Is this probability answer correct?

[island-boy]

given a joint probability distribution f(x,y) which exists for every x, y > 0, and is 0 elsewhere. Is the conditional probability P(x<y | x<2y) always equal to 1 for any values of f(x,y)? Am I correct in this?

thanks.

 

[0rthodontist]

I'm not quite sure what you're asking of what your answer is supposed to be. It is not true that P(x < y | x < 2y) = 1. x and y are just numbers, not probability distributions, so you can't talk about the probability that they take on particular values. It would make more sense to write P(X < Y | X < 2Y) = 1, but this is not true either. What do you mean "for any values of f(x, y)"? Are you visualizing the PDF correctly?

To answer the question asked, the way to do it is to find a function of x and y that integrates to some finite value over x > 0, y > 0, and normalize that function so it sums to 1 over that domain. Then let your PDF function equal 0 for all x or y < 0, and for x > 0, y > 0, let it be your other function.

 

[HallsofIvy]

I take it that (x<y)|(x<2y) means (x<y) OR (x<2y). Of course, if x< y then x< 2y for y>0 so you are really asking P(X<Y) (Orthodontist's point is correct, although I would say that X and Y {as opposed to the numbers x and y} are "random variables" rather than numbers). Certainly, it is possible for some probability distribution, f(x,y), that Y> X!

 

[0rthodontist]

Usually P(A|B) means the probability of A given B, or
$$\frac {P(A \cap B)}{P(B)}$$

although I would say that X and Y {as opposed to the numbers x and y} are "random variables" rather than numbers

Good point

 

[island-boy]

I'll pose the original question:
if f(x,y) = e^(-x-y) for every x, y > 0 and is 0 otherwise, what is the conditional probability P(X<Y|X<2Y)?

my take is that since the area of X<2Y is inside the area of X<Y for x,y>0 then the probability is 1.

or said another way
P(X<Y|X<2Y)
=P(X<Y interection X<2Y)/P(X<2Y)
=P(X<2Y)/P(X<2y)
=1

so whatever the joint distribution given in the problem, the answer asked is always 1.

Is this correct?

 

[0rthodontist]

No. Use the formula I gave. You have to integrate over the region where X < Y, and then integrate over the region where X < 2Y, and then divide one by the other. Intuitively you should know that your answer doesn't make sense. What if X = 3 and Y = 2?

 

[island-boy]

if I did that, I would get 3/2, but all probabilities should be between 0 and 1...

ETA... DOH!
I think I figured out what I did wrong, instead of graphing X<2Y, I am using 2X<Y
let me try to solve that again...

EETA: got it, the answer is 3/4 right?

On a related note, in doing double integration for example:
solving for:
$$\int_{0}^{\infty} \int_{1}^{\infty} e^{-(x+y)} dx dy$$
we get
$$\int_{0}^{\infty} \lim_{b\rightarrow\infty} [-e^{-y-x}]_{1}^{b}dy$$
now my question is:
for this part: $$\lim_{b\rightarrow\infty} [-e^{-y-x}]_{1}^{b}$$ can we equate $$\lim_{b\rightarrow\infty}-e^{-y-b}$$ to 0 even if y is still a variable yet ot be integrated?

 

[Gagle The Terrible]

I think you got your answer right.
Now for your other question, i would just integrate your variables separately.
i.e. separate Exp(-(x+y)) in Exp(-x)*Exp(-y) and consideder Exp(-y) as a constant when you integrate Exp(-x). The subtle limit problem that arises is therefore avoided.

 

[island-boy]

hey, yeah, that's a way to avoid the limit. thanks!