# Error propagation for non-normal errors

1. Jan 17, 2015

### SamanthaYellow

I have several measurements taken over a time series. Each data point has a standard error value. I need to sum up the data points, and determine the error associated with that sum. The error values across the time series are non-normal, so I'm assuming that I can't use the usual error propagation rules (i.e., SE_total = √(SE1^2 + SE2^2 +....) ). A log-transform of the errors shows that the errors are log-normal though. I'm not sure how to approach this. Is there a way to sum log-normal errors?

2. Jan 17, 2015

### mathman

The formula for standard error does not require a normal distribution. The only condition is that the measurement errors be independent.

3. Jan 19, 2015

### SamanthaYellow

I thought that since the formula for error propagation is derived for a Gaussian distribution, the typical summation of errors in quadrature is not applicable when errors are not normally distributed.

4. Jan 19, 2015

### mathman

The typical derivation is based on the the following (two variables - generalizable to n):
Simplify writing by assuming means are 0. E((X+Y)²)=E(X²)+2E(XY)+E(Y²). IF X and Y are independent (uncorrelated is enough), E(XY)=E(X)E(Y)=0.

5. Jan 19, 2015

### SamanthaYellow

Thank you for your replies. I've done more searching on this topic and it seems I'm not alone in my confusion about this. I want to make sure I've got this right: even though the distribution of the errors is non-normal, the usual error propagation rules are applicable since the function that I'm propagating errors for is a simple sum (i.e., linear). Does the distribution of errors come into play for propagating errors for more complex, nonlinear functions?

6. Jan 20, 2015

### mathman

Not necessarily. E((XY)^2)=E(X^2)E(Y^2), so you have variances multiplying, not adding.

7. Jan 29, 2015

### DrDu

Statisticians call this the delta method, and another important assumption is that you can use a lowest order Taylor expansion. So $Var(f(x))=\langle f^2(x)-\langle f(x)\rangle^2\rangle=\langle (f(0)+f'(0)x+f''(0)x^2/2)^2+\ldots -\langle f(0)+f'(0)x+f''(0)x^2/2+\ldots)\rangle^2 \rangle=f'(0)^2 \langle x^2 \rangle$+ higher order terms.
The higher order terms depend not only on the statistics of x but also on the Taylor series. It might be that they all disappear for a Gaussian, as higher order correlation functions can all be expressed in terms of $E(x^2)$.

You say that your error is lognormal distributed?
So why don't you use error propagation for the logarithmized independent variable , i.e. replacing Var(f(y) by $Var(f(e^x))$?