Uniform Distributions.. plz help

instantaneous · #1 $Old$ October 25th, 2009, 04:23 PM

My first post on this forum.. sorry, this problem is giving me problems the whole day

Infinite number of (x) with properties (a) and (b)

(a) and (b) are uniformly distributed between 0 and 1

What is the fraction of (x) with properties:

a > b
3b > a

How do you calculate this stuff??? Btw, I am biology undergraduate...

Thank you in advance!!!

douber · #2 $Old$ October 25th, 2009, 07:40 PM

Welcome to the forums.

This is a uniform distribution so there is an equal chance that (x) will be near any value from 0 to 1 as it will any other.

This is because it's distribution function is constant.

So if you are looking for the probability that (x) lies between b and a where a > b then it is just the difference between b and a

P{b < (x) < a} = a - b

You should check this out yourself look up uniform distribution in your text or on google. its a fairly simble derevation.

Now I am not sure if I am interpreting your question correctly, because it seems like a weird question to me... but that is how i would answer.

for the other P{a < (x) < 3b} = 3b - a

instantaneous · #3 $Old$ October 26th, 2009, 02:55 AM

thanks for the answer, regarding the question:

you have two variables a and b that are uniformly distributed between 0 and 1

a and b are independent of each other

you chose randomly a and b

what is the probability to chose a and b so that

a > b
3b > a

My suggestion:

a' = a/(a+b)
b' = 1-a'

3 > a'/(1-a') > 1

so that 3/4 > a' > 1/2

giving a probability of 25%... is that correct? is the transformation to a' possible? if so, how would you prove that?

ty

CaptainBlack · #4 $Old$ October 26th, 2009, 03:11 AM

Quote:

Originally Posted by instantaneous $View Post$

thanks for the answer, regarding the question:

you have two variables a and b that are uniformly distributed between 0 and 1

a and b are independent of each other

you chose randomly a and b

what is the probability to chose a and b so that

a > b
3b > a

My suggestion:

a' = a/(a+b)
b' = 1-a'

3 > a'/(1-a') > 1

so that 3/4 > a' > 1/2

giving a probability of 25%... is that correct? is the transformation to a' possible? if so, how would you prove that?

ty

In the first case $p(a>b)=1/2$ simply by symmetry, that is $p(a>b)=p(b>a)$ and $p(a=b)=0$ .

For the second suppose $a$ is chosen first then $p(3b>a|a)=p(b>a/3|a)=1-a/3$ which is:

$p(b>a/3)=\int_{a=0}^{1}p(b>a/3|a)p(a)\;da=\int_{a=0}^{1} (1-a/3)\; da$

CB

instantaneous · #5 $Old$ October 26th, 2009, 04:00 AM

wow!

eehhm, so the probability of both happening 3 > a/b > 1 is

p(a>b) * p(3b>a) = 1/2 * 5/6 = 5/12?

CaptainBlack · #6 $Old$ October 26th, 2009, 04:58 AM

Quote:

Originally Posted by instantaneous $View Post$

wow!

eehhm, so the probability of both happening 3 > a/b > 1 is

p(a>b) * p(3b>a) = 1/2 * 5/6 = 5/12?

So that is one question rather than two?

(Those evenys are not independent so no its not the product).

Assume $a$ is known, then:

$p(b<a \text{ and }b>a/3|a)=2a/3$

(draw a diagram showing the range of $b$ that satisfies the condition)

Then

$p(b<a \text{ and }b>a/3)=\int_{a=0}^1 2a/3\; da$

CB