I have a question on the difference between uniform and pointwise convergence in probability. We know that pointwise convergence in probability does not imply uniform convergence in probability. In more details, let $X_1,...,X_n$ be a sample of real valued i.i.d. random variables. Let $$ f_n(X_1,...,X_n;\beta): \mathbb{R}^n \times B\rightarrow \mathbb{R} $$ be a function of $X_1,...,X_n$ and of a parameter $\beta\in B$. Let $$ f(\beta): B\rightarrow \mathbb{R} $$ be a function of a parameter $\beta\in B$.

Pointwise convergence states that $$ f_n(X_1,...,X_n;\beta)\rightarrow_P f(\beta) \text{ }\forall \beta\in B $$ i.e. $$ \forall \beta\in B, \forall \epsilon>0,\forall \delta>0 \text{ } \exists \bar{n}_{\delta, \epsilon,\beta}\in \mathbb{N} \text{ s.t. } \forall n> \bar{n}_{\delta, \epsilon,\beta}\text{ } P(|f_n(X_1,...,X_n;\beta)-f(\beta)|>\epsilon)<\delta $$ Uniform convergence states that $$ \sup_{\beta\in B}|f_n(X_1,...,X_n;\beta)-f(\beta)|\rightarrow_P 0 $$ i.e. $$ \forall \epsilon>0,\forall \delta>0 \text{ } \exists \bar{n}_{\delta, \epsilon}\in \mathbb{N} \text{ s.t. } \forall n> \bar{n}_{\delta, \epsilon}, \forall \beta\in B\text{ } P(|f_n(X_1,...,X_n;\beta)-f(\beta)|>\epsilon)<\delta $$


What is wrong in the following argument: suppose pointwise convergence holds. Take $\sup_{\beta\in B} \bar{n}_{\delta, \epsilon,\beta}$ and show that uniform convergence is implied by setting $ \bar{n}_{\delta, \epsilon}:=\sup_{\beta\in B} \bar{n}_{\delta, \epsilon,\beta}$.

  • 1
    $\begingroup$ If you have a pointwise but no uniform convergent sequence you will get $\bar{n}_{\delta, \epsilon}:=\sup_{\beta\in B} \bar{n}_{\delta, \epsilon,\beta} = \infty$ so no $\bar{n}$ independent of $\beta$ can be found $\endgroup$ – Gono Dec 5 '16 at 11:42
  • $\begingroup$ Thanks. Why? Can't there be all finite? $\endgroup$ – STF Dec 5 '16 at 12:31
  • $\begingroup$ Ok, I see. I may have a sequence of natural numbers that is not bounded. Thank you $\endgroup$ – STF Dec 5 '16 at 13:00
  • 1
    $\begingroup$ OFC they CAN be finite, but then the convergence is uniform… but if you have an unbounded sequence it's not… $\endgroup$ – Gono Dec 5 '16 at 13:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.