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

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?

• You should be able to formulate this as a system of equations, and then the task is to solve it in terms of the parameters you are allowed to vary. Which of the two steps do you find problematic? Commented Nov 27, 2016 at 18:23
• Please explain what you mean by "MAE" and "MSE". Ordinarily the former refers to "mean absolute error" (around the mean), but for your data that value is $88$, not $80$. Ordinarily the latter refers to "mean squared error" (also about the mean), but for your data that value is $10400$, not $20000$. And to what "model" do you refer? What do you mean by "estimated values"?
– whuber
Commented Nov 27, 2016 at 18:41
• @RichardHardy I thought about that but I have 2 equations for 5 unknowns. PD I edited the question to clarify Commented Nov 27, 2016 at 19:32
• @whuber I updated the question. Thank you for the advices. If you think I must clarify anything more, please, let me know. Commented Nov 27, 2016 at 19:33
• If you spelled these equations out in your post, it would already be a step forward, IMHO. Commented Nov 27, 2016 at 19:36

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.

• This seems to make sense. Would you mind provide some source (link to a article or a book or similar) where I can find the two conditions for feasibility? I haven't heard about them before Commented Dec 8, 2016 at 10:04
• I checked the method and it works like a charm. Using this, you get one value for each z and they are right. But there are several combinations for z values that work, for example z={100,0,0,300,0}. Is there a way to get all this values for z? Commented Dec 8, 2016 at 10:30
• I don't have a reference for this. It's a simple result that depends on norms of vectors. Also, as @whuber mentioned in his comment, there are infinite solutions for $z$, unless you restrict it by imposing other conditions. Commented Dec 8, 2016 at 13:55

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).

• You forgot the squares for the MAE in your inequality. In the OP's data it is definitely true that MSE = 20000 < 5 * 80^2 = n MAE^2. Commented Dec 7, 2016 at 22:16
• I noticed after posting the above comment that I had a typo in my answer where I omitted the square on MAE myself. Since fixed. Commented Dec 8, 2016 at 13:57