Assuming there's a variable I want to measure, but I have only very noisy instrument to do so. So I want to take multiple measurements so that I have a better chance to recover the state of the variable. Hopefully, with each measurement, my instrument can report the result as a Gaussian distribution , with the mean to be the most likely state of variable and the standard deviation suggests a rough possible region of the state.
My problem now is that I don't know how to combine these multiple measurements to get a sensible answer. My guess is that it would be nice if I can get a new gaussian from these results, with the mean centered at the expectation value of the state of the variable, and a standard deviation to reflect how confident I am about the result...
I tried to teach myself about gaussians, and probabilities, but I just couldn't get my head around...please can someone help me?
Actually, you are measuring something of the form \begin{equation} X_i = v + b_i \, , \end{equation} where $v$ is the deterministic value you want to measure, and $b_i$ is the value of a Gaussian noise at the $i$th measurement. If the measurements are independent from each other, then simply take the arithmetic mean \begin{equation} \overline{X}_n = \frac{1}{n} \sum_{i=1}^n X_i \, , \end{equation} which has a normal distribution, by linear combination of Gaussian variables.
If your system is ergodic (broadly speaking, the system and the noise do not change behavior over time, i.e. $v$ and the distribution of $b_i \sim\mathcal{N}(\mu,\sigma^2)$ do not change over time), then the expected value and the variance of $\overline{X}_n$ are \begin{equation} E(\overline{X}_n) = \frac{1}{n} \sum_{i=1}^n E(X_i) = v+\mu \, ,\\ V(\overline{X}_n) = \frac{1}{n^2} \sum_{i=1}^n V(X_i) = \frac{\sigma^2}{n} \, . \end{equation} The random variable $\overline{X}_n$ has a normal distribution $\mathcal{N}(v+\mu, \sigma^2/n)$. If you have a reference measurement where the value $v$ is known, e.g. deduced from another measurement technique, then you can estimate $\mu$ and deduce by how much the noise modifies the mean of $\overline{X}_n$.
If the noise distribution changes at each measurement, $b_i\sim \mathcal{N}(\mu_i,{\sigma_i}^2)$ for each $i$, then the arithmetic mean $\overline{X}_n$ has a normal distribution $\mathcal{N}(v + \overline{\mu}_n, {\overline{\sigma^2}}_n/n)$. Alternatively, one can compute the weighted and centered mean \begin{equation} \widetilde{X}_n = \sum_{i=1}^n w_i \left(X_i - \mu_i\right) \quad\text{with the weights}\quad w_i = \frac{{\sigma_i}^{-1}}{\sum_{j=1}^n {\sigma_j}^{-1}} \, , \end{equation} which reduces to the arithmetic mean $\overline{X}_n$ when $\mu_i = 0$ and $\sigma_i = \sigma$ for all $i$. The expected value and the variance of $\widetilde{X}_n$ are \begin{equation} E(\widetilde{X}_n) = v \sum_{i=1}^n w_i = v \, ,\\ V(\widetilde{X}_n) = \sum_{i=1}^n {w_i}^2 {\sigma_i}^2 = n \left(\sum_{j=1}^n {\sigma_j}^{-1}\right)^{-2} . \end{equation}