4.4 Relationship to the method of least squares  4. The Method of Maximum Likelihood  4.2.6 Estimating exponential-decay time constants

4.3 Application to experimental design

A common problem of experimental design is to determine in advance if a proposed experimental configuration will be able to achieve a specified precision. In the preceding sections experimental results were used to estimate errors in parameters. For experimental design, it is useful to reformulate those expressions in ways suited to a priori estimation of experimental uncertainty.^4.2

The expectation value for an element of the information matrix H is

$$\langle H_{jk}\rangle =\left\langle{{\partial^2W}\over{\partial a_j\partial a_k}}\right\rangle\ .$$ (4.63)

 

If the observations are expected to occur according to some known probability distribution function $\phi$, this form may be reduced to a function of $\phi$. For N observations,

$$W(x,a)=\ln{\cal L}(x,a) = N \ln\phi(x,a) \ .$$ (4.64)

 

$$\langle H_{jk}\rangle = N\int {{\partial^2\ln\phi}\over{\partial a_j\partial a_k}}\phi dx$$ (4.65)

 

$$~~~ = N\int {{\partial}\over{\partial a_j}}\left({{1}\over{\phi}}{{\partial\phi}\over{\partial a_k}}\right)\phi dx$$ (4.66)

 

$$~~~ = N\left[-\int{{1}\over{\phi}}{{\partial\phi}\over{\parti...... a_k}} dx + \int{{\partial^2\phi}\over{a_ja_k}} dx\right] \ .$$ (4.67)

 

If the integration is performed before differentiation in the last term, the integral is over the probability distribution which must give a constant (unity), so the last term in (4.67) vanishes. The expectation value for an element in the information matrix is then

$$\langle H_{jk}\rangle =-N\int {{1}\over{\phi}} {{\partial\ph......er{\partial a_j}}{{\partial\phi}\over{\partial a_k}} dx \ \ .$$ (4.68)

 

In particular, for the case without correlations,

$$V_{a_ja_j} = {{1}\over{N\int {{1}\over{\phi}}({{\partial\phi}......angle({{\partial\ln\phi}\over{\partiala_j}})^2\right\rangle}}$$ (4.69)

 
 

Example 4.3: Suppose that a cloud droplet spectrum is expected to have a mean diameter $\mu$ of 15 $\mu$m and a standard deviation in diameter $\sigma$ of 5 $\mu$m. How many droplets must be measured, if the measurement error is negligible, to permit determination of the standard deviation with 5% precision, if the droplet sizes are distributed approximately in a Gaussian distribution?

From (3.2),

$$\ln\phi = -\ln\sigma - {{(d-\overline{d})^2}\over{2\sigma^2}}+ C$$ (4.70)

 

$${{\partial\ln\phi}\over{\partial \sigma}} = - {{1}\over{\sigma}}+ {{(d-\overline{d})^2}\over{\sigma^3}}$$ (4.71)

 

$$\left ({{\partial\ln\phi}\over{\partial \sigma}}\right )^2= ......e{d})^4 - 2\sigma^2(d-\overline{d})^2+\sigma^4}\over{\sigma^6}}$$ (4.72)

 

For a Gaussian distribution,

$$\langle(d-\overline{d})^2\rangle=\sigma^2$$ (4.73)

 

and

$$\langle(d-\overline{d})^4\rangle=3\sigma^4$$ (4.74)

 

(obtained by integration of the probability distribution), so the expected uncertainty in measurement of $\sigma$, $\sigma_\sigma$, is given by

$$\sigma_\sigma^2 = {{1}\over{N \left({{2\sigma^4}\over{\sigma^6}}\right)}} ={{\sigma^2}\over{2N}} .$$ (4.75)

 

To obtain $\sigma_\sigma=0.05(5)=0.25~\mu$m, N must equal 200 droplets.

   
 4.4 Relationship to the Method of Least Squares  4. The Method of Maximum Likelihood  4.2.6 Estimating exponential-decay time constants