How to check if specific MAE and MSE are feasible given only the real data?

Question

I have as data some real measures, let's say: 1000, 800, 900, 1100, 900 and I have the Mean Absolute Error (MAE) and Mean Squared Error (MSE) 80 and 20000, but I don't know which are the estimated data. \begin{align} MAE_s(h) &= \frac{1}{n}\sum_{x \in s} |f(x)-h(x)| \\[5pt] MSE_s(h) &= \frac{1}{n}\sum_{x \in s} (f(x)-h(x))^2 \end{align} In the above equations, $f(x)$ is the real measure(1000,800, 900, 1100 and 900) and $h(x)$ is the estimated measure (which is unknown to me), while $n$ is the measure amount (which is 5 in our case).

I have to prove if this situation is possible or not, so I took the MAE formula and I replace the values I knew, getting the equations which you can see below. I used $x,y,z,w,t$ to denote the values that I don't know, which are the the values for the estimated data. \begin{align} 80 &= \frac{1}{5} ((1000-x)+(800-y)+(900-z)+(1100-w)+(900-t)) \\[5pt] 20000 &= \frac{1}{5} ((1000-x)^2+(800-y)^2+(900-z)^2+(1100-w)^2+(900-t)^2) \end{align} I tried different values for $x,y,z,w,t$ until I found one estimated values combination that has the requested MAE and MSE. These values are: 900,800, 900, 800 and 900. So I got something like this: \begin{align} 80 &= \frac{1}{5} ((1000-900)+(800-800)+(900-900)+(1100-800)+(900-900)) \\[5pt] 80 &= 80 \\[5pt] 20000 &= \frac{1}{5} ((1000\!-\!900)^2+(800\!-\!800)^2+(900\!-\!900)^2+(1100\!-\!800)^2+(900\!-\!900)^2) \\[5pt] 20000 &= 20000 \end{align}

So I could prove that the situation was possible, but this doesn't seem a very handy way of doing that because there are a lot of possible combinations.

Is there any other way to check if a situation is possible or not given the MAE, MSE and the real data?

Answer 1

I interpret your question as being about the feasibility of a solution rather than its computation.

Let your "real data" be $\{y_i\}_{i=1,\ldots, n}$ and the "estimated data" be $\{\hat{y}_i\}_{i=1,\ldots, n}$. Define $z_i = |y_i - \hat{y}_i|$.

Using the fact that $Var(z_i) \geq 0$, we can show that $MSE(y, \hat{y}) = mean[z_i^2] \geq mean[z_i]^2 = MAE(y, \hat{y})^2$.

We also have $n MAE^2 \geq MSE$ because the $L_1$ norm of a vector is larger than its $L_2$ norm.

Your MSE and MAE values satisfy both these properties, therefore satisfying the necessary conditions for feasibility.

These two properties are also sufficient for feasibility as long as $n \geq 2$.

Proof: We need to find non-negative $z_i$ ($i = 2, \ldots, n$) such that $\sum_i z_i = n MAE$ and $\sum_i z_i^2 = n MSE$.

Set $z_1 = MAE + a$ and $z_i = MAE - \frac{a}{n-1}$ for $i = 2, \ldots, n$, where $a = \sqrt{(n-1) (MSE - MAE^2)}$.

It's easy to check that the the MSE and MAE values work out to what we want, and we can use the fact that $n MAE^2 \geq MSE$ to show that all the $z_i$s are non-negative.

Bottomline The two inequalities $MAE^2 \leq MSE \leq n MAE^2$ are both necessary and sufficient to be able to find an "estimated" vector that results in the particular MSE and MAE values.

Answer 2

The condition MAE≤MSE≤nMAE may be sufficient but is not necessary. It does not hold in the the case of the original example where MSE=20000 and is not <= 5x80=400. It seems that a better condition is (n MAE) squared >= n MSE . In the original example, (n MAE) squared = 160,000 and n MSE = 100,000 and therefore a solution can be found. This condition is based on the inequality: [sum abs(Z_i)] squared >= sum (Z_i squared). Z_i is the difference between f(i) and h(i).