#Probability question
rigid sunBOT
late summit
yes, but you're integrating over x
It's like if you wanted to know the area of a set A subset of R^2, so you took an integral over x from -infty to +infty of the area of the cross section for that x
if you take x infinitely thin, then you get infinitely many summands (uncountably so). Think about dx as taking a small partition of the x-axis
(I think. It's been a while since I did this in terms of probability.)
late summit
x is inside the integral
each P(....) is for a fixed x, but x ranges from -infty to infty
woven mason
I hope this is clear but explain on this, are we integrating on a new range smaller than a to b but finxing a x within that new areA?
@late summit
late summit
You should look at something like P( (x,Y) in the unit circle | x=0.5) where the domain is [-2,2]x[-2,2].
inside of [-2,2]x[-2,2], the probability of the given slice is 0.
But within the slice x=0.5, the probability is sqrt(2)/4.
(I think. The actually value doesn't matter though.)
late summit
I hope this is clear but explain on this, are we integrating on a new range smal...
I somehow misread your question as a statement, but I think this explanation should still make sense. Does it?
woven mason
its building a better picture
@late summit
why is y varying only over x isn't x a input to dtermine a y ? if there is one x there should be only one y?
@late summit
late summit
why is y varying only over x
What does this mean?
woven mason
like x is fixed and we are integrating y wrt x but if there is only one x shouldnt there only be one height
late summit
you are not integrating y wrt x
y is not being integrated
For each fixed x, you are asking the probability that (x,Y) is in B
This is like asking, for each x in [0,1], what is the probability that Y uniformly distributed in [0,1] will make (x,Y) in the unit circle.
You get a different probability for each x.