Comparing the values of functions

Z

zeroes00

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Sep 20, 2008
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Now this may be very simple, but I just can't get my head around it.

Let's say we have two functions \(\displaystyle f(x)\) and \(\displaystyle g(x)\) and both of these functions are continuous between 0 and 1 and always above the x-axis.
Then we pick two random values \(\displaystyle x\)[sub:3cx3enqp]1[/sub:3cx3enqp] and \(\displaystyle x\)[sub:3cx3enqp]2[/sub:3cx3enqp] that are both between 0 and 1.
Now, what are the odds that...
\(\displaystyle P( f(x\)[sub:3cx3enqp]1[/sub:3cx3enqp]\(\displaystyle ) > g(x\)[sub:3cx3enqp]2[/sub:3cx3enqp]\(\displaystyle ) )\)

How would you go about solving this (if we knew what the functions are that is)? Is it comparing the areas the functions define or what? Any help is appreciated.


EDIT:
Maybe it would help if I just say the functions are:
\(\displaystyle f(x)=x\)
\(\displaystyle g(x)=1/(2/x-1)\)

Afterthought:
Maybe it's the ratio of the areas of the probability integrals of these functions? It's all just too complicated for me to see the logic of it.
(By the way I can't figure out how do you get nice formatting here)
 
Maybe it's something like
\(\displaystyle h(x,y) = f(x)-f(y) = x - 1/(2/y-1)\)

and then the probability is the area of positive values of \(\displaystyle h(x,y)\) on the 0 to 1 xy-plane? How the **** would you solve that then?
 
If look at the graphs of f(x) and g(x), you should conclude that:

For every x[sub:1spl8cmu]1[/sub:1spl8cmu], there is a point x[sub:1spl8cmu]3[/sub:1spl8cmu] - such that f(x[sub:1spl8cmu]1[/sub:1spl8cmu]) = g(x[sub:1spl8cmu]3[/sub:1spl8cmu])

Then

for all x[sub:1spl8cmu]2[/sub:1spl8cmu] > x[sub:1spl8cmu]3[/sub:1spl8cmu] we would have f(x[sub:1spl8cmu]1[/sub:1spl8cmu]) < g(x[sub:1spl8cmu]2[/sub:1spl8cmu])

and

for all x[sub:1spl8cmu]2[/sub:1spl8cmu] < x[sub:1spl8cmu]3[/sub:1spl8cmu] we would have f(x[sub:1spl8cmu]1[/sub:1spl8cmu]) > g(x[sub:1spl8cmu]2[/sub:1spl8cmu])

The value of x[sub:1spl8cmu]3[/sub:1spl8cmu] will determine the probability.
 
If I understand correctly, you're saying x[sub:4saqnzcn]3[/sub:4saqnzcn] = P[ f(x[sub:4saqnzcn]1[/sub:4saqnzcn]) > g(x) ]
...where x is a random value between 0 and 1

So would I then get the requested probability P[ f(x[sub:4saqnzcn]a[/sub:4saqnzcn]) > g(x[sub:4saqnzcn]b[/sub:4saqnzcn]) ]
(...where both x[sub:4saqnzcn]a[/sub:4saqnzcn] and x[sub:4saqnzcn]b[/sub:4saqnzcn] are random values between 0 and 1)

...by making a function h(x) where
h(x[sub:4saqnzcn]1[/sub:4saqnzcn]) = x[sub:4saqnzcn]3[/sub:4saqnzcn]
and then solving the mean of this function between 0 and 1 ?

Like so perhaps...
f(x[sub:4saqnzcn]1[/sub:4saqnzcn]) = g(x[sub:4saqnzcn]3[/sub:4saqnzcn])
x[sub:4saqnzcn]1[/sub:4saqnzcn] = 1 / (2/x[sub:4saqnzcn]3[/sub:4saqnzcn] - 1)
x[sub:4saqnzcn]3[/sub:4saqnzcn] = 2 / (1/x[sub:4saqnzcn]1[/sub:4saqnzcn] + 1)

...so
h(x) = 2 / (1/x + 1)
and then I'd get the mean of the function h(x) over the interval (0,1) by using that
integral-kind-of formula which I'm not able to write here, and that would be the requested probability?
 
Did you plot the functions

and

tried to interpret the problem (and my response) graphically?
 
Yes I plotted the functions and I think I understood your response.
But I figured you described the probability that f(x[sub:1lm6dqne]1[/sub:1lm6dqne]) is bigger than g(x)
where x[sub:1lm6dqne]1[/sub:1lm6dqne] is some known value and x is some random value,
and I need the probability that f(x[sub:1lm6dqne]a[/sub:1lm6dqne]) is bigger than g(x[sub:1lm6dqne]b[/sub:1lm6dqne]) when both x[sub:1lm6dqne]a[/sub:1lm6dqne] and x[sub:1lm6dqne]b[/sub:1lm6dqne] are independent random values.
 
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