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 July 21st, 2010, 02:53 PM #1 Newbie   Joined: Jul 2010 Posts: 4 Thanks: 0 want to show that show that two infinite summations R equal ---------------------------------------------------------------------------------------- DESCRIPTION OF THE PROBLEM AND INFORMATION I AM SEEKING ---------------------------------------------------------------------------------------- I would like to show that the following two infinite summations are equal for any two bijective functions f and g, which are defined from a set of natural numbers to a subset of real numbers. Both the set of natural numbers and the set of subset of reals are countably infinite. And the Bijection condition on these functions suggests that there exists a one to one correspondence between the elements of the set of natural numbers and the subset of reals. Ok, so the expression I want to prove to be true is as follows: sum (0 to infinity) [f(n) * prob {x| x = f(n)}] = sum (0 to infinity) [g(n) * prob {x| x = g(n)}] In the above expression the function prob takes as input a set and returns a real value between 0 and 1. It is a probability measure function that returns a real value corresponding to each of its set argument. The two functions are bijective and are defined on the same two sets, that is they both map all elements of set of natural numbers to unique elements of a countably infinite subset of set of real numbers. I am not quite sure how to proceed with the proof and need help. This is what I know and why I think the two infinite summations should be equal. Each summation is over the product of two terms. Both terms in the product are functions of n. Both functions map the same set of natural numbers to the same subset of reals. So if I build a set of all values of f(n) and another set that consists of all values of g(n), these sets will be equal. The probability function just assigns a real value to its set argument. Therefore, for all values of n which is a natural number, a set that consists of all possible real values of type ( f(n)*porb {x|x = f(n)} ) will be equal to the set that consists of all possible real values of type ( g(n)*prob {X|x = g(n)} ). Lets call these sets A and B respectively. Now since both sets A and B are countably infinite and are equal, the sum (0 to infinity) of all elements of these sets will also be equal. I am not a mathematics major and am not quite sure how to proceed with the proof and need help. I want to state above reasoning mathematically and as unambigously as possible. Can some one please comment on my reasoning and point out if there are any mistakes? may be suggest a better method, thanks
 July 22nd, 2010, 01:58 PM #2 Senior Member   Joined: Jun 2010 Posts: 618 Thanks: 0 Re: want to show that show that two infinite summations R equal Hello notnaeem. It seems that for the most part, you have a solid grasp of those parts of the problem which you expand upon, except the last few remarks trying to justify the conclusion. This is, I am sorry to say, because the statement you are trying to prove is, as you have stated it, not true in general. It may be the case that you have omitted some condition on the functions involved (maybe the functions f and g should both take positive values only?), or that there is some notational misrepresentation somewhere in the descriptions of the functions. In any case, let me try and expose some of the reasons why the statement is not true, and then maybe we can come to grips with how to salvage things. Let me first restate the problem, using a bit of mathematical notation to boot, so that we have an unambiguous starting point for our analysis. First, I notice from the index on the summations that you consider zero to be a member of the set of natural numbers. That's fine, though usually we consider them to be only the positive integers. This is of no importance to us here, however. Let me use the following notation for the set of nonnegative integers: $\mathbb{N}_0= \{0,1,2,3,\dots\}$. Now, the functions $f$ and $g$ are both bijections onto the same subset $A \subset \mathbb{R}$. We denote this as follows: $f,g \ : \ \mathbb{N}_0 \leftrightarrow A$, where the double arrow indicates bijection. A second function, prob, which I will denote by $p$ for brevity, takes as its argument a subset of the real numbers (I assume, it's not entirely clear, and doesn't seem to make a difference, as we will see) and returns a number in the open interval (0,1). If we want to be really pedantic, we can denote this by: $p \ : \ \mathscr{P}(\mathbb{R}) \rightarrow (0,1)$, where the notation $\mathscr{P}(\mathbb{R})$ denotes the power set of the reals, that is, the set of all subsets of the reals. I say this only to introduce you to notation, in case you are unfamiliar with it. By the way, since I don't know what level of familiarity you have with mathematical notation\vocabulary, I am being purposefully descriptive, so please don't take offense if this is obvious stuff for you! We wish to prove that the following equality holds: $\sum_{n=0}^{\infty}{f(n)p(\{x | x = f(n)\})} = \sum_{n=0}^{\infty}{g(n)p(\{x | x = g(n)\})}$ Let's note one thing right off. The sets $\{x | x= f(n)\}$ are all one-member sets, since $f$ is a function, so they are really just the sets $\{f(n)\}$. This means that $p(\{x | x= f(n)\}) = p(\{f(n)\})$, and we can thus interpret $p$ on these single-member sets to be just a function of that member. Since we are only concerned with values of $p$ on such sets, and since $f$ and $g$ take values only in $A$, we can, for the sake of this problem, recast $p$ as a function from $A$ into $(0,1)$. Another thing to note is that since $f$ and $g$ are both bijections from $\mathbb{N}_0$ onto $A$, then any bijection $h \ : \ \mathbb{N}_0 \leftrightarrow \mathbb{N}_0$ will simply 'rearrange' the terms of the sequence $(f(n))$. This is fairly obvious, but quite useful. Next, let me note a famous result of Riemann on the rearrangement of a conditionally convergent series (of real numbers). For background (in case you need it), an infinite series (of real numbers) $\sum_{n=0}^{\infty}a_n$ is said to converge absolutely if the series $\sum_{n=0}^{\infty}|a_n|$ converges. A series is said to converge conditionally if it converges but does not converge absolutely. Riemann's result says that a conditionally convergent series of real numbers can be 'rearranged' (in the sense described in the previous paragraph) to converge to any real number r, and it can even be rearranged to diverge to $\pm\infty$, or to oscillate without limit! Now, the same is not true for absolutely convergent series of real numbers. Any such series, however it might be rearranged, will converge to the same number. You may now start to realize how I am going to construct a counterexample to your claim. In broad terms, I am going to choose some conditionally convergent series, and then rearrange it to produce another series that converges to a different real number. So, let's get down to the details of the example. Let me define $f(n)= (-1)^{n+1}(n+2)$ for $n \in \mathbb{N}_0$. Then, we see that the image of $f$, $\mathrm{Im}(f)= \{-2,3,-4,5,-6,7,\dots\}$ which is by definition the set $A$. Further, let me define $p(x)= \frac{1}{x^2}$ for all $x \in \mathbb{R}- [-1,1]$. Note that $p$ then necessarily takes values in $(0,1)$. Also, I don't care how $p$ is defined on $[-1,1]$, as long as it takes values in $(0,1)$ as well. Now, let me, as we discussed earlier, interpret $p(x)$ to be the same as $p(\{y | y= x\})$, and, if need be, I can define $p$ to take some constant value in between 0 and 1 for all the other subsets of $\mathbb{R}$. Next, let me define a bijection $h \ : \ \mathbb{N}_0 \leftrightarrow \mathbb{N}_0$ as follows: $\begin{equation*} h(n)= \begin{cases} \frac{2}{3}n &\text{if n \equiv 0 (mod 3)}\\ \frac{4}{3}(n-1)+1=&\text{if n \equiv 1 (mod 3)}\\ \frac{4}{3}(n-2)+3=&\text{if n \equiv 2 (mod 3)} \end{cases} \end{equation*}=$ You can check that this is in fact a bijection, and you will see that $h(\{0,1,2,3,4,5,6,\dots\})= \{0,1,3,2,5,7,4,\dots\}$. Finally, let me define $g(n)= f(h(n))$ for $n \in \mathbb{N}_0$. Clearly then, both $f$ and $g$ satisfy the requirement of being bijections from $\mathbb{N}_0$ onto $A$. Let us now write down the series in the problem statement. First, we have: $\sum_{n=0}^{\infty}{f(n)p(f(n))} = \sum_{n=0}^{\infty}{\frac{(-1)^{n+1}(n+2)}{((-1)^{n+1}(n+2))^2}} = \sum_{n=0}^{\infty}{\frac{(-1)^{n+1}}{n+2}}$ which is a sum that is well-known to converge (conditionally) to $\ln{2} -1$. Next, we write down the other sum: $\sum_{n=0}^{\infty}{g(n)p(g(n))} = \sum_{n=0}^{\infty}{f(h(n))p(f(h(n)))} = \sum_{n=0}^{\infty}{\frac{(-1)^{h(n)+1}}{h(n)+2}}$ which is a rearrangement of the former series, and which converges to $\frac{3}{2}\ln{2} -1$. You of course have to take on faith that the two series I quoted do in fact converge to the numbers I state, but the discussion on Riemann's result was meant to make this plausible. This shows that the result you sought cannot be true as stated. However, let's hope there was some omission or misstatement in the problem, so that we can get a similar result! Now, I leave it to you to come up with some fix to the problem so that the result holds. In fact, this may not be terribly hard. Best of luck!
 July 26th, 2010, 06:00 AM #3 Senior Member   Joined: Dec 2008 Posts: 306 Thanks: 0 Re: want to show that show that two infinite summations R equal Why can your p be extended to a nontrivial measure?
July 26th, 2010, 10:53 AM   #4
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Re: want to show that show that two infinite summations R equal

