Is there a formula to calculate Standard Deviation from the given value of Mean Squared Error ?
For a given data set, I have the mean squared error value(s) calculated, and Standard Deviation is calculated based on this value. But I am not sure of the formula my system is using. Is there any general statistical formula for this?
This calculation is for Safety Stock value. Demand Standard deviation is used as an input for Safety Stock calculation which in turn is calculated from MSE. I need to find the formula to derive demand standard deviation from MSE.
Thanks in Advance.
Related
I am trying to implement VGGNet-16 for depth map prediction from single image. In the training the RMSE loss comes out to be 0.1599.
That loss value, is it in percentage or not?
No, if you want a percentage of a correctly classified data you can look at a value of accuracy.
Definition of RMSE from Wikipedia:
The root-mean-square deviation (RMSD) or root-mean-square error (RMSE) is a frequently used measure of the differences between values (sample and population values) predicted by a model or an estimator and the values actually observed.
It's always non-negative, and values closer to zero are better.
I’m having some issue with calculating the standard deviation of a game. In the game you can get several different discrete scores. The scores have a fixed probability which is given. There is also a 5% chance that your score is randomly generated. You do not know the distribution of the random variable you are only given the mean and variance.
I’ve calculated the variance of the main game (ignoring the random variable) to be 5.2. The variance of the random variable is 137. From this I get a standard deviation of
sqrt(5.2 + 5% *137) = 3.47
Is this the correct method?
I have some data in which a feature (height) is correlated with output variable (price). How to restore missing data (nulls) in height feature based on existing dependancy (correlation) between these variables?
To be more clear:
Input and output variables have clear correlation. I guess that predicting missing values for excel is not a difficult procedure. But I need some directions how to implement it.
If you got the slope (m) and intercept (c) of the regression line in E2 and E3 (say):-
=SLOPE(C2:C9,B2:B9)
=INTERCEPT(C2:C9,B2:B9)
you could re-arrange the simple regression equation y=mx+c to predict the x-values
x=(y-c)/m
So your predicted heights would be:-
=IF(ISBLANK(B2),(C2-E$3)/E$2,B2)
starting in D2.
You might try the FORECAST¹ function. The first blank does not have enough preceding data to generate a forecast result so a simple ratio will have to suffice but the remaining values can be generated and take previously generated FORECAST results into consideration for their own result(s).
The formula in E2 is,
=IF(ISBLANK(B2), FORECAST(C2, B$2:B$9, C$2:C$9), B2)
¹ See Forecasting functions for alternative algorithms in data prediction.
I have several weighted values for which I am taking a weighted average. I want to calculate a weighted standard deviation using the weighted values and weighted average. How would I modify the typical standard deviation to include weights on each measurement?
This is the standard deviation formula I am using.
When I simply use each weighted value for 'x' and the weighted average for '\bar{x}', the result seems smaller than it should be.
I just found this wikipedia page discussing data of equal significance vs weighted data. The correct way to calculate the biased weighted estimator of variance is
,
though the following, on-the-fly implementation, is more efficient computationally as it does not require calculating the weighted average before looping over the sum on the weighted differences squared
.
Despite my skepticism, I tried both and got the exact same results.
Note, be sure to use the weighted average
.
How to specify excel linest weighted polynomial fit formula, something like
LINEST(y*w^0.5,IF({1,0},1,x)*w^0.5,FALSE,TRUE), but this is for linear fit. I'm looking for similar formula for 2nd order and 3rd order polynomial regression fit.
In a reply to the other post in Weighted trendline an approach was already suggested for weighted polynomials. For example for a cubic fit try with CTRL+SHIFT+ENTER in a 4x1 range:
=LINEST(y*w^0.5,(x-1E-99)^{0,1,2,3}*w^0.5,FALSE)
(-1e-99 ensures that 0^0=1). Similar to the linear case for R^2 try:
=INDEX(LINEST((y-SUMPRODUCT(y,w)/SUM(w))*w^0.5,(x-1E-99)^{0,1,2,3}*w^0.5,FALSE,TRUE),3,1)
Derivation
In standard least squares we find the vector b that minimises:|y-Xb|²=(y-Xb)'(y-Xb)
In the weighted case b is chosen to minimise instead: |W(y-Xb)|²=(y-Xb)'W'W(y-Xb)
So the weighted regression is Wy on WX where W'W = W² is the diagonal matrix of the weights.