Probability one observation is larger than the other

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iocal

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Hi guys,

How would you go ahead to do the following? Suppose a random sample of size 2 is obtained from a distribution with density function \(\displaystyle f(x)=2(1-x),0<x<1\). Compute the probability that one sample observation is at least twice as large as the other.

My first thought was to evaluate the joint density function, the observations being independent. That is \(\displaystyle \displaystyle\int_0^1\int_{2x}^14(1-x)(1-y)dydx\) and then multiply the result by two as for the other case the computation is the same.I do not get the right answer that way, however. According to the book it is 7/12. So have I gone wrong somewhere here? Any help is greatly appreciated, thanks.
 
iocal said:
Hi guys,

How would you go ahead to do the following? Suppose a random sample of size 2 is obtained from a distribution with density function \(\displaystyle f(x)=2(1-x),0<x<1\). Compute the probability that one sample observation is at least twice as large as the other.

My first thought was to evaluate the joint density function, the observations being independent. That is \(\displaystyle \displaystyle\int_0^1\int_{2x}^14(1-x)(1-y)dydx\) and then multiply the result by two as for the other case the computation is the same.I do not get the right answer that way, however. According to the book it is 7/12. So have I gone wrong somewhere here? Any help is greatly appreciated, thanks.
Click to expand...
Same as you, I will suppose the larger of the two random samples is x, and then multiply the result by 2 for the other case. Consider the cumulative distribution,

\(\displaystyle \displaystyle F(x) = \int_0^x 2(1 - t) dt = 2x - x^2 \)

The integrated (cumulative) probability that another sample is less than half of x is

\(\displaystyle F(x/2) = x - \frac{1}{4}x^2 \).

Integrate over x (and double):

\(\displaystyle \displaystyle P = 2\int_0^1 F(x/2)\ f(x)\ dx = \ \cdot \ \cdot \ \cdot\)
 
DrPhil said:
Same as you, I will suppose the larger of the two random samples is x, and then multiply the result by 2 for the other case. Consider the cumulative distribution,

\(\displaystyle \displaystyle F(x) = \int_0^x 2(1 - t) dt = 2x - x^2 \)

The integrated (cumulative) probability that another sample is less than half of x is

\(\displaystyle F(x/2) = x - \frac{1}{4}x^2 \).

Integrate over x (and double):

\(\displaystyle \displaystyle P = 2\int_0^1 F(x/2)\ f(x)\ dx = \ \cdot \ \cdot \ \cdot\)
Click to expand...

Thank you. Evaluating the last integral I got 7/12 as I was supposed to.
What I do not understand however is why the double integral at the top did not give me the same result. Also what is the intuition behind your strategy? I get what the cumulative distribution means but I have some trouble figuring out the last integral. You integrate from 0 to 1 in order to deal with the second observation?
 
iocal said:
Thank you. Evaluating the last integral I got 7/12 as I was supposed to.
What I do not understand however is why the double integral at the top did not give me the same result. Also what is the intuition behind your strategy? I get what the cumulative distribution means but I have some trouble figuring out the last integral. You integrate from 0 to 1 in order to deal with the second observation?
Click to expand...
Your integral from 2x to 1 is not correct because if you let x go to 1 then 2x is outside of the defined function f(x). Instead, I would say you should integrate y from 0 to (x/2). Try it again with those limits!

The integral of f(y)dy is the cumulative distribution, so that is precisely what I did. The F(x/2) in my integration is essentially a conditional probability:

P(y < (x/2) | x) = F(x/2)
P(x) = f(x) dx
P(y < x/2) = \(\displaystyle \int F(x/2)\ f(x)\ dx\), where the integral is over all x.
[Doubled of course, to account for y>x.]
 
DrPhil said:
Your integral from 2x to 1 is not correct because if you let x go to 1 then 2x is outside of the defined function f(x). Instead, I would say you should integrate y from 0 to (x/2). Try it again with those limits!

The integral of f(y)dy is the cumulative distribution, so that is precisely what I did. The F(x/2) in my integration is essentially a conditional probability:

P(y < (x/2) | x) = F(x/2)
P(x) = f(x) dx
P(y < x/2) = \(\displaystyle \int F(x/2)\ f(x)\ dx\), where the integral is over all x.
[Doubled of course, to account for y>x.]
Click to expand...

The limits you suggested work nicely thank you. And the conditional probability you tried is really cool as well. I will definetely make a note of it.
Thanks again.
 
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