# Positivity of linear combination of gaussian variables

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.

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?
• (I found involving $p$ is misleading, I have delete $p$ in my original question.) Commented May 24, 2023 at 10:20
• I still don't understand your question. Maybe you are working with an erroneous equation: $\mathbb{P}(f > 0) = 1 - \mathbb{P}(f > 2b)$? Commented May 24, 2023 at 10:21
• $\mathbb{P}(f<0)$ is equal to $\mathbb{P}(f>2b)$, since $f$ is $\mathcal{N}(b,p^2c^2)$ and its symmetric to $x=b$ am I right? Commented May 24, 2023 at 10:27
• yes you are right. But I still don't know why are you working with estimates and not with the exact expression. $\Phi$ is monotone with the consequence thart you can easily convert any bound for the probability into bounds for the coefficients $a,b,c,\ldots$. Commented May 24, 2023 at 10:51