Question

Equation $\mathbf{(}\mathbf{10}\mathbf{.12}\mathbf{)}$isonly an approximation (but usually satisfactory). Show, however, that if you keep the second order termsin,$\mathbf{\left(}\mathbf{10.10}\mathbf{\right)}$then

role="math" localid="1664364127028" $\overline{\mathbf{w}}\mathbf{=}\mathit{w}\mathbf{(}\overline{\mathbf{x}}\mathbf{,}\overline{\mathbf{y}}\mathbf{)}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\left(}\frac{{\mathbf{\partial }}^{\mathbf{2}}\mathbf{w}}{\mathbf{\partial }{\mathbf{x}}^{\mathbf{2}}}\mathbf{\right)}{\mathit{\sigma }}_{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\left(}\frac{{\mathbf{\partial }}^{\mathbf{2}}\mathbf{w}}{\mathbf{\partial }{\mathbf{y}}^{\mathbf{2}}}\mathbf{\right)}{\mathit{\sigma }}_{\mathbf{y}}^{\mathbf{2}}$.

Short answer

Required expression is:$\overline{w}=w(\overline{x},\overline{y})+\frac{1}{2}\left(\frac{{\partial }^{2}w}{\partial x^2}\right){\sigma }_x^2+\frac{1}{2}\left(\frac{{\partial }^{2}w}{\partial y^2}\right){\sigma }_y^2$

Step-by-step solution

Given Information

The equation,$\left(10.10\right)$ i.e.,$w(x,y)\cong w({\mu }_x,{\mu }_y)+\left(\frac{\partial w}{\partial x}\right)(x-{\mu }_x)+\left(\frac{\partial w}{\partial y}\right)(y-{\mu }_y)$

The equation$\left(10.12\right)$, i.e.,$\overline{w}=w(\overline{x},\overline{y})$

Definition of Expectation

The expected value is a generalisation of the weighted average, commonly known as expectation, in probability theory.

Expand using Taylor’s series.

Write the expression using Taylor’s Series.

$\begin{array}{c}w(x,y)\cong w({\mu }_x,{\mu }_y)+\left(\frac{\partial w}{\partial x}\right)(x-{\mu }_x)+\left(\frac{\partial w}{\partial y}\right)(y-{\mu }_y)\\ +\frac{1}{2}[\frac{{\partial }^{2}w}{\partial x^2}{(x-{\mu }_x)}^2+2\frac{{\partial }^{2}w}{\partial x\partial y}(x-{\mu }_x)(y-{\mu }_y)+\frac{{\partial }^{2}w}{\partial y^2}{(x-{\mu }_y)}^2]\\ \left(w\right(x,y\left)\right)\cong w({\mu }_x,{\mu }_y)+\left(\frac{\partial w}{\partial x}\right)\left(E\left(x\right)-{\mu }_x\right)+\left(\frac{\partial w}{\partial y}\right)\left(E\left(y\right)-{\mu }_y\right)\\ +\frac{1}{2}[\frac{{\partial }^{2}w}{\partial x^2}E{(x-{\mu }_x)}^2+2\frac{{\partial }^{2}w}{\partial x\partial y}E\left[(x-{\mu }_x)(y-{\mu }_y)\right]+\frac{{\partial }^{2}w}{\partial y^2}E{(x-{\mu }_y)}^2]\end{array}$

Calculate expected value. 

Solve for expected value.

$\begin{array}{c}\left(E\left(x\right)-{\mu }_x\right)=\left(E\left(y\right)-{\mu }_y\right)\\ =0\\ E{(x-{\mu }_x)}^2={\sigma }_x^2\\ E{(x-{\mu }_y)}^2={\sigma }_y^2\end{array}$

Simplify further.

$\begin{array}{c}E\left[(x-{\mu }_x)(y-{\mu }_y)\right]=0\\ \overline{w}=w(\overline{x},\overline{y})+\frac{1}{2}\left(\frac{{\partial }^{2}w}{\partial x^2}\right){\sigma }_x^2+\frac{1}{2}\left(\frac{{\partial }^{2}w}{\partial y^2}\right){\sigma }_y^2\end{array}$

Hence, required expression is:.$\overline{w}=w(\overline{x},\overline{y})+\frac{1}{2}\left(\frac{{\partial }^{2}w}{\partial x^2}\right){\sigma }_x^2+\frac{1}{2}\left(\frac{{\partial }^{2}w}{\partial y^2}\right){\sigma }_y^2$