The MMSE of an Equivalent Scalar Channel with a Mixtrue Gaussian Prior

Posted on | In | 阅读量 次

Mixtrue Gaussian

We consider mixtrue Gaussian distribution
\begin{align}
p(h)=\sum_{k=1}^K \rho_k \mathcal{N}_c(h|0,\sigma_k^2)
\end{align}
Followings are two conditions of mixture Gaussian distribution

  1. Normalization
    \begin{align}
    \int p(h)\text{d}h=1 \quad \Rightarrow \quad \sum_{k=1}^K\rho_k=1
    \end{align}
  2. Power
    \begin{align}
    \mathbb{E}[h^2]
    &=\int |h|^2 \sum_{k=1}^K \rho_k \mathcal{N}_c(h|0,\sigma_k^2)\text{d}h=1 \quad \Rightarrow \quad \sum_{k=1}^K \rho_k\sigma_k^2=1
    \end{align}

The mean and variance of distribution
\begin{align}
&\frac{p(h)\mathcal{N}_c(h|m,v)}{\int p(h)\mathcal{N}_c(h|m,v) \text{d}h}\\
=&\frac{\sum_{k=1}^K \rho_k\mathcal{N}_c(h|0,\sigma_k^2)\mathcal{N}_c(h|m,v)}{\int \sum_{k=1}^K \rho_k\mathcal{N}_c(h|0,\sigma_k^2)\mathcal{N}_c(h|m,v) \text{d}h}\\
=&\frac{\sum_{k=1}^K \rho_k\mathcal{N}_c\left(m|0,\sigma_k^2+v\right)\mathcal{N}_c\left(h|\frac{m\sigma_k^2}{\sigma_k^2+v},\frac{\sigma_k^2v}{\sigma_k^2+v}\right)}
{\sum_{k=1}^K \rho_k\mathcal{N}_c\left(m|0,\sigma_k^2+v\right)}
\end{align}
are given
\begin{align}
f_a(h|m,v)
&\overset{\triangle}{=}\mathbb{E}[h|m,v]\\
&=\frac{\sum_{k=1}^K\rho_k \frac{m\sigma_k^2}{\sigma_k^2+v}\mathcal{N}_c\left(m|0,\sigma_k^2+v\right)}{\sum_{k=1}^K \rho_k \mathcal{N}_c(m|0,\sigma_k^2+v)}\\
f_b(h|m,v)&\overset{\triangle}{=}\mathbb{E}[|h|^2|m,v]\\
&=\frac{\sum_{k=1}^K\rho_k \left[\frac{\sigma_k^2v}{\sigma_k^2+v}+\left(\frac{m\sigma_k^2}{\sigma_k^2+v}\right)^2\right]\mathcal{N}_c\left(m|0,\sigma_k^2+v\right)}{\sum_{k=1}^K \rho_k \mathcal{N}_c(m|0,\sigma_k^2+v)}\\
&=\frac{\sum_{k=1}^K\rho_k \left[\frac{\sigma_k^2v(\sigma_k^2+v)+|m|^2\sigma_k^4}{(\sigma_k^2+v)^2}\right]\mathcal{N}_c\left(m|0,\sigma_k^2+v\right)}{\sum_{k=1}^K \rho_k \mathcal{N}_c(m|0,\sigma_k^2+v)}\\
\end{align}
where the expectation is over $\frac{p(h)\mathcal{N}_c(h|m,v)}{\int p(h)\mathcal{N}_c(h|m,v) \text{d}h}$. Based on above, we get
\begin{align}
f_c(h|m,v)&\overset{\triangle}{=}\text{Var}[h|m,v]\\
&=f_b(h|m,v)-|f_a(h|m,v)|^2
\end{align}

