf{y}\right\Vert |h| \sqrt{\rho n}}{\sigma_w^2}\right)$$
where $I_n(.)$ modified Bessel function with paramater $n$.
Hint 1
Given $h$, the random variable $(2/\sigma_w^2)\left\Vert\mathbf{y}\right\Vert^2$ follows a noncentral $\chi$-squared distribution with $2n$ degrees of freedom and noncentrality parameter $|h|^2\rho n \times 2/\sigma_w^2$.
My doubt
Note that $h$ is scalar and other random variables are vectors. We have \begin{align} ||\mathbf{y}||^2 &= |h|^2 ||\mathbf{x}||^2 + \mathbf{x}^*h^*\mathbf{w} + \mathbf{w}^*h\mathbf{x} + ||\mathbf{w}||^2 \\ &=|h|^2 \rho n + \mathbf{x}^*h^*\mathbf{w} + \mathbf{w}^*h\mathbf{x} + ||\mathbf{w}||^2\\ 2/\sigma_w^2||\mathbf{y}||^2 &=2/\sigma_w^2 |h|^2 \rho n + 2/\sigma_w^2 \left(\mathbf{x}^*h^*\mathbf{w} + \mathbf{w}^*h\mathbf{x} + ||\mathbf{w}||^2\right) \end{align}
Of course $\big( |h|^2 ||\mathbf{x}||^2 + ||\mathbf{w}||^2 \big)$ follows $\chi$-squared distribution, but what about the part $\big( \mathbf{x}^*h^*\mathbf{w} + \mathbf{w}^*h\mathbf{x} \big)$ ?
Hint 2
$\mathbf{y}$ is also isotropically distributed. We can derive $P_{\mathbf{Y}|H}(\mathbf{y}|h)$ by using the formula of $n$-dimensional complex sphere of radius $\sqrt{\rho n}$
$$S = \frac{2\pi^n (\sqrt{n \rho})^{2n-1}}{\Gamma(n)}$$
My doubt
Is it simply $P_{\mathbf{Y}|H}(\mathbf{y}|h) = \frac{1}{S} \times P_{\left\Vert\mathbf{Y}\right\Vert^2|H}(\left\Vert\mathbf{y}\right\Vert^2|h)$ thanks to the "isotropically distribution" property ?
For 1, the trick is to notice that for fixed $X$ we can split $W$ into $wX+W’$ where $w$ is normally distributed and $W’$ is a normally distributed vector orthogonal to $X$ ($2n-1$ degrees of freedom). You then get an expression for $\|\mathbf y\|^2$ as a sum of squares of normal variables one of which has nonzero mean.
For 2: yes, exactly.