Statistics question

[KnickNut3]

Hello. I need to calculate the SD of my error given the SD in 3 directions (assuming no covariance).

So, dealing with variances, I want:

Var(error) = Var(sqrt(x^2+y^2+z^2))

does that equal

sqrt[var(x)^2+var(y)^2+var(z)^2] ?

I'm not always sure about variances of functions of variables. Couldn't find anything to direct me on Mathworld, etc.

Thanks.

 

[bonkers325]

no, it doesnt.

var(f(x)) = { f'[E(X)]^2 * var(X) }

 

[KnickNut3]

Hmm, by calculating it, assuming all three variables are centered at 0, hence Var(x) = E(x^2) - E(x)^2 --> Var(x) = E(x^2)

Var(sqrt(x^2+y^2+z^2)) = E((sqrt(x^2+y^2+z^2))^2) - E(sqrt(x^2+y^2+z^2))^2

Now the last term goes it 0 since the expected position is 0 and the first tirm just becomes E(x^2+y^2+z^2)

Because of linearity that becomes E(x^2) + ... = Var(x) + Var(y) + Var(z)

Am I right?

 

[KnickNut3]

Originally posted by: bonkers325
no, it doesnt.

var(f(x)) = { f'[E(X)]^2 * var(X) }

Huh?

The derivative of my function would be 1/2 (x2+y2+z2)^(-1/2) * 2x
Plugging in E(x) = 0 would make everything go to 0 every time, making your whole equation go to zero, and the variance definitely isn't zero. Am I misunderstanding something?

 

[KnickNut3]

Looking for confirmation...

 

[JohnCU]

Originally posted by: KnickNut3
Hello. I need to calculate the SD of my error given the SD in 3 directions (assuming no covariance).

So, dealing with variances, I want:

Var(error) = Var(sqrt(x^2+y^2+z^2))

does that equal

sqrt[var(x)^2+var(y)^2+var(z)^2] ?

I'm not always sure about variances of functions of variables. Couldn't find anything to direct me on Mathworld, etc.

Thanks.

the variance is not a linear operator. Var(aX + b) = a^2Var(X)

 

[KnickNut3]

I know that.

But what about square roots of added values, each of which has variances?

 

[JohnCU]

i have no idea. we didn't go that indepth into variances... (didn't take a statistics class though).

 

[chuckywang]

Originally posted by: KnickNut3
Hmm, by calculating it, assuming all three variables are centered at 0, hence Var(x) = E(x^2) - E(x)^2 --> Var(x) = E(x^2)

Var(sqrt(x^2+y^2+z^2)) = E((sqrt(x^2+y^2+z^2))^2) - E(sqrt(x^2+y^2+z^2))^2

Now the last term goes it 0 since the expected position is 0 and the first tirm just becomes E(x^2+y^2+z^2)

Because of linearity that becomes E(x^2) + ... = Var(x) + Var(y) + Var(z)

Am I right?

You can't say the last term goes to 0.

 

[KnickNut3]

Right, because I'm now finding displacement from zero, not net position. Damn.

Any other ideas?

 

[chuckywang]

Originally posted by: KnickNut3
Right, because I'm now finding displacement from zero, not net position. Damn.

Any other ideas?

Compute E(sqrt(x^2+y^2+z^2))^2 directly?

But it probably isn't the way you want to go. I would suggest trying to do the problem another way.

 

[KnickNut3]

Originally posted by: chuckywang

Originally posted by: KnickNut3
Right, because I'm now finding displacement from zero, not net position. Damn.

Any other ideas?

Compute E(sqrt(x^2+y^2+z^2))^2 directly?

But it probably isn't the way you want to go. I would suggest trying to do the problem another way.

Are you assuming this is part of a larger problem? It's not.

A research paper reported that an object could be positioned using a new technology in 3D with these SDs. I'm trying to approximate an overall SD from actual position, assuming no covariance.

 

[KnickNut3]

Paper's due tomorrow, google revealed nothing, would appreciate some assistance. Is there any way I can calculate this without knowing the mean x y and z?