Hi notnaeem.

I see, you want prob to be a probability measure in the sense that its integral over its underlying measure space is 1, and it is a measure in the sense of measure theory. This was certainly not clear from your description of the probability function a bit further down:

Quote:
 Originally Posted by notnaeem The probability function just assigns a real value to its set argument.
My counterexample with p is certainly not a probability measure in this sense.

I may be able to get back to you with a proof of what you want, but please describe completely all the properties you want prob to satisfy.

Best of luck.

 August 16th, 2010, 12:32 PM #5 Newbie   Joined: Jul 2010 Posts: 4 Thanks: 0 Re: want to show that show that two infinite summations R equal Dear Ormkarr Thank you so much for your detailed response, I really appreciate it. Unfortunately I could not write back any sooner. Your re description of the problem was accurate and very helpful for me, I am now better able to understand the problem. I am sorry about the missing information. Here is some other information I know about the problem. The probability function in my problem is a probability measure such that it satisfies all the basic axioms of probability, that is, its value is between 0 and 1 both inclusive. The probability of empty set is 0 and that of the universe is 1. The probability of union of two disjoint sets is the sum of their individual probabilities, etc. I guess the clue to the proof is that if a series is absolutely convergent then any possible rearrangement of it would sum to a unique value. I was wondering if you could help me and guide me to get a detailed proof for this problem. I will be checking the post regularly and would be able to provide more information if needed and also update the post If I make any progress on my own. Thanks in advance

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