Let be some random variables converging to in probability, and which are uniformly integrable.

Inituitively, the convergence in probability says that they converge nicely `most of the time’ and the uniform integrability says that where they don’t converge nicely, they don’t converge too nastily. This is the intuitive reason why we can deduce that these two conditions suffice for convergence in , and I attach what I hope is an intuitive proof (though it differs wildly from that in the official notes).

Let .

Firstly, fix such that for every , whenever (using uniform integrability).

Secondly, fix such that whenever , (possible by convergence in probability.

We deduce that for all ,

So we are done, for exactly the reasons we expected. In both official sets of course notes I’ve been using, the proof takes an entirely different and what feels to me much more obscure route of picking an a.s. convergent subsequence, using Fatou’s lemma and another lemma about uniform integrability which seem to be just missing the point, but if there is a flaw in my above proof I’d be glad to know about it.

EDIT: Thanks to Spencer for pointing out that the flaw was assuming that X is uniformly integrable. Fortunately, this can be proved relatively cleanly, and essentially the proof I want is documented well in Martin Orr’s notes, also posted below. Thanks for the feedback guys.

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May 10, 2011 at 9:29 am

ThomasHey,

I can’t spot any mistakes, but maybe it fails for general measurable spaces. You’d definitely have to modify the last line, maybe using the second condition for uniform integrability in the general case (all functions are concentrated up to some given bound on a set of finite measure).

Cheers,

Thomas

May 11, 2011 at 3:14 pm

Martin OrrThis is the same proof as in my notes on Probability and Measure (http://www.martinorr.name/2008/probability/) which suggests that it was probably the proof I heard in lectures.

May 11, 2011 at 4:53 pm

tloveringBrilliant. Title edited accordingly. What wonderful notes by the way! Do you have any more? I wonder if every year several people recompile new vaguely comprehensible sets of P&M notes out of a feeling they should understand Lebesgue integration properly… I have my own PDF near completion at the moment based mainly a lot of work over Christmas.

May 11, 2011 at 4:58 pm

tloveringOh and Thomas, thanks. That’s a decent observation… yeah, I guess maybe the proof in the original notes was to allow this more general case to work…. I can’t see an instant way to fix it by chopping ends off without proving the lemma about uniformity of decay to infinity.

May 12, 2011 at 9:29 pm

Martin OrrNo, Probability and Measure was the only course that motivated me to type up notes. It seems to require more digestion than most pure courses in Part II.

May 13, 2011 at 5:59 pm

tloveringYeah. Excepting perhaps algebraic geometry, which is giving me indigestion as I type.

May 14, 2011 at 12:26 am

SpencerI’m not too convinced by your “intuitive reason”. And, more importantly, you do make it look neater than is perhaps fair by using the fact that X is UI. If all you are given is X_n and told that they are UI and converge in P then this needs justifying, no?

May 15, 2011 at 9:59 pm

tloveringHey! What’s wrong with my intuitive reason?…. To be fair there could easily be something wrong with it… I’ve self-taught quite a lot of this course.

I have missed out the proof that a single variable is UI, which requires some nontrivial Borel-Cantellidge, it’s true. This feels like a more fundamental fact than the one I’m trying to prove though (like, if a single variable isn’t, taken alone, ‘uniformly’ integrable, what on earth are we doing?).

May 16, 2011 at 9:23 pm

SpencerWell it’s very obvious that a single random variable *in L^1* should be UI but that isn’t part of your assumption, is it? You just get given the sequence and told they converge in P…

May 18, 2011 at 4:28 pm

tloveringAh, I see…. yes, in general is it even true….? That’s a good point. Maybe I shouldn’t have deleted the `possibly false’ from the title of this post so quickly. So I need to add in the condition that X is integrable, or see if the proof in the notes works in the X not integrable case…. thanks.

EDIT: Oh, ok, so in Martin’s notes this is dealt with. All is well with the world.

May 21, 2011 at 10:08 pm

Martin OrrI think there is an error in this step (proving that X is L^1) in my notes: the sequence to which I apply MCT is not increasing. By the time you fix this you have basically reproduced the proof of Fatou’s lemma, applied to an a.s. convergent subsequence, so maybe this comes back to the proof you disliked.