Dividing a Number with a Vector Results in a new Vector with wrong mean

Hey All,

I have an interesting problem;

I have a vector (VectorInitial). I want to calculate another vector (VectorTarget) with the same size such that the elements in VectorTarget(i)=60/VectorInitial(i).

I have tried several ways;

VectorTarget = rdivide(60,VecInitial)

VectorTarget = 60./VectorInitial

Writing a for loop to go through all the elements and divide 60 by each element 1 by 1 and write it into a new vector.

All of these methods create the same VectorTarget as expectedly. However there is a huge problem.

In my case the mean(VectorInitial) = 25.0192

60/25.0192 = 2.3982

One would expect mean(VectorTarget) to be 2.3982

However mean(VectorTarget) = 2.4863

I am really baffled as to why this is happening. Any suggestions would be highly appreciated.

Answer 1

[Luna]

Because they are not equal mathematically.

Suppose that:

VectorInit = [a,b,c]

meanVectorInit = (a+b+c)/3

VectorTarget = [60/a,60/b,60/c]

meanVectorTarget = (60/a+60/b+60/c)/3 = 20/a + 20/b + 20/c

Then check;

60/meanVectorInit = 60/( (a+b+c)/3 ) = 180/(a+b+c)

So;

20/a + 20/b + 20/c ~= 180/(a+b+c) -> they are not always equal. Depends on what a,b,c is.

meanVectorTarget ~= 60/meanVectorInit

I recommend you to do the symbolic math operations on a paper with your hand, you will see they are not equal.

[Ferhat Buke]

You are right and I feel kind of stupid.

I am working with a large dataset where it would make a lot of sense for the value to be closer to 2.4 than it would be to 2.5.

I see now that for this to work, the spread around the average cannot be a normal distribution. You need a distribution heavier on one side with a longer tail on the other side.

Thanks a lot!!

[Luna]

Your welcome :)