From the above two assumptions, the accuracy of the extracted watermark depends mainly on the variation of the image pixels. For example, the two neighbor pixels having highly different values have a high chance of obtaining an error prediction for $w\u2032(i,j)$. In fact, the variation of the $Y$ component after being watermarked using the above scheme always increases, and hence, unavoidably causes a lower accuracy on watermark extraction. To enhance the performance of $w\u2032(i,j)$ estimation of the $Y$ component, we consider using a new prediction technique for $Y(i,j)$, taking into account the different values between two nearby components, i.e., the center and its neighbor. That is, instead of using the true value of the neighbor component around ($i,j$) in the prediction process, we first apply a weighting factor to every neighbor component around ($i,j$), so that all neighbor components get closer to the center pixel. Conceptually, the weighting factor is determined based on the different values between the predicting component and its neighbors. Since, as mentioned earlier, a component value at coordinates ($i,j$) is assumed to be predicted from its neighbors, and the neighbor component value should be close to the predicting one. Also, since the range of values for the $Y$ component varies from 16 to 235 and the different values between the two components can be varied from 0 to 219, the weighting factor is applied directly to the nearby component in accordance with the difference between that component and the predicting one. Based on this concept, the weighted neighbor component $Y\xaf\u2032(i,j)$ around $Y\u2032(i,j)$ of an area $3\xd73$ pixels can be represented by the following equation: Display Formula
$Y\xaf\u2032(i+m,j+n)=Y\u2032(i+m,j+n)+\alpha [Y\u2032(i,j)\u2212Y\u2032(i+m,j+n)],$(21)
where $\alpha $ is a constant value used to adjust the weighted component, $m$ and $n=\u22121$, 0, 1. Finally, a new prediction of $Y(i,j)$, which we denote as $Y\xaf\u2032\u2032(i,j)$, is given by: Display Formula$Y\xaf\u2032\u2032(i,j)=18{[\u2211m=\u221211\u2211n=\u221211Y\xaf\u2032(i+m,j+n)]\u2212Y\xaf\u2032(i,j)}.$(22)