I am having trouble to understand why an alternate derivation of mine related to the Eq.(10) of this paper is not matching with the result in the paper. First, let me describe the problem.
The paper considers the random vectors $\mathbf{X},\mathbf{Y}\in \mathbb{R}^n$, such that $\mathbf{X}\sim \mathcal{N}(\mathbf{0},\mathbf{I}_n),\ \mathbf{Y}\vert \mathbf{X}=\mathbf{x}\sim\mathbf{N}(\mathbf{x},\sigma^2\mathbf{I}_n)$. Then they evaluate $f_{\mathbf{Y},\|\mathbf{X}\|}(\mathbf{y},t)$ as below: \begin{align} f_{\mathbf{Y},\|\mathbf{X}\|}(\mathbf{y},t) & = \int_{\mathbb{R}^n} f_{\mathbf{Y}\vert \mathbf{X}}(\mathbf{y}\vert \mathbf{x})f_{\|\mathbf{X}\|\vert \mathbf{X}}(t\vert \mathbf{x})f_{\mathbf{X}}(\mathbf{x})d\mathbf{x}\\ \ & = \int_{\mathbb{R}^n} \frac{e^{-\|\mathbf{y}-\mathbf{x}\|^2/2\sigma^2}}{(2\pi\sigma^2)^{n/2}}\delta(\|\mathbf{x}\|-t)\frac{e^{-\|\mathbf{x}\|^2/2}}{(2\pi)^{n/2}}d\mathbf{x}.\tag{1} \end{align} In the above $\|\cdot\|$ denotes the $l_2$ norm. Beyond the above step they utilize a spherical coordinate transformation, utilize the fact that $\|\mathbf{x}\|=t$, and utilize the rotational invariance of the Normal distribution to conclude that \begin{align} f_{\mathbf{Y},\|\mathbf{X}\|}(\mathbf{y},t) & = \frac{1}{(2\pi)^{n/2}}e^{-t^2/2}\frac{1}{\sqrt{\sigma^2}}f_{\mathrm{Chi}}\left(\frac{t}{\sqrt{\sigma^2}};n,\frac{\|\mathbf{y}\|}{\sqrt{\sigma^2}}\right)\tag{2}, \end{align} where $f_{\mathrm{Chi}}(x;n,\lambda)$ is the density of the non-central Chi distribution with parameters $n,\lambda$. Please see the paper mentioned above to find details about the derivation.
Now I have tried to express Eq.(1) in an alternate way as below: \begin{align} f_{\mathbf{Y},\|\mathbf{X}\|}(\mathbf{y},t) & = \frac{e^{-\|\mathbf{y}\|^2/2(\sigma^2+1)}}{(2\pi(\sigma^2+1))^{n/2}}\int_{\mathbb{R}^n} \frac{e^{-\|\mathbf{x}-\frac{\mathbf{y}}{\sigma^2+1}\|^2/\left(\frac{2\sigma^2}{\sigma^2+1}\right)}}{(2\pi\frac{\sigma^2}{\sigma^2+1})^{n/2}}\delta(\|\mathbf{x}\|-t)d\mathbf{x}\tag{3}, \end{align} where I have used $$\frac{\|\mathbf{y}-\mathbf{x}\|^2}{2\sigma^2}+\frac{\|\mathbf{x}\|^2}{2}=\frac{\sigma^2+1}{2\sigma^2}\left\|\mathbf{x}-\frac{\mathbf{y}}{\sigma^2+1}\right\|^2+\frac{\|\mathbf{y}\|^2}{2(\sigma^2+1)}.$$ I think the integral in Eq.(3) is nothing but the density of the random variable $\|\mathbf{Z}\|$ where $\mathbf{Z}\sim \mathcal{N}(\frac{\mathbf{y}}{\sigma^2+1},\frac{\sigma^2}{\sigma^2+1}\mathbf{I}_n)$. However, in that case, the right hand side (RHS) of Eq.(3) seems to me to be proportional to $$\frac{e^{-\|\mathbf{y}\|^2/2(\sigma^2+1)}}{(2\pi(\sigma^2+1))^{n/2}}f_{\mathrm{Chi}}\left(\frac{t}{\sqrt{\sigma^2/(\sigma^2+1)}};n,\frac{\|\mathbf{y}\|}{\sqrt{\sigma^2(\sigma^2+1)}}\right).$$ This is different than the RHS of Eq.(2) in that the first argument inside the Chi density are not quite the same, and more importantly, there is a term containing $\|\mathbf{y}\|$, and not $t$ in the exponential outside the density.
This disparity is troubling me a lot, but I don't see why this has happened, although I am sure I have missed something. Can someone kindly help? Thanks in advance.