Basic stats help.

[Anonymous No.2698628]

Hey guys, I'm not too sure how to structure this post, but here it is.
This is part of my stats 1 assignment (the ENTIRE rest of it was piss easy) which I don't have any idea how to solve (you'll see why in a second)...

Can anyone help out??

Select two numbers randomly from interval [0, 1] call them x and y, what is the probability that x(the first number) is greater that 2y ?

The reason why this looks hard (I'm pretty sure it is extremely simple and I'm just looking into it too much) is because there is an infinite amount f numbers between zero and one...

Help?

[Anonymous No.2698704]

Bump.
Pic related, it's me right now.

[Anonymous No.2698747]

Isn't it just <span class="math">1-2y[/spoiler]?

[Anonymous No.2698786]

whos the chick in the photo

[Anonymous No.2698814]

>>2698786
Looks a lot like Michelle Trachtenberg...

[Anonymous No.2698829]

>>2698747
I don't think so...
But the equation is equivalent to that.
P(x) <=> P(1-2y)>0

[Anonymous No.2698846]

>>2698747
Don't think so...
It is equivalent to the original question though.
P(x) <=> P(1-2y)>0

[Anonymous No.2698975]

Bump, have a sweet as water trick.

[Anonymous No.2699067]

Bump...

[Anonymous No.2699101]

think of the joint distribution as a surface plotted above the xy plane. in this case (uniform distribution) the question can basically be reduced to:
what is the area of the region (where x is between zero and one and y is between zero and one) such that x is greater than 2y

[Anonymous No.2699155]

>>2699101

That surface isn't exactly trivial to construct though. The area of surface needs to represent the probability distribution of the values (x, y) being chosen, and moreover the double integral with 0<x<1 and 0<y<1 has to be 1.
That keeps the question as complex as it was to begin with!

[Anonymous No.2699233]

bump, again

[Anonymous No.2699306]

Alright, I'm just a highschoolfag so I apologize if this is totally wrong. I took stats and I'm in calculus, but we've never done anything close to this.

x could be anywhere from 0 to 1.
There are an infinite amount of values that x or y could be, but let's say there's a range of 1 value, which is true. I just don't know how to word it.

Also, to make things easier, let's say x can be greater or equal to 2y. We can fix this later.

The chance of x being greater than 2y is (x/2)/1
If x is 1/2 then y must be (1/2)/2 or lower
So it must be 1/4 or lower (or else 2y will be greater).
Which means there's a 1/4 chance of that happening.
I'm just giving you an idea of why my formula works

So the chance of x being greater than 2y for any given value of x assuming they have the same possible range of values is x/2

That's because y and x may only be from 0 to 1, but whatever value x is, y must be less than half of that. So y really only has half as many options for the condition to be satisfied. The probability of a certain value of x. Whatever the value of x is, y must be half that or less. It must be anywhere from 0 to x/2 out of a possible range of 1.
The chance of that is x/2/1 or x/2.

So the chance of x being greater than 2y is x/2, and x can be anywhere from 0 to 1. We need to find the average value of x/2 from 0 to 1.

Yay calculus, I can do this. Use the average value function.

1/(b-a) * integral from a to b of f(x)
so
1/(1-0) * int of x/2
1 * x^2/4 from 0 to 1
x^2/4 from 0 to 1

1^2/4 - 0^2/4
or
1/4
Did I do well /sci/?

Also that's only for x being greater than or equal to y. I'm not sure how it works for x being jut greater than y. The difference is pretty much negligible though, I think, for a range of all real numbers.

[Anonymous No.2699388]

If I have a hat full of numbers in [0, 1], and I pull out the number y, I have 1-2y numbers for which x can be to win. For example,
when y=0.1, I have (1) - (0.2) = 80% of win
when y= 0.5, I have (1) - (1) = 0% of win

Let's make this a function of x.

y(x) = (1-x)/2

Now let's integrate on [0, 1] and evaluate to get the probability!

Y(x) = -(1/4)(x²) + (1/2)x
Y(1) - Y(0) = -(1/4) + (1/2) = 1/4

looks like this scientist got it right, too:

>>2699306

[Anonymous No.2699415]

>>2699388

No, you most definitely don't have 1-2y numbers left to chose. Say y = .8. you can't pick -.6 numbers.
your logic is only valid for the actual probability function p(x), which means p(x) = p(1-2y), as previously stated. the notion that 1-2y is your probability function is horribly wrong, unless we're restricting y to .5

[Anonymous No.2699454]

>>2699415

good point; I just happened to get the right answer haha. How's this:

p(x) = (0.5)x
P(x) = (0.25)x²
P(1) - P(0) = 0.25 - 0 = 1/4

[Anonymous No.2699476]

>>2699454
you're getting an answer, but your probability function p(x), over it's maximal domain [0, 1] doesn't sum to 1, so it's still not a valid answer - a probability function _must_ sum to 1. i appreciate the effort though!
hmm, so i know for a fact that the uniform distribution function over the domain [0, 1] is p(x) = 1 for 0 <= x <= 1 and p(x) = 0 otherwise. (http://en.wikipedia.org/wiki/Probability_density_function#Further_details)) - which definitely means you can't be right here. sorry!

[Anonymous No.2699502]

>>2699476

Doesn't it add to 1?

The probability of x<2y =
integral of 1 - (1/2)x, [0,1] =
x - (1/4)x², [0, 1]
1 - (1/4) = 3/4

[Anonymous No.2699660]

>>2699476
No, I'm sorry. You're wrong. The probability function should not sum to 1.
Does the probability of getting a certain amount of numbers in a row from a number cube sum to 1?
No.
P(x) = (1/6)^x

So shut the fuck up.

[Anonymous No.2699698]

>>2699660

You're right, grammar-fail. the integral over the entire domain of the function should be one, however. any elementary probability textbook will tell you that. and if you're telling me the integral of (x/2) from 0 to 1 is 1, well.....
it's just not.