calculus probability quesiton

[Semidevil]

so X and Y have joint pdf f X Y(x, y) = 1/2 for 0 < y < x < 2, and 0 otherwise.

find P(0< Y < 2/5 | X = 3/4)

first, I dont understand how to read the function. So the function is equal to 1/2 when 0 < y < x < 2. And it is equal 0 otherwise.

So I need to find Probabilty Y is between 0 and 2/5, given that X = 3/4? is that what it is asking?

I think this has to do with integration, but what am I integrating? It's a single integeral right? and what do I do w/ the X = 3/4 part?

 

[Goosemaster]

joing?

 

[RaynorWolfcastle]

I'll let you squirm a bit before I solve it for you

 

[Goosemaster]

Originally posted by: RaynorWolfcastle
I'll let you squirm a bit before I solve it for you

:evil:

 

[MrScott81]

reading your textbook might have the answer.....or the teachers lecture notes

 

[Random Variable]

Find the marginal pdf of X

 

[Random Variable]

then the conditional pdf g(y|x) is the the joint pdf f(x.y) divided by the marginal pdf of X

EDIT: I had it backwards

 

[Semidevil]

Originally posted by: Vespasian
then the conditional pdf of g(x|y) is the the joint pmf f(x.y) divided by the marginal probabilty of Y

thanx.

ok, so first off. when they list x< y < x < 2, are the limits from 0 to 2 in both cases when I do the marginals. if so, I got the marginals to be 1 for both f(x) and f(y) right?

now what do I do next? so what do I do with the 0 < y < 2/5 and X = 3/4 part?

btw, if anyone is good at probability, and has some time to tutor a fellow atot, msn messenger is semidevilz@hotmail.com

edit, nevermind, I got it...but I still get 1/2 as my answer....for the whole thing

 

[Semidevil]

ok, im not understanding expected values. so lets say f(xy) = k(x + y) for 10< x < 20, and 10 < y < 20. what is E(Z) when Z = 1/X + 1/Y.

now how do you find the expected values? I thought the exected value is always like n*p....but when they give you a function, do I just integrate it? and put 1/(what I get for x) + 1/(what I get for y)?

 

[cchen]

E(x) = integral from -inf to +inf x * f(x) dx

apply that to the joint pdf

 

[Semidevil]

Originally posted by: cchen
E(x) = integral from -inf to +inf x * f(x) dx

apply that to the joint pdf

ok, so I dont know how to do it "jointly" and the z = 1/x + 1/y thing. i know how to integerate seperatly and with the formula, but how do I do it jointly?

 

[Random Variable]

Integrate f(x.y) from 0 to x to get the marginal p.d.f. of X. Then g(y|x) is f(x,y) divided by the marginal p.d.f. of X. Then to find the P(0<y<2/5|x=3/4), plug 3/4 into g(y|x) for x and then integrate from 0 to 2/5.

Does that make sense?

 

[Yossarian]

update acrobat reader so you can open the pdf.

 

[Semidevil]

ok, after thinking this thouroughly, can I do this?

so I need to find E(Z) where Z = 1/X + 1/Y....so is it suppose to be lower case x and y?

anyways, so can I do this? I see this formula in my book.

where it says integeral from -inf to inf of (x*f(x)) + integeral of -inf to inf (y*f(y)), so can I just put 1/x instead of x, and I think f(x) and f(y) are the marginals right?

 

[Random Variable]

Originally posted by: Semidevil
ok, after thinking this thouroughly, can I do this?

so I need to find E(Z) where Z = 1/X + 1/Y....so is it suppose to be lower case x and y?

anyways, so can I do this? I see this formula in my book.

where it says integeral from -inf to inf of (x*f(x)) + integeral of -inf to inf (y*f(y)), so can I just put 1/x instead of x, and I think f(x) and f(y) are the marginals right?

Look at the what I posted like 5 minutes ago.

 

[Semidevil]

Originally posted by: Vespasian

Originally posted by: Semidevil
ok, after thinking this thouroughly, can I do this?

so I need to find E(Z) where Z = 1/X + 1/Y....so is it suppose to be lower case x and y?

anyways, so can I do this? I see this formula in my book.

where it says integeral from -inf to inf of (x*f(x)) + integeral of -inf to inf (y*f(y)), so can I just put 1/x instead of x, and I think f(x) and f(y) are the marginals right?

Look at the what I posted like 5 minutes ago.

im actually doing a different problem now...so..

does it apply too?

 

[Random Variable]

Wouldn't it just be the double integral of [(1/x)+(1/y)] *k(x+y)] using the limits you stated? Am I missing something?

 

[Semidevil]

yea, thats what I was thinking...but im not sure...so....I think so..

 

[Random Variable]

Or you could find the expected values of 1/x and 1/y separately and add them together.

 

[Semidevil]

Originally posted by: Vespasian
Or you could find the expected values of 1/x and 1/y separately and add them together.

ok, i'm getting confused...but would this work?

integral from 10 to 20 of 1/x * f(x) + integeral from 10 to 20 of 1/y * f(y). so f(x) and f(y) are the marginal values...

 

[Random Variable]

To the find the expected value of 1/x and 1/y you have to do double integrals.

EDIT: I think I'm starting to confuse myself.

 

[Semidevil]

Originally posted by: Vespasian
To the find the expected value of 1/x and 1/y you have to do double integrals.

aw man, i'm getting confused..there's too many formulas...ok, so double integral from 10 to 20, of (1/x + 1/y) dy dx right? but whatabout the original function?

this is hard....

 

[Random Variable]

(1/x)*k(x+y)dxdy + (1/y)*k(x+y)dxdy. Is there an answer to this problem?

 

[Semidevil]

ok, that works...thanx....

but if I multiply the 1/x or 1/y through, i'm gonna end up integerating 1 + y/x.......and that will give me natural logs right?

so 1 +y/x dx = 1x + ylnx..

 

[Random Variable]

yes but you forget the k