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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.

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    $\begingroup$ (+1) Welcome to the site, Sven. First, you might note that $X-Y$ and $X+Y$ are actually iid $\mathcal N(0,2\sigma^2)$ random variables and $\exp z$ is a monotonic function, so your problem reduces to finding tail bounds on $\beta \sigma^2 Z_1^2 / \sqrt{2} + \beta \sigma Z_2$ where $Z_1$ and $Z_2$ are iid standard normal. (Here $\beta = \alpha / \sqrt{2}$ and $Z_1^2$ is, of course, a $\chi^2$ random variable with one degree of freedom, independent of $Z_2$.) $\endgroup$
    – cardinal
    Nov 22, 2012 at 15:28
  • $\begingroup$ I find it interesting that you have $\alpha$ in front of both terms. Is that correct, or should the first term have $\alpha^2$? Also, for tail bounds, one is usually interested in achieving at least some particular rate of convergence. What would be "good enough" for the problem you are considering? $\endgroup$
    – cardinal
    Nov 22, 2012 at 15:29
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    $\begingroup$ First of all, thank you for the welcome and your answer. The $\alpha$ (no squares) is correct in both cases. I also head the idea to split it into chi-squared and normal, actually that is how i computed the characteristic function. But unfortunately I couldn't do any more. Regarding your question what would be 'good enough', I will have to think about that, first. I will post something about that later. But in general, the sharper, the better, obviously... $\endgroup$
    – Sven Stodtmann
    Nov 22, 2012 at 15:45
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    $\begingroup$ Instead of considering $X-Y$ and $X+Y$ as iid normal random variables, perhaps something like $$P\{Z>z\}=P\{aX^2+bY^2+cXY+dX+eY > \ln z\}$$ where the right side is the probability that the random point $(X,Y)$ lies outside an ellipse might work. A lower bound on the tail probability is thus the probability that $(X,Y)$ is outside the rectangle bounding the ellipse might work. An upper bound would be the probability of being outside an inscribed rectangle. Note that because of the circular symmetry, the ellipse can always be taken as having major and minor axes parallel to the $x$ and $y$ axes. $\endgroup$
    – Dilip Sarwate
    Nov 22, 2012 at 17:09
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    $\begingroup$ I like the idea of approaching the problem from geometry. However one problem is, that in the above case, $b^2-4ac = 0$ and we have a parabola instead of an ellipse - bounding by rectangles is probably not an option, maybe by perturbing the problem $b\rightarrow b-\varepsilon$ and then let $\varepsilon\rightarrow 0$, but somehow, I doubt this works. I will still give it a try. To answer cardinal's question what I need it for: I want to sandwich the expression $Z\in [K^{-1},K]$ "in distribution" between two i.i.d. Bernoulli variables. $\endgroup$
    – Sven Stodtmann
    Nov 23, 2012 at 9:18

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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.

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