Suppose $(Z_1,Z_2)\sim N_2(\mu,0; 2,2; \frac{1}{2})$, i.e., a bivariate normal distribution with mean vector $(\mu,0)$ with each component having variance $2$ and covariance between the variables being $1$. Suppose $\{(Z_{1i},Z_{2i})\}_{i=1}^d$ is an independent and identically distributed sample on $(Z_1,Z_2)$. I am interested in finding a tight upper bound for the probability, $$P\{\sum_{i=1}^d(Z_{1i}^2-Z_{2i}^2)\leq 0\}.$$
Intuitively, $\sum_{i=1}^d Z_{1i}^2$ is a noncentral chi-square distribution with noncentrality parameter $d\mu^2$ and $\sum_{i=1}^dZ_{2i}^2$ is a central chi-square distribution. Hence $\sum_{i=1}^d(Z_{1i}^2-Z_{2i}^2)$ is more likely to take positive values than negative values. It should be less than $\frac{1}{2}$. But by what amount? Can an upper bound be established like, $$P\{\sum_{i=1}^d(Z_{1i}^2-Z_{2i}^2)\leq 0\} \leq \frac{1}{2}-f(\mu),$$ for some function $f$ ?
Any help or directions will be highly appreciated.
By properties of multivariate normal RV's, \begin{align*} \begin{pmatrix} X_i \\ Y_i \end{pmatrix} \overset{\text{def}}{=}\begin{pmatrix} Z_{1i} - Z_{2i} \\ Z_{1i} + Z_{2i} \end{pmatrix} \sim N\left(\begin{pmatrix} 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 2 & 0 \\ 0 & 6 \end{pmatrix} \right) \end{align*} and so $Z_{1i}^2 - Z_{2i}^2 = X_iY_i$. Notice that $X_iY_i$ has symmetric densities about 0, and since sum of symmetric RV's is symmetric, we indeed have \begin{align*} P\left(\sum_{i=1}^{d}\{Z_{1i}^2 - Z_{2i}^2\} \le 0\right) = \frac{1}{2} \end{align*}