means that the set is infinite. For example, if and then . The last two sums are both infinite.
I'm working through a basic proof in a probability theory book, and I'm stuck trying to work out whether one of the statements is a typo, or I'm missing something:
is an arbitrary indexed set
then since for every , , for all there exists some such that and therefore
And at this point I'm confused: it seems to be that by construction, the indicator function in the second sum should always return 1, but the first does not necessarily. Therefore while they are both infinite sums, the first one may be smaller than the second, not the other way around.
Thanks for you reply.
So I don't think my question was clear enough. I absolutely understand that both sums are infinite. However I don't agree with the direction of the inequality between them: if I've understood the construction, the indicator of the second sum should always equal 1, whereas the first sum's indicator, whilst having an infinite number of 1's, can also return zeros. Is my understanding correct and therefore, this aspect of the proof wrong?
I'm still not quite sure I understand.
I read the sum on the RHS to mean a sum over all - since we've stated there exists a for every . Thus both sums are over the same domain (i.e. all positive integers).
Moreover, the actual index could very well be the same for multiple terms of the (RHS) sum - whatever it takes to ensure that for all .
Thus the RHS indicator always returns ones, the LHS may or may not, but since they're over the same domain, the LHS is smaller.
Sorry if I'm being dumb on this - if I've misunderstood the notation, could you point to the specific thing I've misread?
Hi, sorry I still understand how your explanation resolves this:
- I don't see how equates to your RHS sum. Doesn't mean sum over all ?
- Even if my sum is the same as your RHS sum, to me your sum very clearly seems to be an equality: both indicators return an exactly equal number of 1's.
The comparison would be correct if we defined to be some index such that and , ..., . Then , , ... is a subsequence of 1, 2, ..., so .
Edit: Changed to .
As an aside, could you help me understand why your correction example (above) is not simply always an equality? As far as I see it, you've defined the subsequence of to be exactly the indexes that the indicator in the left sum returns 1, and excluded the indices that the left indicator return zero. Thus, (I think) your two sums are equal.
By the way, I made a mistake defining ; it should be . Then ; , etc.
Wonderful, I get it now, thanks. So if you added that was the lowest index, and each was the next highest with we should have equality.
And well spotted, I had indeed missed that mistake.