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I have recently encountered variants of the following expression: \begin{equation} S = H(a,b,c,d)-H(a+b,c+d) \end{equation} where $H$ is the Shannon entropy function, that is $H(X)=\sum_{x\in X}-x\log x$. And restrictions: \begin{eqnarray} a + b &=& t\\ a + c &=& t\\ a + b +c + d &=& 1 \end{eqnarray} for some given $t$. It is supposed to be easy to maximize this expression with Lagrange multipliers but I am unfamiliar with them so any hint in how $S$ can be optimized would be welcome.

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up vote 3 down vote accepted

Presumably $a,b,c,d > 0$ are required. In order for this to be possible, $0 < t < 1$. You should actually check the limits as one or more of $a,b,c,d$ go to $0$, but I will not do so. Putting in a Lagrange multiplier $\lambda_i$ for each constraint, the Lagrangian is $$L = -a\ln \left( a \right) -b\ln \left( b \right) -c\ln \left( c \right) -d\ln \left( d \right) + \left( a+b \right) \ln \left( a+b \right) + \left( c+d \right) \ln \left( c+d \right) +\lambda_{{1}} \left( a+b-t \right) +\lambda_{{2}} \left( a+c-t \right) +\lambda_{{3 }} \left( a+b+c+d-1 \right)$$ Now solve the system of equations obtained by setting to $0$ the derivative of $L$ with respect to each of the variables $a,b,c,d,\lambda_1, \lambda_2,\lambda_3$: $$ \eqalign{ a+b-t &= 0\cr a+c-t &= 0\cr a+b+c+d-1 &= 0\cr -\ln(d)+\ln(c+d)+\lambda_3 &=0\cr -\ln(b)+\ln(a+b)+\lambda_1+\lambda_3 &= 0\cr -\ln(c)+\ln(c+d)+\lambda_2+\lambda_3 &=0\cr -\ln(a)+\ln(a+b)+\lambda_1+\lambda_2+\lambda_3 &= 0\cr}$$ obtaining $$ a={t}^{2},b=c=t-{t}^{2}, d=(t-1)^2,\lambda_1 = 0,\lambda_{{2 }}=\ln(t/(1-t)) , \lambda_{{3}}=\ln (1-t) $$

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Another approach is to avoid Lagrange multipliers. Solve the system of constraints for $b$, $c$, and $d$ to find $$\begin{eqnarray*} b &=& t-a \\ c &=& t-a \\ d &=& a-2t+1. \end{eqnarray*}$$ We have the restrictions $0\le a,b,c,d \le 1$, so $\textrm{max}(0,2t-1)\le a\le t$, where $0\le t\le 1$.

Thus, $$\begin{equation*} S(a) = H(a,t-a,t-a,a-2t+1)-H(t,1-t), \end{equation*}$$ so the extremum is the solution to $$\frac{\partial}{\partial a} H(a,t-a,t-a,a-2t+1) = 0,$$ or $a = t^2$, reproducing the solution found by @RobertIsrael. The maximum is $S_{\mathrm{max}} = S(t^2) = (t-1)\log(1-t) - t\log t$.

Note that $b=0 \iff c=0 \iff a=t$, but $S(t) = H(t,0,0,1-t)-H(t,1-t) = 0$. Also $d=0 \iff a=2t-1.$

entropy plot for a=0 and d=0

Figure 1. Plot of $S_{\mathrm{max}}$ (black), and $S$ for $a=0$ (red) and $d=0$ (blue).

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Ok, so a purely analytic approach works as well. I will sure find use of this alternative, much lighter, approach. Thanks! – Euclean Jun 26 '12 at 12:56
@Euclean: Glad to help. Lagrange multipliers start to show their power when the constraints are more complicated than a linear system. – user26872 Jun 26 '12 at 16:20

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