RMSE between original and predicted values.

Hi,

If I have thousand samples of my signal in a vector form like 1*1000, and I will predict my signal at each iteration that results into 1*1000 also. Then In this case, how will I find the RMSE of my model?

Many Thanks

Answer 1

[Star Strider]

Please see my Comment replying to your Comment.

[MAT-Magic]

Thank you very much Sir again for your kind help.

[Star Strider]

As always, my pleasure!

[MAT-Magic]

@ Star Strider, writing below code in for loop for RMSE. Is it correct way? waiting for your reply. Thanks

close all; clear all; clc;

v1 = [0.3 0.6 0.9];

v2 = [0.8 0.9 0.7];

for k = 1:length(v1)

y = v1-v2;

y1 = y.^2;

sumy1 = sum(y1);

end

RMSE = sqrt(sumy1/numel(v1));

[Star Strider]

It is correct, however you can write it much more simply:

v1 = [0.3 0.6 0.9];
v2 = [0.8 0.9 0.7];

RMSE = sqrt(mean((v1-v2).^2))

producing:

RMSE =
         0.355902608401044

Remembering that ‘RMSE’ means the ‘root of the mean of the squares’.

[MAT-Magic]

Thanks. But actually, I am accumulating the error inside the loop, so after that, I can take the mean and square root outside the loop to get RMSE of my model.

[Star Strider]

The RMSE calculation remains the same. You need to take the diferences, square them, accumulate them, take the mean, and the the square root of that.

[MAT-Magic]

OK. Thanks alot

[Star Strider]

As always, my pleasure!

Answer 2

[Image Analyst]

Without any other information, the maximum likelihood prediction for every element would be the mean of the entire signal. But it seems you'd rather have the rms, so you'd have

RMSE = rms(yourVector)
predictionVector = RMSE * ones(length(yourVector));

[MAT-Magic]

Thanks for the reply.