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

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?

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.
• Thanks! But what do you mean by rotational invariance of the standard normal? How can we write $\mu = b/2c = (m,0,\ldots,0)$? May 22, 2020 at 17:15
• The vector $z\sim N(0,I_n)$ has the same distribution as the vector $w=Rz$, when the matrix $R$ is orthogonal, that is, obeys $R^TR=I_n$. Because the joint density function of $z$ actually only depends on $\|z\|^2$. As for the $\mu$ and $m$ claim: for any 2 unit vectors, there is an orthogonal matrix mapping the one to the other. May 22, 2020 at 18:17