The background for this question is inferential statistics. I am trying to derive some characteristics about estimators in the context of design-based inference. Let us assume I have a population of $N$ elements. Each element is associated to a value $\beta_i$. I do not know the individual values of $\beta_i$ for each population element, but I know their mean (denoted as $\mu_\beta$) and variance (denoted as $\sigma_\beta^2$). To each population element, I apply the function:
$$\int_{2-\beta_i}^{\infty} \, e^{-t^2/2} \, dt. $$
Now, I want to sum the result of this function over each population element. I believe I can formally write this problem in the following format:
$$ \sum_{i=1}^{N} \int_{2-\beta_i}^{\infty} \, e^{-t^2/2} \, dt. $$
Do you think this can be simplified, expressed, or solved as a function of the only valid information I have, namely the mean and variance of $\beta_i$?
EDIT: After the very good answers from Henry and Eduardo, there doesn't appear to be any further simplification possible. However, what would happen if I knew the distribution of values of βi, in addition to their mean and variance? Could this potentially lead to a multivariable calculus problem?
BTW I hope this edit in within the forum policy, I can create a new post if necessary....
I don't think you can simplify this expression further. You can rewrite the integral as: $$\int_{2- \beta_i}^\infty e^{-t^2/2} dt= \sqrt{\frac{\pi}{2}}\left(\text{erf}\left(\frac{\beta_i - 2}{\sqrt{2}}\right) + 1\right)$$
and then your sum becomes: $$\sqrt{\frac{\pi}{2}} \sum_{i = 1}^N \left(\text{erf}\left(\frac{\beta_i - 2}{\sqrt{2}}\right) + 1\right)=N\sqrt{\frac{\pi}{2}}+\sqrt{\frac{\pi}{2}}\sum_{i = 1}^N \text{erf}\left(\frac{\beta_i - 2}{\sqrt{2}}\right)$$
and I don't see any way of simplifying this.
Edit: As @Henry said in his comment, you can rewrite this in terms of the CDF of a standard normal, as: $$\Phi(x)=\frac{1}{2}\left(1+\text{erf}\left(\frac{x}{\sqrt 2}\right)\right)$$ thus: $$\text{erf}\left(\frac{x}{\sqrt 2}\right)=2\Phi(x)-1$$ So we can simplify the expression as: $$N\sqrt{\frac{\pi}{2}}+\sqrt{\frac{\pi}{2}}\sum_{i = 1}^N [2\Phi(\beta_i - 2)-1]=\sqrt{2\pi}\sum_{i=1}^N \Phi(\beta_i - 2)$$