Is the bias form an estimator a real number or we need to define it as a positive number/negative number? It seems to me that it can be any real number from the following.
From the proof of the variance and bias relationship as in here, we have that
$$\text{MSE}= \mathbb{V}ar_{\theta}(\hat{\theta})+\text{Bias}^2(\theta,\hat{\theta})$$
By setting $\beta=\text{Bias}(\theta,\hat{\theta})$ and since $\text{MSE}=\mathbb{E}[(\hat{\theta}-\theta)^2]$ we end up with the following
$$\beta=-\sqrt{\mathbb{V}ar_{\theta}(\hat{\theta})-\mathbb{E}[(\hat{\theta}-\theta)^2]},\quad\text{or}\quad \beta=\sqrt{\mathbb{V}ar_{\theta}(\hat{\theta})-\mathbb{E}[(\hat{\theta}-\theta)^2]}$$
it holds that $\mathbb{V}ar_{\theta}(\hat{\theta})\geq\mathbb{E}[(\hat{\theta}-\theta)^2]$. In other words the bias is a real number, right? Or am I missing something?
Is there any suggestion that bias,namely $\beta$, shohld be defined as a positive number and if yes why?
The actual definition of the bias is: $$\mathrm{Bias}(\hat\theta,\theta) = \mathbb{E}(\hat\theta-\theta)=\mathbb{E}(\hat\theta) - \theta$$ (Note that nothing is squared here.) So, it is a real number that can be either positive, negative, or zero. It is positive, if the expected value of the estimator is larger than the the estimand (i.e., the estimator has a positive bias) or negative if it's smaller. It is zero if the estimator is unbiased, in the sense that its expected value is exactly equal to the estimand.