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I derived ML estimator for Rician distributed data and I am trying to show Rao-Cramer lower bound of $\hat{A}$ estimator variance.

$$f(x_k|A,\sigma) = \frac{x_k}{\sigma^2}\exp\left(-\frac{x^2_k+A^2}{2\sigma^2}\right)I_0\left(\frac{xA}{\sigma^2}\right) \tag{Rician distribution}$$

where $I_n(x) = \left( {1 \over 2}x \right )^{n} \sum_{k=0}^{\infty} \frac{\left( {1 \over 4}x^2 \right )^k}{k! (n+k)!}$ is modified Bessel function of the first kind of $n$-th order ($n \in N$)

The ML estimator, that I derived:

$$\hat{A} = \frac{1}{N} \sum_{k=1}^{N} x_k  \frac{  I_1 \left( \frac{x_k A}{\sigma^2} \right )}{   I_0 \left( \frac{x_k A}{\sigma^2} \right )}$$

Unfortunately, variance derivation looks like challenging problem. Anyone could suggest a trial?

$$\operatorname{Var}_\hat{A} = \frac{1}{N^2}E_{\hat{A}} \left( \sum_{k,l=1}^N\left( x_k-E_{\hat{A}}\left(x_k\frac{I_1}{I_0}\right) \right) \left( x_l-E_{\hat{A}}\left(x_l\frac{I_1}{I_0}\right) \right) \right)$$

where $I_0, I_1$ is of course $I_0 \left( \frac{x_k A }{\sigma^2} \right )$, $I_1 \left( \frac{x_k A }{\sigma^2} \right )$, respectively.

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    The formula of the ML estimator is actually an implicit equation, isn't it? ($A$ in the RHS should be rather $\hat{A}$)2012-11-17
  • 0
    @leonbloy - Yes, It's no closed form of $\hat{A}$2012-11-17

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