wu :: forums - Print Page


    
      
        wu :: forums
        (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi)
      

        riddles >> medium >> Interval Inequality
        
(Message started by: ThudanBlunder on Aug 23^rd, 2010, 7:27am)

Title: Interval Inequality
Post by ThudanBlunder on Aug 23^rd, 2010, 7:27am

Two numbers, a and b, are randomly chosen, the first (a) from the interval [0, 0.5] and the second (b) from the interval [0.5, 1].
For n http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/geqslant.gif 2, what is the probability that b http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/geqslant.gif na?

Title: Re: Interval Inequality
Post by Hippo on Aug 23^rd, 2010, 7:32am

Expecting uniform distribution the question can be reformulated as a is chosen from interval [0,n/2] .... or scaled twice more a from [0,n] b from [1,2].
There are 3 cases ... a<1, a belongs to [1,2] and a>2 ...

Title: Re: Interval Inequality
Post by TenaliRaman on Aug 23^rd, 2010, 8:25am

[hide]0.1875 * 4 = 0.75 for n = 2[/hide]

Edit : Reconsidering after towr's post. Realized that I had done it for a specific case of n = 2. Also fixed a bug (had divided by the wrong area)

-- AI

Title: Re: Interval Inequality
Post by towr on Aug 23^rd, 2010, 8:37am

Uh.. what? Shouldn't it depend on n, Tenali?

I get:
[hide]
a=x/2
b=y/2+1/2

a < b
n/2*x < y/2+1/2
x < (y+1)/n
int_0^1 (y+1)/n dy
(y^2/2+y)/n|_0^1
1.5/n [/hide]
It seems to fit with experimental results, but it can't be right for n=1.5, even though I think it should (not that it's asked)... [edit]Ah no wait, I think I figured it out.. then we should [hide]integrate max( (y+1)/n, 1)[/hide][/edit]

Title: Re: Interval Inequality
Post by ThudanBlunder on Aug 23^rd, 2010, 9:07am

The original problem has n an integer http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/geqslant.gif 2, but n real and http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/geqslant.gif 3/2 is sufficient, I think.

Title: Re: Interval Inequality
Post by towr on Aug 23^rd, 2010, 9:10am

You need n >= 2 to keep it simple. Otherwise any n would work, although it's just 1 for n<=1.

[edit]For 1<n<2 [hide]P(b > an) = 2-(n²+1)/(2n) [/hide][/edit]

Title: Re: Interval Inequality
Post by ThudanBlunder on Aug 23^rd, 2010, 9:38am

on 08/23/10 at 08:25:05, TenaliRaman wrote:

[hide]0.1875 * 4 = 0.75 for n = 2[/hide]

[hide]That is correct.

And for n > 2 the same sort of shapes are involved, right?[/hide]

Title: Re: Interval Inequality
Post by TenaliRaman on Aug 23^rd, 2010, 12:51pm

on 08/23/10 at 09:38:36, ThudanBlunder wrote:

[hide]That is correct.

And for n > 2 the same sort of shapes are involved, right?[/hide]

[hide]Yeah, a rectangle at the bottom with a triangle on top[/hide]

Edit : ~~Ok, I tried to generalize the idea for any n. I might be doing something wrong here because I am getting a different answer than towr.~~

Edit2 : Found the error

-- AI

Title: Re: Interval Inequality
Post by TenaliRaman on Aug 23^rd, 2010, 1:56pm

Just documenting the method I used here -
[hide] The method is same as what gets used in meeting probability solution (which is not much different from what towr has done).

Let us represent the two random variables on the two axis, just as was done for the Meeting Probability (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1066732266). However, the intervals on the x and y axis this time would be (0,1). This gives us a unit square. Since, a takes values in (0, 0.5) and b takes values in (0.5,1), our sample space is the fourth quadrant (i.e if you divide the unit square into four equal halves).

Now, given an 'n', shade the event space. The shaded space would be :
1] a rectangle of height (0.5/n) and
2] a triangle on top of the rectangle of height ((1/n) - (0.5/n))

Computing the area of this shaded region would give you : (0.25) * (3/(2n))

The area of the sample space is 0.25, so the probability comes out to 3/(2n) or 1.5/n.
[/hide]

-- AI
P.S. -> Yeah, a diagram like Wu's would have been really useful here :-/

Title: Re: Interval Inequality
Post by ThudanBlunder on Aug 23^rd, 2010, 2:03pm

on 08/23/10 at 08:37:31, towr wrote:

It seems to fit with experimental results, but it can't be right for n=1.5, even though I think it should (not that it's asked)... [edit]Ah no wait, I think I figured it out.. then we should [hide]integrate max( (y+1)/n, 1)[/hide][/edit]

[hide]Can't we avoid nuking a gnat by considering the square formed by the points
(0, 0.5)
(0, 1.0)
(0.5, 0.5)
(0.5, 1.0)
and its intersection with the line y = nx[/hide]?

Title: Re: Interval Inequality
Post by towr on Aug 23^rd, 2010, 2:57pm

I don't think that's really different from what I end up doing. If you split up the parts of the integral right*, the only significant difference is that using an integral is a silly way to calculate the area of a triangle; for the rectangular parts it makes no difference.

*[hide](You can just split the integral at the point where (y+1)/n=1 to do away with the max function.)[/hide]