PDF of Sum of $\chi^2$ and Normal Distributions

Question

I am interested in calculating the p.d.f of the random variable

$$ Y = a + b^\intercal z + c \|z\|_2^2, $$ where $a,c$ are constants, $b \in\mathbb{R}^n$, and $z \sim \mathcal{N}(0,I)$ is a standard normal distribution, where $z \in \mathbb{R}^n$. It is known that the norm squared of a standard normal distribution is a $\chi^2$ distribution, which has p.d.f

$$ f(z) = \frac{1}{2^{n/2}\Gamma(n/2)}z^{n/2 - 1} e^{-z/2}, $$ and $b^\intercal z$ is a normal distribution such that $b^\intercal z \sim \mathcal{N}(0,b^\intercal b)$. I can always rewrite this in terms of a standard normal distribution as $b^\intercal z = \sqrt{b^\intercal b} \ \tilde{z} = \|b\|_2 \ \tilde{z}$, where $\tilde{z} \sim \mathcal{N}(0,1)$ is a standard normal random variable (scalar). This then gives

$$ Y = a + \|b\|_2 \tilde{z} + c \|z\|_2^2, $$ which is the sum of a standard normal random variable and $\chi^2$ random variable with $n$ degrees of freedom. My question is what would be the p.d.f of $Y$, i.e., what is the p.d.f of the sum of a $\chi^2$ distribution and a standard normal distribution?

Answer 1

Complete the square to write $Y=a+b^Tz+c\|z\|^2$ in the form $$\begin{align*}\frac{Y-a}c&=\frac{b^Tz}c +\|z\|^2\\ &=\|z+\frac{b}{2c}\|^2-\frac{\|b\|^2}{4c}\\&=\|z+\mu\|^2-\frac{\|b\|^2}{4c} \end{align*}$$ where $\mu=b/2c.$ By rotational invariance of $N(0,I_n)$ we may assume $\mu=(m,0,0,\ldots,0)$, so $\|z+\mu\|^2=(z_1+m)^2+z_2^2+\cdots+z_n^2$, which has the well-known noncentral chi squared distribution. The density of $Y$ is thus obtainable from the noncentral chi-squared density by an affine transformation of variables.