sum of chi-square and normal distributed variables Given $Y = aX_1 + bX_2^2$, where $(X_1, X_2)$~Normal (zero vector, Identity matrix). 
What is the distribution of $Y$? 
$a,b$ are constants.
 A: As Whuber points out the final result in this post is not valid. I can't delete the post because it is an accepted answer. Just don't trust this answer
Let $W=X_2^2$ be a chi square random variable
The distribution for a Chi square variable with one degree of freedom is $\frac{1}{\sqrt{2\pi}} e^{-0.5w} w^{-0.5}$
The pdf for a normal variable with $0$ mean and unit variance is $\frac{1}{\sqrt{2\pi}} e^{-0.5x_1^2}$
Suppose that the random variable $W$ has a value $w$, in order for $aX_1+bw=y$ we can see that we need $X_1=\frac{y-bw}{a}$
The probability that $W=w$ is just $\frac{1}{\sqrt{2\pi}} e^{-0.5w} w^{-0.5}$.
The probability that $X_1=\frac{y-bw}{a}$ is found by putting this expression into the probability density function for $X_1$. Therefore the probability is $\frac{1}{\sqrt{2\pi}} e^{-0.5(\frac{y-bw}{a})^2}$
The probability of both of these values of $X_1$ and $W$ to occur is just the two probabilities multiplied together which is:
$$\left(\frac{1}{\sqrt{2\pi}} e^{-0.5(\frac{y-bw}{a})^2} \right) \left(\frac{1}{\sqrt{2\pi}} e^{-0.5w} w^{-0.5}\right)$$
Simplifying a bit gives:
$\frac{1}{2\pi}w^{-0.5}e^{-0.5\frac{y^2+(a-2yb)w+b^2w}{a}}$
Considering all possible values of $w$ and the required value of $x_1$ which achieves $Y=y$ we can integrate over the probability space. 
$$\int_{0}^{\infty}\frac{1}{2\pi}w^{-0.5}e^{-0.5\frac{y^2+(a-2yb)w+b^2w}{a}} dw$$
This integral gives a result for the probability of $y$ in terms of $a$ and $b$. Now you just need to solve the integral, it should have been obvious from the start that the answer wasn't going to be pretty.
