E(X_n)->E(X) implies E(YX_n)->E(YX)

fmfreshman · February 22nd, 2013, 5:21 pm

Hi, I am considering a convergent problem. Suppose X_n is a sequence of random variable, and we know that when n goes into infinitylimE(X_n)->E(X) and limX_n=X almost surely, does it implies that limE(YX_n)->E(YX) ? where Y is a random variable. Thanks very much.

quantiquequant · February 22nd, 2013, 5:28 pm

Hi,If you have L2 convergence it is direct application of Cauchy-Swartz inequality. In the general case I am thinking about an example.

fmfreshman · February 24th, 2013, 10:58 am

Thanks for your answer! It seems to be correct in many continous cases, but I still suspect it is not right in general

EscapeArtist999 · February 27th, 2013, 11:25 am

I would say you could start with the condition E[|XY|] is finite, and E[|YX_n|] is finite for all n. then create a series X_k_n = X_n * Y_{|Y| <= K}, you could then use dominate convergence and the assumptions on most of the swapping limits and expectations the only issue would be about swapping Lim_k and Lim_n - think about that - if you do that, you probably have the most general version - that said, I have not done this in a VERY long time, and there are smarter ppl on these forums, so happy to be corrected or improved on.

quantiquequant · February 28th, 2013, 11:20 am

QuoteOriginally posted by: fmfreshmanThanks for your answer! It seems to be correct in many continous cases, but I still suspect it is not right in generalWell if you are interested in an application to finance second moment exist for sure ! or how the heck you can talk about volatility (Taleb has nice speechs on this)

fmfreshman · March 4th, 2013, 10:24 pm

Thanks, very interesting!