Template:Normal Distribution fisher matrix confidence bounds

Bounds on the Parameters
The lower and upper bounds on the mean, $$\widehat{\mu }$$, are estimated from:


 * $$\begin{align}

& {{\mu }_{U}}= & \widehat{\mu }+{{K}_{\alpha }}\sqrt{Var(\widehat{\mu })}\text{ (upper bound),} \\ & {{\mu }_{L}}= & \widehat{\mu }-{{K}_{\alpha }}\sqrt{Var(\widehat{\mu })}\text{ (lower bound)}\text{.} \end{align}$$

Since the standard deviation, $${{\widehat{\sigma }}_{T}}$$, must be positive,  $$\ln ({{\widehat{\sigma }}_{T}})$$  is treated as normally distributed, and the bounds are estimated from:


 * $$\begin{align}

& {{\sigma }_{U}}= & {{\widehat{\sigma }}_{T}}\cdot {{e}^{\tfrac{{{K}_{\alpha }}\sqrt{Var({{\widehat{\sigma }}_{T}})}}}}\text{ (upper bound),} \\ & {{\sigma }_{L}}= & \frac\text{ (lower bound),} \end{align}$$

where $${{K}_{\alpha }}$$  is defined by:


 * $$\alpha =\frac{1}{\sqrt{2\pi }}\int_^{\infty }{{e}^{-\tfrac{2}}}dt=1-\Phi ({{K}_{\alpha }})$$

If $$\delta $$  is the confidence level, then  $$\alpha =\tfrac{1-\delta }{2}$$  for the two-sided bounds and  $$\alpha =1-\delta $$  for the one-sided bounds. The variances and covariances of $$\widehat{\mu }$$  and  $${{\widehat{\sigma }}_{T}}$$  are estimated from the Fisher matrix, as follows:


 * $$\left( \begin{matrix}

\widehat{Var}\left( \widehat{\mu } \right) & \widehat{Cov}\left( \widehat{\mu },{{\widehat{\sigma }}_{T}} \right) \\ \widehat{Cov}\left( \widehat{\mu },{{\widehat{\sigma }}_{T}} \right) & \widehat{Var}\left( {{\widehat{\sigma }}_{T}} \right) \\ \end{matrix} \right)=\left( \begin{matrix} -\tfrac{{{\partial }^{2}}\Lambda }{\partial {{\mu }^{2}}} & -\tfrac{{{\partial }^{2}}\Lambda }{\partial \mu \partial {{\sigma }_{T}}} \\ {} & {} \\   -\tfrac{{{\partial }^{2}}\Lambda }{\partial \mu \partial {{\sigma }_{T}}} & -\tfrac{{{\partial }^{2}}\Lambda }{\partial \sigma _{T}^{2}}  \\ \end{matrix} \right)_{\mu =\widehat{\mu },\sigma =\widehat{\sigma }}^{-1}$$

$$\Lambda $$ is the log-likelihood function of the normal distribution, described in Chapter 3 and Appendix C.

Bounds on Reliability
The reliability of the normal distribution is:


 * $$\widehat{R}(T;\hat{\mu },{{\hat{\sigma }}_{T}})=\int_{T}^{\infty }\frac{1}{{{\widehat{\sigma }}_{T}}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{t-\widehat{\mu }} \right)}^{2}}}}dt$$

Let $$\widehat{z}(t;\hat{\mu },{{\hat{\sigma }}_{T}})=\tfrac{t-\widehat{\mu }},$$  then  $$\tfrac{dz}{dt}=\tfrac{1}.$$  For  $$t=T$$,  $$\widehat{z}=\tfrac{T-\widehat{\mu }}$$ , and for  $$t=\infty ,$$   $$\widehat{z}=\infty .$$  The above equation then becomes:


 * $$\hat{R}(\widehat{z})=\int_{\widehat{z}(T)}^{\infty }\frac{1}{\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{z}^{2}}}}dz$$

The bounds on $$z$$  are estimated from:


 * $$\begin{align}

& {{z}_{U}}= & \widehat{z}+{{K}_{\alpha }}\sqrt{Var(\widehat{z})} \\ & {{z}_{L}}= & \widehat{z}-{{K}_{\alpha }}\sqrt{Var(\widehat{z})} \end{align}$$


 * where:


 * $$Var(\widehat{z})={{\left( \frac{\partial z}{\partial \mu } \right)}^{2}}Var(\widehat{\mu })+{{\left( \frac{\partial z}{\partial {{\sigma }_{T}}} \right)}^{2}}Var({{\widehat{\sigma }}_{T}})+2\left( \frac{\partial z}{\partial \mu } \right)\left( \frac{\partial z}{\partial {{\sigma }_{T}}} \right)Cov\left( \widehat{\mu },{{\widehat{\sigma }}_{T}} \right)$$


