Tail bounds on a function of normally distributed variables I am looking for tail bounds (both at $0$ and at $\infty$) for
$$ Z:=\exp \left(\frac{\alpha}{4}(X-Y)^2+\frac{\alpha}{2}(X+Y)\right)$$
where $\alpha$ is a positive real and $X,Y$ are i.i.d. normal with mean $0$ and variance $\sigma^2 >> 1$. I would like to control the probability of $Z$ being outside an interval $[a,b]$ in the limit of small $a$ and large $b$.
My first approach was characteristic functions, I managed to compute
$$ E[\exp(i\omega \log Z)] = \frac{1}{\sqrt{1+i\alpha\sigma^2\omega}}\exp\left(-\frac{1}{4}\alpha^2\sigma^2\omega^2\right)$$
but i could not find an inverse Fourier transform or do something useful with it.
 A: Using the idea suggested by @cardinal, let $a$ and $b$ denote positive numbers and 
consider a random variable $Z$ defined as $Z = \exp(aX^2 + bY)$ where $X$ and $Y$
are independent standard normal random variables.  Then, for $K > 1$,
$$\begin{align*}
P\{Z > K\} &= P\{\exp(aX^2 + bY) > K\}\\
&= P\{aX^2 + bY > \alpha\} & \text{where}~\alpha = \ln K\\
&= \int_{-\infty}^\infty \phi(x)P\{aX^2 + bY > \alpha\mid X = x\}\,\mathrm dx\\
&= \int_{-\infty}^\infty \phi(x)P\left\{Y > \frac{\alpha-ax^2}{b}\right\}\,\mathrm dx\\
&= \int_{-\infty}^\infty \phi(x)Q\left(\frac{\alpha-ax^2}{b}\right)\,\mathrm dx\\
&= E\left[Q\left(\frac{\alpha-aX^2}{b}\right)\right]
\end{align*}$$
where $\phi(\cdot)$ is the standard normal density function and $Q(\cdot)$ is the
complementary cumulative probability distribution function of a standard normal
random variable. I suspect that this integral cannot be computed
analytically, but its value might be computable very fast by numerical integration 
(cf. this answer by whuber
for a different problem. 
Now, $Q((\alpha -ax^2)/b)$ is an even function of $x$
with asymptotic value $1$ as $x \to \pm\infty$ and a minimum value of 
$Q(\alpha/b) <\frac{1}{2}$ at
$x=0$. So, we get the obvious lower and upper bounds
$$Q(\alpha/b) < P\{Z > K\} < 1.$$
Tighter upper bounds can be obtained by bounding $Q((\alpha - ax^2)/b)$ from
above by $1$ for $|x| > \beta$ for some suitable $\beta$; and by straight
lines through $(-\beta,1)$ and $(0,Q(\alpha/b)$, and through
$(0,Q(\alpha/b)$ and $(\beta, 1)$ for $|x| \leq \beta$. Since $x\phi(x)$ is
a perfect integral, the expected value of this upper bound can found
as something like $2Q(\beta) + f(\beta)$ where $f(\cdot)$ is an
exponential function of $\beta$, and one could even choose the value
of $\beta$ to minimize this upper bound.
Alternatively, note that $Q(t) \leq \frac{1}{2}\exp(-t^2/2)$ for $t \geq 0$,
and so for $-\sqrt{\alpha/a} \leq x \leq \sqrt{\alpha/a}$, we have
$$Q((\alpha - ax^2)/b)\leq \frac{1}{2}\exp(-((\alpha - ax^2)/b)^2/2)$$
which might lead to a better bound since $\phi(x)$ is large only when
$|x|$ is small and that is exactly where we have a better upper bound
that the straight-line bounds mentioned above.
