You are given a sequence of random variables defined on some probability space, say with measure P. Find an equivalent (probability) measure Q such that under Q, all of the random variables are integrable.

Is the sequence finite or infinite?If it's infinite, I don't know if it can be done. Here is an attempt at a counterexample that seems to be extremely difficult to regulate in the manner you propose. Take your underlying sample space be the real numbers with measure P being whatever reasonably well behaved function you want (say, a standard normal distribution). Let the random variables also live in the reals. Since the rationals are countable there exists a bijection f: N -> Q. Now make the following definitions:The main point, of course, is that under the original probability measure P there is some member of the sequence X_i which is nonintegrable around each rational number in the underlying sample space, and the rationals are dense in the reals.I'm not quite comfortable enough with measure theory to prove that this is a counterexample, but I don't quite see how you would be able to define an equivalent measure which would make every X_i integrable.

If the sequence is finite, the problem is easier to solve, but it is still true in the infinite case.

Hmmm. Can you give me a hint as to what measure would make my "counterexample" integrable for all X_i? If I have that it might be obvious how to solve in general.

Each of the random variables has a trouble spot where it blows up. But outside of an arbitrarily small interval around that one bad point, the random variables are already integrable. By choosing the "right" intervals around the bad points you can make sure that you don't land in them too much.

For finitely many here is a simple solution. First let's find this measure for one random variable X, passing to positive/negative part of X, we can assume WLOG that X is nonnegative, and since probability space is of finite measure, we can further assume X>=1 (this is just for convenience...). The entire probability space, Omega, can be written as a union of mutually disjoint open sets, A_i, preimages of X of intervals [i,i+1), where i runs through all positive integers (and a set of zero measure, B, where X=+infinity). The basic idea is to modify the measure on each A_i, namely divide it by i, so "symbolically" Q(A_i)=(1/i)P(A_i). This is not quite yet a definition of Q on all measurable subsets of Omega but can easily be made such since A_i, i>0 define a countable partition of Omega (up to the set B mentioned above...). So for any measurable A\subset Omega, we can intersect A with all the A_i and define the measure as the sum of the intersected parts each of which looks by definition as Q(A\intersection A_i)=(1/i)P(A\intersection A_i). I will leave it to you that this measure is equivalent to P and also 0<Q(Omega)<infinity. So by further dividing this measure by Q(Omega) we get the right probability measure (still equivalent to the original one). It's easy to verify that X is integrable under such measure (just split \Integral_{Omega}XdQ into a sum of integrals over A_i and use that X<i+1 there and the fact that the space itself is of finite measure...). Once we have this construction done for one variable, we can continue the same way (starting now from the "new" measure constructed for X_1) until we make all (finitely many) random variables integrable. Note that in this inductive step we use the consequence of the above construction which implies that , symbolically, "Q<alpha*P" for some constant alpha>0; this makes sure the newly constructed measures leave all the previous random variables integrability intact. For infinitely many random variables one has to refine the above (define the partition of Omega a little differently or something along those lines...let's see if someone wants to chip in here).

QuoteOriginally posted by: RDKEach of the random variables has a trouble spot where it blows up. But outside of an arbitrarily small interval around that one bad point, the random variables are already integrable. By choosing the "right" intervals around the bad points you can make sure that you don't land in them too much.The bad points are dense in the reals, though.This is my problem with things.For finitely many X_i I think I can do it pretty easily. But when you're allowed (by infinity of series) to come up with random variables which blow up at points which are dense, I don't see how it's possible.This is not to say it's not possible; I'm just a physicist with more real analysis skills than the average bear. I don't claim to be an expert in measure or probability theory.

For the inductive step jurowilmott11 mentions, you can also use the fact that the supremum of a finite collection of real valued random variables is again a real valued random variable. But you can't use this reasoning for infinitely many.

Yes, that is my problem.You can't take X(x) = max (X_i(x)) for infinitely many X_i

Solution for the infinite case: Note that my first solution for the finite case defines dQ_i/dP as a fininte "step" function, equal to 1/i on A_i, let's call this density f_i. By almost sure finiteness of X_i, there exists a constant a_i such that P(X_i>a_i)<1/2^i for all i>0. Denote B_i={X_i>a_i}. Now "Sum_{i=1}^{i=infinity} P(B_i) < infinity", therefor by Borel-Cantelli lemma the set of those points in Omega which belong to infinitely many B_i is of zero measure. Therefore X_i<=a_i almost surely for all sufficiently large i>0 and since we already solved it for finitely many, let's assume this holds for all i>0. Now define a new "grand" density f=Sum_{i=1}^{i=infinity} b_i*f_i", where b_i>0 are constants such that "Sum_{i=1}^{i=infinity} b_i*a_i < infinity". This new density defines (up to a constant multiple) a new probability measure equivalent to the original one and such that (by construction) all X_i are integrable.

QuoteTherefore X_i<=a_i almost surely for all sufficiently large i>0???Can you please explain this statement?For any finite i, the way you have constructed a_i implies that there is the possibility that P(X_i > a_i) > 0. The maximal probability goes down as i increases, but this is no guarantee that it is 0 for all i greater than some number.Quote Now define a new "grand" density f=Sum_{i=1}^{i=infinity} b_i*f_i", where b_i>0 are constants such that "Sum_{i=1}^{i=infinity} b_i*a_i < infinity". This new density defines (up to a constant multiple) a new probability measure equivalent to the original one and such that (by construction) all X_i are integrable.I don't understand this bit at all.X_1 is integrable under f_1 by construction and X_2 is integrable under f_2 by construction. Why is it that X_1 is integrable under f_2 and X_2 is integrable under f_1? EDIT: please note once more that I am extremely unfamiliar with measure theory, and I might be asking a dumb question as a consequence

Let me respond a little later when I have time.

I'm not sure about the last part of this argument. Here is the solution I came up with. It might be similar, it definitely starts the same way. The game is to reduce things to the case of just one random variable which was solved above. Without loss of generality, assume that all of the random variables in question are nonnegative. Choose, for each n, an such that . Note that this is possible because for any (positive) finite valued random variable X, . By the Borel-Cantelli Lemma, it follows that P-almost surely, for suitably large n. This n, of course, depends on , but the point is that there is an above which the inequality holds.Now choose positive constants such that . Create the random variable , and note that by construction it is finite almost surely. Apply the reasoning given above to construct an equivalent probability measure under which this random variable is integrable. Each summand of the above random variable must also be integrable, and so we are done.

QuoteOriginally posted by: RDKBy the Borel-Cantelli Lemma, it follows that P-almost surely, for suitably large n.This is not the statement of Borel-Cantelli as I understand it.In this context, Borel-Cantelli makes the statement that the probability that infinitely many of the X_n exceed alpha_n is 0. This is not the same as saying that the probability of any given X_n exceeding alpha_n is 0. Can you please explain to me how you get your result from the Borel-Cantelli lemma, because I don't understand that step. What Borel-Cantelli ACTUALLY says, as I understand it, is that for any GIVEN omega in the sample space which is NOT an element of some measure-zero set there is an M s.t. for all n > M, X_n < alpha_n. This is not a statement about the probability of X_n > alpha_n being 0 for all n > M.Again, if I've misunderstood something please let me know.

QuoteOriginally posted by: jurowilmott1The entire probability space, Omega, can be written as a union of mutually disjoint open sets, A_i, preimages of X of intervals [i,i+1)How is this? Only if X: Omega -> R is continuous (topologically speaking) is it true that the preimage (under X) of an open set in R is an open set in Omega