Interesting problem

[Vinay M]

Let's say we choose a number randomly from the interval (0,1). If X is the number of times we have to pick numbers from the interval so that the sum of all those picked numbers just goes more than 1. What is the expectation of X? 

[Himanshu]

Yep, it is intersting. I will try it.

[Himanshu]

Suppose (S)=(0,1)  

We can see that range(X)=(1,infinity)

Event (X<x)= Event (S>1/x). = Event(1/x<S<1)

P(X<x)=P(S>1/x)=1-P(S<1/x)=1-1/x.   

So, distribution function of X = 1-1/x.

Density function of X /, f(x) = 1/x^2 for x in (1,infinity) and 0 otherwise. 

Expectation of X = integral_(over X) (xf(x)dx) = integral_{1}^{infinity}(1/x)dx = ln(x) from 1 to infinity.  

But it is not defined.  

Is my way of doing and answer correct?

[Himanshu]

I think I have made a mistake 

The event (X>x) should be event (1/x<S<1). 

Because if we take any x1 in X, then there is s1 such that s1x1>1, so for any s>s1, x1 and x>x1 will make hold the inequality. 

But then distribution function of X=1/x. 

But still the expectation value can't be evaluated as it will divereges.

[Himanshu]

But the density function 1/x is giving a sort of contradiction that if we find the probability of whole space we get -1 as the density function will be -1/x^2.   

I am not able to figure out how to proceed it.  

Really, it is an interesting problem.

[Vinay M]

Very nice, showing interest in this problem, I think I did not understand your idea here, but maybe I have say that X is the set ={2,3,4,5,....), because we can pick 2 numbers whose sum just crosses 1, and 3, 4, 5, so on all are possible but 1 is not there in this set because, we cannot pick a number only once and say this crosses 1, we need to pick 2 at least, so that there is chance for the sum to cross 1. So this is the problem. I have an idea which connects the same length notion for probability, but here for this problem, I am facing computation difficulty. My idea is just consider let's say number 2 in set X, which means we pick 2 numbers whose sum just crosses 1, now what are all the possibilities, it's infinite, but there is length to save us from that, now using that length we can compute the probability of selecting 2 numbers from (0,1) which just crosses 1. I got it to be 0.58333... but this could be wrong, and Similarly calculate for X=3, which means we have to pick 3 numbers, only then their sum crosses 1, and then for X=4, 5, 6,..   so on. I think the probabilities decreases as number of picks increase, that's just a guess. After getting some handful terms then compute the expectation value and see where it is going, or converging. This is my plan, but I could be wrong, I have stopped at 3 and I'm lazy to do further computation. Let me try to show you what I'm doing with a picture, please find it attached to this email. Thank you. 

[Himanshu]

@Vinay, this post might be somewhat long, but please read this post and the attachment completely.     

Initially when I read this problem, I thought that X is a function from (0,1) to range(X) where Range(X)=(1,infinity). And if we select any number so from (0,1), then by archimedian property we get a number x from (1, infinity) such that xso>1.     

Now we can take set A={x|x in (1,infinity) s.t. xso>1}, This set has a infimum and that we take xo.     

Now we can easily see that if we take any element s from the set (so,1), xo.s>1.     

It ia also easy to see that so=1/xo    

So corresponding to any element of x>=xo, we get an interval in (0,1) [which is (1/xo,1)].   [Ex- consider 2 in (1,infinity), then consider any element s in (1/2,1),  2.s>1 also 3.s>1,4.s>1,....] 

So, in short corresponding to any interval [xo,infinity] we get an interval (1/xo,1) over which we know the probability is just length.   

Thus P(x>=xo)= 1-1/xo      

So, P(x<xo)=1/xo     

Distribution function on X is 1/x   

Density function on X is derivative of distribution function which is -1/x^2     

E[X]=integral(x.(-1/x^2)dx)  but it can't be evaluated.       

But problem with this reasoning is that I get -1 as the total probabilty over (1,infinity) as the density function we get is 1/x^2. I don't know why I get this?         

That's what I am trying to explain in my previous reply.    

But in your previous reply, I come to know that Range(X)=set of natural numbers greater than 1 instead of (1,infinity) which I previously thought. If I take any element of range(X), I have to take that much element from S=(0,1) such that their sum is greater than 1.    

I think that this X is then not a function from  (0,1) to {2,3,4,...}.   

Like if we take 2 in range(X),    

Then we get a tuple like (1.8,1.8),(1.8,0.3),(1/8,0.44), {union{x in (1/2,1)}(1.8,x)}............    

We won't get an interval as such because there is also a possibility of taking one x<1/2 and other x>1/2.  

So, it is better a function from S^n to n for n>1. {S^n=SXSX.....(n times) - cartesian product}

     

Please take a look at the pdf file which I have attached -    

But the problem is that if we sum up the probability of all the elements of range(X) to get the probability of whole sample space (we can do it as range(X) is a countable set), the sum gets diverges not equal to 1.   

[Himanshu]

For n=4, if s4 tend to zero, we get a 3D structure as shown in the above pdf and when s4 tends to 1, then the structure tends to become  cube of dimension 1X1X1, with translation across the s4 axis (which is orthogonal to other 3), which we can't imagine.   

Same for higher dimensions.  

[Vinay M]

Hi it's been a while since I've commented here. Sorry for not reading your comments. This problem was in my mind for long now and today I tried it again and I seem to have got a glimpse into the answer but the proof is not complete. Before telling you about that, let me reply to your comments, I seem to think that you have not understood the question properly, maybe I should have stated it more clearly( I'm sorry). I'll state or maybe explain the problem here again. The problem goes like this, consider we have a closed interval 0 to 1, I randomly pick numbers from this interval, meaning I get anything between 0 to 1. Now how many time should I pick numbers so that the sum of all those I picked just goes or crosses 1, for instance I pick a number, let's say it is 0.78, now its not more than 1, so I pick a number again, let it be 0.53, now if I add both it crosses 1. For THIS case the number of times I had to pick so that I cross 1 is 2, but had I picked 0.15 instead of 0.53, then I have to pick again, and even after picking thrid time, if the sum of all of picked numbers is not more than 1 then I have to keep picking. There are infinitely many ways to pick. For example (0.5, 0.74)-2times, (0.2,0.1,0.8)-3times, (0.6,0.002,0.2,0.4) -4 times,...so on. Call X as the number of times I had to pick so that I cross 1. Now, what is the expectation of this random variable X? 

So I seem to get the answer as the transcendental number "e" minus 1, e-1. I still have not proved it.