How to Simplify the Mean of 3D Variables to 1D?

  • Thread starter Thread starter pangyatou
  • Start date Start date
  • Tags Tags
    1d 3d Mean
pangyatou
Messages
4
Reaction score
0
Hi,

There are three variables ax, ay and az, my question is:
How to simplify the mean value <(ax^2+ay^2+az^2)^(1/2)> to <|ax|> ?
What assumptions are required during the simplification?

The statistical property of ax, ay and az is <ax^2>=<ay^2>=<az^2>.
The assumption of the propability is: pdf(ax), pdf(ay) and pdf(az) are independent to each other: p(ax,ay,az)=p(ax)p(ay)p(az)

Thanks
 
Physics news on Phys.org
No further assumptions are needed to carry out the calculation. It is messy because you are taking the square root before calculating the integral.
 
Thanks Mathman!
What theory or property can be applied to this problem? I don't even have a clue.

Really appreciate.
 
It is a 3-d integral where the integrand is the product of the 3 density functions multiplied by the expression (square root etc.).
 
pangyatou said:
Hi,

There are three variables ax, ay and az, my question is:
How to simplify the mean value <(ax^2+ay^2+az^2)^(1/2)> to <|ax|> ?
What assumptions are required during the simplification?

Are you asking if the mean value of r = \sqrt{a_x^2 + a_y^2 + a_z^2} must be equal to the mean value of the absolute value of a_x ?
 
Namaste & G'day Postulate: A strongly-knit team wins on average over a less knit one Fundamentals: - Two teams face off with 4 players each - A polo team consists of players that each have assigned to them a measure of their ability (called a "Handicap" - 10 is highest, -2 lowest) I attempted to measure close-knitness of a team in terms of standard deviation (SD) of handicaps of the players. Failure: It turns out that, more often than, a team with a higher SD wins. In my language, that...
Hi all, I've been a roulette player for more than 10 years (although I took time off here and there) and it's only now that I'm trying to understand the physics of the game. Basically my strategy in roulette is to divide the wheel roughly into two halves (let's call them A and B). My theory is that in roulette there will invariably be variance. In other words, if A comes up 5 times in a row, B will be due to come up soon. However I have been proven wrong many times, and I have seen some...

Similar threads

Back
Top