Does Convergence in Distribution Guarantee Probability Equality for Events?

[rukawakaede]

Hi,

Here is my question: Given that X_n\xrightarrow{\mathcal{D}}Z as n\rightarrow\infty where Z\sim N(0,1).
Can we conclude directly that \lim_{n\rightarrow\infty}P(|X_n|\leq u)=P(|Z|\leq u) where u\in (0,1)?
Is this completely trivial or requires some proof?

Also what is the differences between convergence in distribution and weak convergence?
I found both of them quite confusing as I was given a distinct definition for both concepts while some other books (including wikipedia) say they are the same.

Thanks!

 

[micromass]

Hi rukawakaede!

Hi,

Here is my question: Given that X_n\xrightarrow{\mathcal{D}}Z as n\rightarrow\infty where Z\sim N(0,1).
Can we conclude directly that \lim_{n\rightarrow\infty}P(|X_n|\leq u)=P(|Z|\leq u) where u\in (0,1)?
Is this completely trivial or requires some proof?

I don't find this to be completely trivial, as a direct proof is annoying. The easiest proof uses the continuous mapping theorem ( http://en.wikipedia.org/wiki/Continuous_mapping_theorem ). Applying this gets us

X_n\xrightarrow{\mathcal{D}}Z~~\Leftrightarrow~~|X_n|\xrightarrow{\mathcal{D}}|Z|

and by definition this gives us

\lim_{n\rightarrow\infty}P(|X_n|\leq u)=P(|Z|\leq u)

Also what is the differences between convergence in distribution and weak convergence?
I found both of them quite confusing as I was given a distinct definition for both concepts while some other books (including wikipedia) say they are the same.

May I ask you what your books mean with weak convergence (or which books you are using). I guess that the term "weak convergence" is not in standard use in probability and that therefore many conflicting definitions exist, but that's my guess...

 

[rukawakaede]

Hi rukawakaede!
I don't find this to be completely trivial, as a direct proof is annoying. The easiest proof uses the continuous mapping theorem ( http://en.wikipedia.org/wiki/Continuous_mapping_theorem ). Applying this gets us

X_n\xrightarrow{\mathcal{D}}Z~~\Leftrightarrow~~|X_n|\xrightarrow{\mathcal{D}}|Z|

and by definition this gives us

\lim_{n\rightarrow\infty}P(|X_n|\leq u)=P(|Z|\leq u)
May I ask you what your books mean with weak convergence (or which books you are using). I guess that the term "weak convergence" is not in standard use in probability and that therefore many conflicting definitions exist, but that's my guess...

I was given:

Let Q,Q_1,Q_2,\cdots:\mathcal{B}(\mathbf{R}) \rightarrow [0,1] be probability measures. Q_n converges weakly to Q whenever
\lim_{n\rightarrow\infty}I_{Q_n}(f)=I_Q(f),\quad n\rightarrow\infty
for all f\in\mathcal{C}_b(\mathbf{R}).

is weak convergence a measure theoretic equivalent to the idea of convergence in distribution in probability theory since convergence in distribution is the weakest among all four other types?

 

[micromass]

Ah, I see. I'm not quite sure what you mean with I_Q(f) actually.

But, anyways, weak convergence is a generalization of convergence in distribution to arbitrary measure spaces. Of course, measures don't have cdf's, so we can't apply the same definition. The definition I'm used to is that

\mu_n\Rightarrow \mu~\text{if and only if}~\mu_n(A)\rightarrow \mu(A)~\text{for all continuity sets}

The definition you gave seems to be equivalent to the above one.

The difference between convegence in distribution and weak convergence is
1) weak convergence generalizes to arbitrary measure spaces
2) weak convergence is a convergence between measures, while convergence in distribution is a convergence between random variables.

But there is a link between the two convergences: that is, a sequence X_n converges in distribution if and only if

P_{X_n}\rightarrow P_X

converges weakly. With

P_Y(A)=P\{Y\in A\}