Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $(E,\mathscr E)$ be a measure space and $P:E \times\mathscr E\to [0,1]$ be a stochastic kernel - i.e. $$ P(x,A)\in [0,1] $$ for any $x\in E$ and $A\in \mathscr E$. On a set $b\mathscr E$ of bounded measurable functions with a norm $$\|f\| = \sup\limits_{x\in E}|f(x)|$$ define the action of the kernel as a linear operator $$ Pf(x) = \int\limits_E f(y)P(x,dy). $$ Let $\tilde P$ be another probability kernel and consider two norms $$ \|\tilde P - P\|' = \sup\limits_{A\in \mathscr E}\sup\limits_{x\in E}|\tilde P(x,A) - P(x,A)| $$ $$ \|\tilde P - P\|'' = \sup\limits_{f\in b\mathscr E\setminus\{0\}}\frac{\|(\tilde P - P)f\|}{\|f\|}. $$

It is easy to show that $\|\tilde P - P\|'\leq \|\tilde P - P\|''$ since an indicator function $1_A\in b\mathscr E$ for all measurable sets $A$. I wonder if the reverse inequality is true as well.

My idea was to consider a simple function $f(x) = \sum\limits_{i=1}^n f_i1_{E_i}(x)$ where $E_i$ is a partition of $E$. But if I am not missing anything, even for simple function the reverse inequality is not true.

share|cite|improve this question
up vote 1 down vote accepted

Indeed, the reverse inequality is not true. Take $x_0\in X$, and put $P_1(x,A):=\delta_x(A)$, and $P_2(x,A):=\frac 12(\delta_x(A)+\delta_{x_0}(A))$. We have \begin{align*} \lVert P_1-P_2\rVert&=\sup_{A\in\mathcal E}\sup_{x\in E}|P_1(x,A)-P_2(x,A)|\\ &=\sup_{A\in\mathcal E}\sup_{x\in E}|\delta_x(A)-\frac 12\delta_x(A)-\frac 12\delta_{x_0}(A)|\\ &=\frac 12\sup_{A\in\mathcal E, A\neq E}\sup_{x\in E}|\delta_x(A)-\delta_{x_0}(A)|\\ &=\frac 12 \end{align*} (indeed, since $|\delta_x(A)-\delta_{x_0}(A)|\leq 1$ this supremum is $\leq 1$, and it is reached for a set $A$ which contains $x_0$). Now if we assume $\{x_0\}$ measurable, consider the map $f$ defined by $f(x)=\begin{cases} 1&\mbox{ if }x\neq x_0\\\ -1&\mbox{ if }x=x_0. \end{cases}$ $f$ is measurable, bounded, and its norm is $1$. We have $$(P_1-P_2)(f)(x)=f(x)-\frac 12f(x)-\frac 12f(x_0)=1/2f(x)-1/2f(x_0)=1$$ so $\lVert P_1-P_2\rVert''\geq\frac{\lVert (P_1-P_2)(f)\rVert}{\lVert f\rVert}=1>\lVert P_1-P_2\rVert'$.

share|cite|improve this answer
thank you - I was confused about the definition of the total variation norm for a finite signed measure and the total variation distance between two probability measure which does not take factor $2$ into account – Ilya Mar 30 '12 at 11:16

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.