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On this page... (hide) Estimation of the noise covariance map of a smoothed datasets. 1. 1-D Polar covariance maps smoothingWhen smoothing a 1-D n-pixel map m, we convolve it with a Gaussian kernel K_\sigma, with $\sigma$ being the bandwidth. The resulting smoothed map at pixel p becomes \tilde m_p = K_\sigma * m ~~~~~ \text{with} ~~~~~K_\sigma(p) \equiv \frac 1 {\sigma^2} \exp\left(\frac {p^2}{2\sigma^2}\right) with * the convolution product. Since we work in pixel space, the convolution becomes \tilde m_p = \frac 1{ n_{}}\sum_{p'}^{n_{}} K_\sigma(p-p') m_{p'} From it, the covariance of the smoothed maps \tilde x and \tilde y then becomes Cov[\tilde x_p, \tilde y_p] = Cov[K_\sigma * x, K_\sigma * y] Cov[\tilde x_p, \tilde y_p] = K^2_\sigma * Cov[ x, y] Cov[\tilde x_p, \tilde y_p] = \frac12\frac1{\sigma^2 n_{}} K_{\sigma/\sqrt2} * Cov[ x, y]. We conclude that the resulting covariance of two smoothed maps \tilde x and \tilde y by a Gaussian kernel K_\sigma is equivalent to the smoothing of the initial covariance map between x and y by a Gaussian kernel \displaystyle \frac{K_{\sigma/\sqrt 2}}{2\sigma^2 n_{}}. As a more detailed calculation, let \tilde m(p) = \frac1{n} \sum_i^n K_\sigma(p-p') m(p') the value of the $n$-pixel smoothed map at pixel p, with \displaystyle K_\sigma(p) \equiv \frac 1 {\sigma^2} \exp\left(-\frac12 \frac{p^2}{\sigma^2}\right) the Gaussian smoothing kernel. Thus, Var[\tilde m (p)] = E [\tilde m(p)^2] - E [\tilde m(p)]^2 Var[\tilde m (p)] = E \left[\frac1 { n^2} \sum_i^n K_\sigma(p-p_i)m(p_i)\sum_j^n K_\sigma(p-p_j)m(p_j) \right] - \frac1 {n^2} \left( \sum_i^n K(p-p_i) \underbrace{E [m(p_i)]}_{=0} \right)^2 Var[\tilde m (p)] = \frac1 {n^2} \left(\sum_i^n K_\sigma^2(p-p_i) E[m(p_i)^2] + \sum_{i\neq j}^n K_\sigma(p-p_i)K_\sigma(p-p_j)\underbrace{ E[m(p_i)m(p_j)]}_{=0}\right) Var[\tilde m (p)] =\frac1 {n^2} \sum_i^n K_\sigma^2(p-p_i) Var[m(p_i)] Var[\tilde m (p)] =\frac 12 \frac1 {\sigma^2 n} \left(\sum_i^n K_{\sigma/\sqrt2}(p-p_i) Var[m(p_i)]\right). Since K_{\sigma}(p)^2 = \frac 1 {\sigma^4} \exp\left(-\frac{p^2}{\sigma^2}\right) = \frac 12\frac 1 {\sigma^2} K_{\sigma/\sqrt2}(p)
K_{\sigma}(p)^2 = \frac 1 {\sigma^4} \exp\left(-\frac{p^2}{\sigma^2}\right) = \frac 1 {\sigma^2} K_{\sigma/\sqrt2}(p) 2. 2-D Polar covariance maps smoothing
3. Validation simulations
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