Measurement and Propagation of Error (Part 2)

Posted by Amit Rajan on Wednesday, November 14, 2018

#### Linear Combinations of Dependent Measurements :

In the case of dependent measurement, to quantify the uncertainty, we need to know the value of covariance for all the possible pairs of measurements. This is practically not feasible. In this case, an upper bound can be placed on the uncertainty. If $X_1, X_2, …, X_n$ are $n$ dependent measurements and $c_1, c_2, …, c_n$ are constants, then the uncertainty of $c_1X_1 + c_2X_2 + … + c_nX_n$ can be bounded as:

$$\sigma _{c_1X_1 + c_2X_2 + … + c_nX_n} \leq |c_1|\sigma _{X_1} + |c_2|\sigma _{X_2} + … + |c_n|\sigma _{X_n}$$


#### Uncertainties for Functions of One Measurement :

If $X$ is a measurement whose uncertainty $\sigma_X$ is small and if $U$ is a function of $X$, then uncertainty in the measurement of $U$ can be approximated as (where the derivative is computed at the observed measurement $X$):

$$\sigma_U \approx \bigg| \frac{dU}{dX} \bigg| \sigma_X$$

The above approximation can be proved by using a simple mathematical technique. A Taylor Series of a function $f(X)$, which is infinitely differentiable at $a$ is given as:

$$f(a) + \frac{f^{’}(a)}{1!}(x-a) + \frac{f^{’’}(a)}{2!}(x-a)^2 + \frac{f^{’’’}(a)}{3!}(x-a)^3 + …$$

In our case, if $X$ is close to $\mu_X$, then the first-order Taylor Series approximation for $U(X)$ can be given as:

$$U(X) \approx U(\mu_X) + \frac{U^{’}(\mu_X)}{1!}(x-\mu_X) = U(\mu_X) + \frac{dU}{dX}(X-\mu_X)$$

where $\frac{dU}{dX}$ is evaluated at $\mu_X$. It should be noted that for any reasonable precise measurement, $X$ will be close enough to $\mu _{X}$ for the Taylor series approximation to be valid. Rearranging the above expression, we get

$$U(X) \approx \bigg( U(\mu_X) - \frac{dU}{dX} \mu _{X} \bigg) + \frac{dU}{dX}X$$

As $\frac{dU}{dX}$ is measured at $\mu_X$, it is constant and hence the quantity inside the bracket is constant. This gives the uncertainty (standard deviation) as:

$$\sigma_U \approx \bigg| \frac{dU}{dX} \bigg| \sigma_X$$

This is the propagation of error formula and is applied in almost all the back-propagation algorithms in neural-networks as well.


#### Uncertainties for Functions of Several Measurements :

If $X_1, X_2, …, X_n$ are independent measurements whose uncertainties $\sigma _{X_1}, \sigma _{X_2}, …, \sigma _{X_n}$ are small and if $U = U(X_1, X_2, …, X_n)$ is a function of $X_1, X_2, …, X_n$, then

$$\sigma_U \approx \sqrt{\bigg( \frac{\partial U}{\partial X_1}\bigg)^2 \sigma _{X_1}^2 + \bigg( \frac{\partial U}{\partial X_2}\bigg)^2 \sigma _{X_2}^2 + … + \bigg( \frac{\partial U}{\partial X_n}\bigg)^2 \sigma _{X_n}^2}$$

This is the multivariate propagation of error formula and is valid only when the measurements are independent. This formula can be derived in a similar way as the one discussed for the one measurement case.

Example: Two resistors with resistances R1 and R2 are connected in parallel. The combined resistance R is given by R = (R1R2)/(R1+ R2). If R1 is measured to be 100±10, and R2 is measured to be 20±1, estimate R and find the uncertainty in the estimate.

Sol: The estimate of $R$ can be given as $\frac{100 \times 20}{100 + 20}$ = 16.67. First of all, we need to compute the partial derivative of R as:

$$\frac{\partial R}{\partial R1} = \bigg( \frac{R2}{R1+R2}\bigg)^2 = 0.0278$$

$$\frac{\partial R}{\partial R2} = \bigg( \frac{R1}{R1+R2}\bigg)^2 = 0.694$$

Now, $\sigma _{R1} = 10$ and $\sigma _{R2} = 1$, and hence

$$\sigma _{R} = \sqrt{\bigg( \frac{\partial R}{\partial R1}\bigg)^2 \sigma _{R1}^2 + \bigg( \frac{\partial R}{\partial R2}\bigg)^2 \sigma _{R2}^2} = 0.75$$

Hence, the combined resistance is 16.67 ± 0.75.


#### Uncertainties for Functions of Dependent Measurements :

In the case of the dependent measurements, the uncertainty can be calculated accurately if the covariance of each pair of measurements is known. Instead, we can give a conservative estimate for uncertainty as (where the terms have usual meaning):

$$\sigma _{U} \leq \bigg| \frac{\partial U}{\partial X_1}\bigg| \sigma _{X_1} + \bigg| \frac{\partial U}{\partial X_2}\bigg| \sigma _{X_2} + … + \bigg| \frac{\partial U}{\partial X_n}\bigg| \sigma _{X_n}$$

This inequality is valid for almost all the cases of dependent measurements.


#### Reference:

https://www.mheducation.com/highered/product/statistics-engineers-scientists-navidi/M0073401331.html