 * or:


 * $$Var(\widehat{z})=\frac{1}{\widehat{\sigma }_{T}^{2}}\left[ Var(\widehat{\mu })+{{\widehat{z}}^{2}}Var({{\widehat{\sigma }}_{T}})+2\cdot \widehat{z}\cdot Cov\left( \widehat{\mu },{{\widehat{\sigma }}_{T}} \right) \right]$$

The upper and lower bounds on reliability are:


 * $$\begin{align}

& {{R}_{U}}= & \int_^{\infty }\frac{1}{\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{z}^{2}}}}dz\text{ (upper bound)} \\ & {{R}_{L}}= & \int_^{\infty }\frac{1}{\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{z}^{2}}}}dz\text{ (lower bound)} \end{align}$$

Bounds on Time
The bounds around time for a given normal percentile (unreliability) are estimated by first solving the reliability equation with respect to time, as follows:


 * $$\hat{T}(\widehat{\mu },{{\widehat{\sigma }}_{T}})=\widehat{\mu }+z\cdot {{\widehat{\sigma }}_{T}}$$


 * where:


 * $$z={{\Phi }^{-1}}\left[ F(T) \right]$$


 * and:


 * $$\Phi (z)=\frac{1}{\sqrt{2\pi }}\int_{-\infty }^{z(T)}{{e}^{-\tfrac{1}{2}{{z}^{2}}}}dz$$

The next step is to calculate the variance of $$\hat{T}(\widehat{\mu },{{\widehat{\sigma }}_{T}})$$  or:


 * $$\begin{align}

Var(\hat{T})= & {{\left( \frac{\partial T}{\partial \mu } \right)}^{2}}Var(\widehat{\mu })+{{\left( \frac{\partial T}{\partial {{\sigma }_{T}}} \right)}^{2}}Var({{\widehat{\sigma }}_{T}}) \\ & +2\left( \frac{\partial T}{\partial \mu } \right)\left( \frac{\partial T}{\partial {{\sigma }_{T}}} \right)Cov\left( \widehat{\mu },{{\widehat{\sigma }}_{T}} \right) \\ Var(\hat{T})= & Var(\widehat{\mu })+{{\widehat{z}}^{2}}Var({{\widehat{\sigma }}_{T}})+2\cdot z\cdot Cov\left( \widehat{\mu },{{\widehat{\sigma }}_{T}} \right) \end{align}$$

The upper and lower bounds are then found by:


 * $$\begin{align}

& {{T}_{U}}= & \hat{T}+{{K}_{\alpha }}\sqrt{Var(\hat{T})}\text{ (upper bound)} \\ & {{T}_{L}}= & \hat{T}-{{K}_{\alpha }}\sqrt{Var(\hat{T})}\text{ (lower bound)} \end{align}$$

Example 4
Using the data of Example 2 and assuming a normal distribution, estimate the parameters using the MLE method.

Solution to Example 4
In this example we have non-grouped data without suspensions and without interval data. The partial derivatives of the normal log-likelihood function, $$\Lambda ,$$  are given by:


 * $$\begin{align}

\frac{\partial \Lambda }{\partial \mu }= & \frac{1}\underset{i=1}{\overset{14}{\mathop \sum }}\,({{T}_{i}}-\mu )=0 \\ \frac{\partial \Lambda }{\partial \sigma }= & \underset{i=1}{\overset{14}{\mathop \sum }}\,\left( \frac{{{T}_{i}}-\mu }-\frac{1}{\sigma } \right)=0 \end{align}$$

(The derivations of these equations are presented in Appendix C.) Substituting the values of $${{T}_{i}}$$  and solving the above system simultaneously, we get  $$\widehat{\sigma }=29.58$$  hours $$,$$   $$\widehat{\mu }=45$$  hours $$.$$

The Fisher matrix is:


 * $$\left[ \begin{matrix}

\widehat{Var}\left( \widehat{\mu } \right)=62.5000 & {} & \widehat{Cov}\left( \widehat{\mu },\widehat{\sigma } \right)=0.0000 \\ {} & {} & {} \\   \widehat{Cov}\left( \widehat{\mu },\widehat{\sigma } \right)=0.0000 & {} & \widehat{Var}\left( \widehat{\sigma } \right)=31.2500  \\ \end{matrix} \right]$$

Using Weibull++, the MLE method can be selected from the Set Analysis page.



The plot of the solution for this example is shown next.