# Standard Deviation

1. Jun 15, 2007

### daudaudaudau

Hi all.

If I have a function V(x1,x2,x3,x4) and I want to calculate it's standard deviation when x1,x2,x3,x4 are standard normal and their variances are small, then formula (8) on this page

-Anders

Last edited by a moderator: Apr 22, 2017
2. Jun 15, 2007

### EnumaElish

3. Jun 15, 2007

### daudaudaudau

Hi.

Hmm, I think it's the clever part I don't understand then. Must I taylor expand V and then compute the standard deviation? And also, I can't see the connection between "difference" and "variance". I'd appreciate a few formulas. Thanks.

-Anders

4. Jun 15, 2007

### EnumaElish

I thought it would follow from dV = Sum[(partial V/partial xi) dxi, i=1,2,3,4]. The "d" operator is similar to "deviation" (e.g., from the mean). Let's say you have only 2 x's. Then dv = (Dv/Dx1) dx1 + (Dv/Dx2) dx2 ==> dv^2 = (Dv/Dx1)^2 dx1^2 + (Dv/Dx2)^2 dx2^2 + ignored term* approx. equal to (Dv/Dx1)^2 dx1^2 + (Dv/Dx2)^2 dx2^2. (I've used capital D for "partial.")

In fact, you don't even need the link I posted.

*This is the interaction term 2(Dv/Dx1)(Dv/Dx2) dx1 dx2. If you think that for a given "random draw" either of dx1 or dx2 (but not necessarily both, and you don't know which) is likely to be "very small" then you can assume 2(Dv/Dx1)(Dv/Dx2) dx1 dx2 = 0.

Last edited: Jun 16, 2007
5. Jun 15, 2007

### daudaudaudau

Okay. The only part I don't understand now is why dx is the same as a standard deviation ?

6. Jun 15, 2007

### EnumaElish

Last edited: Jun 15, 2007
7. Jun 17, 2007