In the MAT7381 course (graduate course on regression models), we will talk about optimization, and a classical tool is the so-called conjugate. Given a function f:\mathbb{R}^p\to\mathbb{R} its conjugate is function f^{\star}:\mathbb{R}^p\to\mathbb{R} such that f^{\star}(\boldsymbol{y})=\max_{\boldsymbol{x}}\lbrace\boldsymbol{x}^\top\boldsymbol{y}-f(\boldsymbol{x})\rbraceso, long story short, f^{\star}(\boldsymbol{y}) is the maximum gap between the linear function \boldsymbol{x}^\top\boldsymbol{y} and f(\boldsymbol{x}).
Just to visualize, consider a simple parabolic function (in dimension 1) f(x)=x^2/2, then f^{\star}(\color{blue}{2}) is the maximum gap between the line x\mapsto\color{blue}{2}x and function f(x).
x = seq(-100,100,length=6001)
f = function(x) x^2/2
vf = Vectorize(f)(x)
fstar = function(y) max(y*x-vf)
vfstar = Vectorize(fstar)(x)
We can see it on the figure below.
viz = function(x0=1,YL=NA){
idx=which(abs(x)<=3) par(mfrow=c(1,2)) plot(x[idx],vf[idx],type="l",xlab="",ylab="",col="blue",lwd=2) abline(h=0,col="grey") abline(v=0,col="grey") idx2=which(x0*x>=vf)
polygon(c(x[idx2],rev(x[idx2])),c(vf[idx2],rev(x0*x[idx2])),col=rgb(0,1,0,.3),border=NA)
abline(a=0,b=x0,col="red")
i=which.max(x0*x-vf)
segments(x[i],x0*x[i],x[i],f(x[i]),lwd=3,col="red")
if(is.na(YL)) YL=range(vfstar[idx])
plot(x[idx],vfstar[idx],type="l",xlab="",ylab="",col="red",lwd=1,ylim=YL)
abline(h=0,col="grey")
abline(v=0,col="grey")
segments(x0,0,x0,fstar(x0),lwd=3,col="red")
points(x0,fstar(x0),pch=19,col="red")
}
viz(1)
or
viz(1.5)
In that case, we can actually compute f^{\star}, since f^{\star}(y)=\max_{x}\lbrace xy-f(x)\rbrace=\max_{x}\lbrace xy-x^2/2\rbraceThe first order condition is here x^{\star}=y and thusf^{\star}(y)=\max_{x}\lbrace xy-x^2/2\rbrace=\lbrace x^{\star}y-(x^{\star})^2/2\rbrace=\lbrace y^2-y^2/2\rbrace=y^2/2And actually, that can be related to two results. The first one is to observe that f(\boldsymbol{x})=\|\boldsymbol{x}\|_2^2/2 and in that case f^{\star}(\boldsymbol{y})=\|\boldsymbol{y}\|_2^2/2 from the following general result : if f(\boldsymbol{x})=\|\boldsymbol{x}\|_p^p/p with p>1, where \|\cdot\|_p denotes the standard \ell_p norm, then f^{\star}(\boldsymbol{y})=\|\boldsymbol{y}\|_q^q/q where\frac{1}{p}+\frac{1}{q}=1The second one is the conjugate of a quadratic function. More specifically if f(\boldsymbol{x})=\boldsymbol{x}^{\top}\boldsymbol{Q}\boldsymbol{x}/2 for some definite positive matrix \boldsymbol{Q}, f^{\star}(\boldsymbol{y})=\boldsymbol{y}^{\top}\boldsymbol{Q}^{-1}\boldsymbol{y}/2. In our case, it was a univariate problem with \boldsymbol{Q}=1.
For the conjugate of the \ell_p norm, we can use the following code to visualize it
p = 3
f = function(x) abs(x)^p/p
vf = Vectorize(f)(x)
fstar = function(y) max(y*x-vf)
vfstar = Vectorize(fstar)(x)
viz(1.5)
or
p = 1.1
f = function(x) abs(x)^p/p
vf = Vectorize(f)(x)
fstar = function(y) max(y*x-vf)
vfstar = Vectorize(fstar)(x)
viz(1, YL=c(0,10))
Actually, in that case, we almost visualize that if f(x)=|x| then\displaystyle{f^{\star}\left(y\right)={\begin{cases}0,&\left|y\right|\leq 1\\\infty ,&\left|y\right|>1.\end{cases}}}
To conclude, another popular case, f(x)=\exp(x) then{\displaystyle f^{\star}\left(y\right)={\begin{cases}y\log(y)-y,&y>0\\0,&y=0\\\infty ,&y<0.\end{cases}}}We can visualize that case below
f = function(x) exp(x)
vf = Vectorize(f)(x)
fstar = function(y) max(y*x-vf)
vfstar = Vectorize(fstar)(x)
viz(1,YL=c(-3,3))
The first idea (used in the previous post, 
is somehow the new parameter of the distribution. Then, a conjugate prior
for the parameter 
(where 
is the dimension of 
are i.i. with distribution 
, assume that the a priori distribution of 
is 
, , then the a posteriori distribution of 
is still Beta, with parameters
, assume that the a prioridistribution of 
, assume that the a priori distribution of 
, assume that the a priori distribution of 
, then the a posteriori distribution of 
, assume that the a prioridistribution of 
assume that the a prioridistribution of 
, then the a posteriori distribution of 
assume that 
, then
, assume that the a prioridistribution of 
, then the a posteriori distribution of 
, but 
, where 
(for some bijective transformation). The distribution (density) of 
denote Fisher information of parameter 
), it seems natural to consider
The idea is quite simple. We need a prior distribution so that
i.i.d. with distribution 
. Here we note
is a random variable. The idea is to assume that 
. Thus, we need to compute the distribution of 
which is here extremely simple (due to properties of the Gaussian distribution), i.e.
as an estimator of given our sample data (and thus, we also have a confidence interval since we know the distribution of 
. Note, first, that \theta has support 
. So we need a distribution on that support. Why not a beta distribution ? E.g.
