I have a vector of points r where each element has an x and y value.
For every one of those points there is a z value, which corresponds to a measurement taken at position r.
If I have say n measurements taken at n different positions (for example see below) is there a way I can use the data I have to predict the expectation value for z at a random point r_2?
Yes, but for this you need a model.
A model in statistics (econometrics) defines a relation between the dependent variables, control variables and error (due to natural errors, errors of estimation etc.).
The easiest model that I know of is as follows:
$$z=Ax+By+\epsilon\text{ where }\epsilon\text{~}\mathbb{N}(0,\delta)$$
Now, what all that mean? You assume that $z$ depends linearly on $x$ and $y$. For example, with every increase in $x$ by $1$, $z$ increases roughly by $2$ or something like that.
You need this "roughly" because in general you are never going to get all your points $(x,y,z)$ lying in the same plane. If there were not "roughly" here, you would not be able to assume the linear dependence in this case, because you would have to have all points strictly in a plane. In the model we express this "roughly" as the error $\epsilon$. The error is something that is zero on average, but allows the observed values to be not exactly what is predicted by the model.
Now, given the model above, you can predict the values of $z$ for any pair $(x,y)$. Indeed,
$$Ez=Ax+By$$
because $E\epsilon=0$. The procedure should be as follows:
Collect all the data $(x,y,z)$ you have so far.
(optional but usually necessary) Try to plot the data $z$ against $(x,y)$ in a program that allows you to visually estimate possible functional form of $z$ as a function of $x$ and $y$. If you have few observed variables, or if you are fine with linear function, you can use the model above. Otherwise, if you see that the dependence is far from being linear, you might need to do one of two things:
a. Change the model. Find a function $f$ such that $z=f(x,y)$ will be similar to what you observe.
b. Transform data so that it looks linear. Suppose, for example, that increase in $z$ is proportional to the current values of $x$ and $y$, then you can take $\log$ of values to make it linear, etc.
Estimate the model. You basically find $A$ and $B$ such that they minimize the resulting error (the difference between observed data and their prediction). I am not going to go into all details here, because it is quite a standard procedure, and all you need to do is read about "linear regression". Even Excel can do that estimation for you if you load statistical package into it. BTW, Excel can also visualize the data pretty good.
Once you obtain estimated $A$ and $B$, you can substitute them into the above equation for averages to find the prediction of $z$.
Now, you asked about prediction of $z$ at a "random point" $r$. If by this you mean that $(x,y)$ is chosen randomly, then in order to estimate the predicted value of $z$, you need the distribution (or at least the mean value, i.e. average) for $x$ and $y$. Assuming, for example, that $x$ and $y$ are distributed uniformly from $0$ to $5$ as in the picture, their averages will be both equal to $2.5$, and in the linear model (only) you can simply plug them in for $x$ and $y$ to find out the expected value for $z$. If the chosen dependence $f$ is not linear, then you should calculate the expected value of $f(x,y)$ given their (joint) distribution (I am not sure you need/want this).
