Optimal Bit Allocation

Bit Allocation Problem

Say we have a block of RVs X1,,XkX_{1},\dots,X_{k} and they’re scalar quantized. (Note that the XiX_{i} may have different distributions). The main goal we have here is that we want to minimize the overall MSE E[i=1k(XiQi(Xi))2]E\left[ \sum^{k}_{i=1}(X_{i}-Q_{i}(X_{i}))^{2} \right]where QiQ_{i} has NiN_{i} levels, under the condition that overall no more than BB bits are used: i=1klogNiB\sum_{i=1}^{k}\log N_{i}\le Bwhere Ni=2biN_{i}=2^{b_{i}} and logNi=bi\log N_{i}=b_{i}. Another way to interpret this is that for each RV in our random vector we assign it a NiN_{i}-level quantizer such that we meet our constraint and we wish to optimize it (i.e. minimize collective distortion, mini=1kD(Qi)\min \sum_{i=1}^{k}D(Q_{i})) under this constraint (i.e. keeping collective rate under a specified limit, BB).

We define WiW_{i} as the MSE of our constrained and optimal bib_{i}-bit quantizer: Wi(bi)=minQ:r(Q)biE[(XiQ(Xi))2]W_{i}(b_{i})=\min_{Q:r(Q)\le b_{i}}E[(X_{i}-Q(X_{i}))^{2}]We see if bib_{i} bits are used to quantize XiX_{i} optimally then the overall optimal distortion is D(b)=i=1kWi(bi)D(\mathbf{b})=\sum_{i=1}^{k}W_{i}(b_{i})where b=(b1,,bk)T\mathbf{b}=(b_{1},\dots,b_{k})^{T}.

Definition (Bit allocation problem)

Given the constraint i=1kbiB\sum_{i=1}^{k}b_{i}\le B find b=(b1,,bk)T\mathbf{b}=(b_{1},\dots,b_{k})^{T} minimizing D(b)D(\mathbf{b}).

Simplifications…

Before we state the theorem we must state some high-resolution approximations that are made.

  1. High-Resolution Conditions: Each XiX_{i} has pdf fif_{i} and Wi(b)=112fi1322b,  i=1,,kW_{i}(b)=\frac{1}{12}\lVert f_{i} \rVert_{\frac{1}{3}}2^{-2b}, \ \ i=1,\dots,k

  2. Let σi2=Var(Xi)\sigma_{i}^{2}=\text{Var}(X_{i}) and X~i=Xiσi\tilde{X}_{i}=\frac{X_{i}}{\sigma_{i}}. Then X~i\tilde{X}_{i} has unit variance (i.e. Var(X~i)=1\text{Var}(\tilde{X}_{i})=1) and its pdf f~i(x)=σifi(σix)\tilde{f}_{i}(x)=\sigma_{i}f_{i}(\sigma_{i}x) satisfies fi13=f~i13σi2\lVert f_{i} \rVert _{\frac{1}{3}}=\lVert \tilde{f}_{i} \rVert _{\frac{1}{3}}\sigma_{i}^{2}

  3. Hence Wi(b)=112f~i13hiσi222b=hiσi222b\begin{align*} W_{i}(b)&=\underbrace{ \frac{1}{12}\lVert \tilde{f}_{i} \rVert _{\frac{1}{3}} }_{ h_{i} }\sigma_{i}^{2}2^{-2b}\\ &=h_{i}\sigma_{i}^{2}2^{-2b} \end{align*}

    Note

    hih_{i} is invariant to scaling and hence, is the same for all XiX_{i} so we treat it as a constant.

Theorem (Optimal Bit Allocation)

Our optimal distortion for the bit allocation problem is defined as D(b)=i=1khiσi222biD(\mathbf{b})=\sum^{k}_{i=1}h_{i}\sigma_{i}^{2}2^{-2b_{i}}and is minimized subject to our bit constraint i=1kbiB\sum_{i=1}^{k}b_{i}\le B if and only if bi=bˉ+12log2σi2ρ2+12log2hiH,  i=1,,kb_{i}=\bar{b}+\frac{1}{2}\log_{2}\frac{\sigma_{i}^{2}}{\rho^{2}}+\frac{1}{2}\log_{2} \frac{h_{i}}{H}, \ \ i=1,\dots,kwhere bˉ=Bkρ2=(i=1kσi2)1kH=(i=1khi)1/k\bar{b}=\frac{B}{k}\quad\rho^{2}=\left( \prod_{i=1}^{k}\sigma_{i}^{2} \right)^{\frac{1}{k}\quad}H=\left( \prod_{i=1}^{k}h_{i} \right)^{1/k}i.e. ρ2\rho^{2} is the geometric mean of variance and HH is the geometric mean of our shape.

Remark

We see that the optimal b\mathbf{b} must satisfy i=1kbi=B\sum_{i=1}^{k}b_{i}=B

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