Close-Form of MMSE

Given the Gaussian mixture distribution
\begin{align}
p(h)=\sum_{k=1}^K \rho_k \mathcal{N}_c\left(h|0,\sigma_k^2\right)
\end{align}
and equivalent scalar channel
\begin{align}
m=h+n \sim \mathcal{N}_c(n|0,v)
\end{align}
the distribution of $m$ is then expressed by
\begin{align}
p(m)=\sum_{k=1}^K \rho_k\mathcal{N}_c(m|0,v+\sigma_k^2)
\end{align}
$proof$:
$\underline{\text{Step 1} }$: Assume $X\sim \mathcal{N}(x|a,A)$ and $Y\sim \mathcal{N}(y|b,B)$. Define $Z=X+Y$, then its distribution is obtained by convolution formula
\begin{align}
p(z)
&=\int_{-\infty}^{+\infty} p_X(x)p_Y(z-x)\text{d}x\\
&=\int_{-\infty}^{+\infty}\mathcal{N}(x|a,A)\mathcal{N}(z-x|b,B)\text{d}x\\
&=\int_{-\infty}^{+\infty}\mathcal{N}(x|a,A)\mathcal{N}(x|b-z,B)\text{d}x\\
&\overset{(a)}{=}\mathcal{N}(0|a-(b-z),A+B)\\
&=\mathcal{N}(z|b-a,A+B)
\end{align}
where $(a)$ holds by Gaussian product lemma.

$\underline{\text{Step 2} }$: Assume $X\sim \sum_{k=1}^K \mathcal{N}(x|a_k,A_k)$ and $Y\sim \mathcal{N}(y|b,B)$. Define $Z=X+Y$, using the convolution formula we can easy get
\begin{align}
p(z)
&=\int_{-\infty}^{+\infty}p_X(x)p_Y(z-x)\text{d}x\\
&=\int_{-\infty}^{+\infty}\sum_{k=1}^K\rho_k\mathcal{N}(x|a_k,A_k)\mathcal{N}(z-x|b,B)\text{d}x\\
&=\int_{-\infty}^{+\infty}\sum_{k=1}^K \rho_k \mathcal{N}(x|a_k,A_k)\mathcal{N}(x|b-z,B)\text{d}x\\
&=\sum_{k=1}^K \rho_k \mathcal{N}(0|a_k-(b-z),A_k+B)\\
&=\sum_{k=1}^K\rho_k\mathcal{N}(z|b-a_k,A_k+B)
\end{align}
$\underline{\text{Step 3} }$: If $a_k$ and $b$ are equal to zero, we then get
\begin{align}
p(z)=\sum_{k=1}^K \rho_k \mathcal{N}(z|0,A_k+B)
\end{align}

The MMSE of this AWGN model is given by
\begin{align}
\text{MMSE}
&=\mathbb{E}[\text{Var}[h|m,v]]\\
&=\int f_c(h|m,v)\sum_{k=1}^K\rho_k\mathcal{N}_c(m|0,\sigma_k^2+v)\text{d}m\\
&=\int \left[\frac{\sum_{k=1}^K\rho_k \left[\frac{\sigma_k^2v(\sigma_k^2+v)+|m|^2\sigma_k^4}{(\sigma_k^2+v)^2}\right]\mathcal{N}_c\left(m|0,\sigma_k^2+v\right)}{\sum_{k=1}^K \rho_k \mathcal{N}_c(m|0,\sigma_k^2+v)}-\left|\frac{\sum_{k=1}^K\rho_k \frac{m\sigma_k^2}{\sigma_k^2+v}\mathcal{N}_c\left(m|0,\sigma_k^2+v\right)}{\sum_{k=1}^K \rho_k \mathcal{N}_c(m|0,\sigma_k^2+v)}\right|^2\right]\\
&\quad \times \sum_{j=1}^K\rho_j\mathcal{N}_c(m|0,\sigma_j^2+v)\text{d}m\\
&=\sum_{k=1}^K\rho_k\frac{\sigma_k^2v+\sigma_k^4}{\sigma_k^2+v}-\int \left|\frac{\sum_{k=1}^K\rho_k \frac{m\sigma_k^2}{\sigma_k^2+v}\mathcal{N}_c\left(m|0,\sigma_k^2+v\right)}{\sum_{k=1}^K \rho_k \mathcal{N}_c(m|0,\sigma_k^2+v)}\right|^2 \sum_{j=1}^K \rho_j\mathcal{N}_c(m|0,\sigma_j^2+v)\text{d}m
\end{align}
where the inner expectation is over $\frac{p(h)\mathcal{N}_c(h|m,v)}{\int p(h)\mathcal{N}(h|m,v)\text{d}h}$ while the outer expectation is taken over $p(m)=\sum_{k=1}^K \rho_k\mathcal{N}(m|0,\sigma_k^2+v)$.