Compute the bias of $\frac{\sum_{i=1}^N\frac{z_i}{\sigma_i^2}}{\sum_{i=1}^N\frac{1}{\sigma_i^2}}$

Question

I found the maximum likelihood estimator of $x$. Now, how to compute the bias of: $$ \hat{x}_{ML} = \frac{\sum_{i=1}^N\frac{z_i}{\sigma_i^2}}{\sum_{i=1}^N\frac{1}{\sigma_i^2}} $$

Where $\hat{x}_{ML}$ is the maximum likelihood estimator of $x$, the relationship of $z$ and $x$ is: $z_i=x+v_i$ being the error $v_i$ a Gaussian noise with zero mean and variance $\sigma_i^2$. I got stuck here:

$$ Bias(\hat{x}_{ML})=|x-\mathbb{E}\left(\hat{x}_{ML}\right)|=\left|x-\mathbb{E}\left(\frac{\sum_{i=1}^N\frac{z_i}{\sigma_i^2}}{\sum_{i=1}^N\frac{1}{\sigma_i^2}}\right)\right| $$

Now, I do not know how to get inside the expectation, becasue the expectation of the quotient is not the quotient of expectations....

Answer 1

Hint

$$\mathbb{E}\left(\frac{\sum_{i=1}^N\frac{z_i}{\sigma_i^2}}{\sum_{i=1}^N\frac{1}{\sigma_i^2}}\right)=\frac{\sum_{i=1}^N\frac{\mathbb{E}\{z_i\}}{\sigma_i^2}}{\sum_{i=1}^N\frac{1}{\sigma_i^2}}$$