91影视

The equation,$\left(10.10\right)$ i.e.,$w(x,y)鈮�w({渭}_x,{渭}_y)+\left(\frac{鈭�w}{鈭�x}\right)(x鈭�{渭}_x)+\left(\frac{鈭�w}{鈭�y}\right)(y鈭�{渭}_y)$

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

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

Write the expression using Taylor鈥檚 Series.

$\begin{array}{c}w(x,y)鈮�w({渭}_x,{渭}_y)+\left(\frac{鈭�w}{鈭�x}\right)(x鈭�{渭}_x)+\left(\frac{鈭�w}{鈭�y}\right)(y鈭�{渭}_y)\\ +\frac{1}{2}[\frac{{鈭�}^{2}w}{鈭�x^2}{(x鈭�{渭}_x)}^2+2\frac{{鈭�}^{2}w}{鈭�x鈭�y}(x鈭�{渭}_x)(y鈭�{渭}_y)+\frac{{鈭�}^{2}w}{鈭�y^2}{(x鈭�{渭}_y)}^2]\\ \left(w\right(x,y\left)\right)鈮�w({渭}_x,{渭}_y)+\left(\frac{鈭�w}{鈭�x}\right)\left(E\left(x\right)鈭�{渭}_x\right)+\left(\frac{鈭�w}{鈭�y}\right)\left(E\left(y\right)鈭�{渭}_y\right)\\ +\frac{1}{2}[\frac{{鈭�}^{2}w}{鈭�x^2}E{(x鈭�{渭}_x)}^2+2\frac{{鈭�}^{2}w}{鈭�x鈭�y}E\left[(x鈭�{渭}_x)(y鈭�{渭}_y)\right]+\frac{{鈭�}^{2}w}{鈭�y^2}E{(x鈭�{渭}_y)}^2]\end{array}$

Solve for expected value.

$\begin{array}{c}\left(E\left(x\right)鈭�{渭}_x\right)=\left(E\left(y\right)鈭�{渭}_y\right)\\ =0\\ E{(x鈭�{渭}_x)}^2={蟽}_x^2\\ E{(x鈭�{渭}_y)}^2={蟽}_y^2\end{array}$

Simplify further.

$\begin{array}{c}E\left[(x鈭�{渭}_x)(y鈭�{渭}_y)\right]=0\\ \stackrel{炉}{w}=w(\stackrel{炉}{x},\stackrel{炉}{y})+\frac{1}{2}\left(\frac{{鈭�}^{2}w}{鈭�x^2}\right){蟽}_x^2+\frac{1}{2}\left(\frac{{鈭�}^{2}w}{鈭�y^2}\right){蟽}_y^2\end{array}$

Hence, required expression is:.$\stackrel{炉}{w}=w(\stackrel{炉}{x},\stackrel{炉}{y})+\frac{1}{2}\left(\frac{{鈭�}^{2}w}{鈭�x^2}\right){蟽}_x^2+\frac{1}{2}\left(\frac{{鈭�}^{2}w}{鈭�y^2}\right){蟽}_y^2$