Positivity of linear combination of gaussian variables

Question

Consider a collection of independent standard Gaussian variables $w_i$ for $i = 1, 2, \ldots, N$. Define its linear combination $f:=\sum_{i=1}^Na_iw_i+b_i$, where $a_i=pb_i$ ($p$ is a fixed parameter), its expectation $\mu:=\sum_{i=1}^Nb_i>0$. And from this we have $\sigma^2:=p^2\sum_{i=1}^Nb_i^2=\sum_{i=1}^Na_i^2$.

My question is: Derive a lower bound on $$\mathbb{P}(f>0)>g(\mu,\sigma^2) \text{ (or}\geq)$$

I would need the bound $g(\mu,\sigma^2)$ to be sensitive to the mean and variance of $f$. Consequently, a trade-off between the mean and variance arises, leading to the following observations:

(1) When the variance is fixed, a larger mean corresponds to a larger probability $\mathbb{P}(f > 0)$.

(2) When the mean is fixed, a smaller variance corresponds to a larger probability $\mathbb{P}(f > 0)$.

I appreciate any insights, references, or techniques that can shed light on this problem.

Answer 1

Eventually not what you want. I have problems understanding your question. Let $b := \sum_{i=1}^n b_i > 0$ and $c^2 := \sum_{i=1}^n b_i^2$. Then $f \sim \mathcal{N}(b,p^2 \cdot c^2)$, thus $q := \mathbb{P}(f > 0) = 1-\Phi\left(\frac{-b}{pc} \right)$ with $\Phi$ the distribution funtion of $\mathcal{N}(0,1)$. We get $\frac{-b}{pc} = \Phi^{-1}(1-q)$ and $p = \frac{-b}{c\Phi^{-1}(1-q)}$. Since $\Phi$ and $\Phi^{-1}$ are well known, what is the problem?