#1
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who wants to help me?
for my simulation class, I have a review of probability due at 11 but I don't remember probability much, and I can't find this one in the text.
f(x,y) = c(5x^2y) x e [0,1], y e [0,1] otherwise 0 a) P(X + Y > 1) b) E[x] c} E[y] d) E[xy] e) Cov(x,y) any help/answers would be greatly appreciated |
#2
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Re: who wants to help me?
I'm happy to help, but I think you have your paratheses wrong. I think this should be:
f(x,y) = c*(x^2)*y where c = 6/5 to make the density integrate to 1. If not, can you tell me what the function c is and whether y is in the exponent or not? |
#3
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Re: who wants to help me?
[ QUOTE ]
for my simulation class, I have a review of probability due at 11 but I don't remember probability much, and I can't find this one in the text. f(x,y) = c(5x^2y) x e [0,1], y e [0,1] otherwise 0 a) P(X + Y > 1) b) E[x] c} E[y] d) E[xy] e) Cov(x,y) any help/answers would be greatly appreciated [/ QUOTE ] Note: f(x,y) must first be normalized by adjusting the constant if necessary such that the double integral of f(x,y) for x from 0 to 1, and y from 0 to 1, is equal to 1. a) double integrate f(x,y) for x from 0 to 1, and y from 1-x to 1. b) integrate x*f(x,y) for x from 0 to 1. <font color="red">Integrate the result for y from 0 to 1. Note that the integral of f(x,y) for y for all y is the marginal distribution of x or f(x), which can then be multiplied by x and integrated to get E(x). Use this method in general since you may not always get the same result by interchanging the order of the integrals.</font> c) integrate y*f(x,y) for y from 0 to 1. <font color="red">Integrate the result for x from 0 to 1. Note that the integral of f(x,y) for all x is the marginal distribution of y or f(y) which can then be multiplied by y and integrated to get E(y). Use this method in general since you may not always get the same result by interchanging the order of the integrals.</font> d) double integrate xy*f(x,y) for x from 0 to 1, and y from 0 to 1. e) cov(x,y) = E(xy) - E(x)*E(Y) |
#4
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Re: who wants to help me?
hey, we got an extension until tomorrow and these responses have been really helpful. thanks
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#5
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Re: who wants to help me?
[ QUOTE ]
I'm happy to help, but I think you have your paratheses wrong. I think this should be: f(x,y) = c*(x^2)*y where c = 6/5 to make the density integrate to 1. If not, can you tell me what the function c is and whether y is in the exponent or not? [/ QUOTE ] you're right. I'm an idiot for writing it the way I did I'm getting 6 and not 6/5 for some reason, if you're around, which I doubt. I'll check back before I hand it in in about 8 hours in case someone is so kind as to steer me in the right direction [img]/images/graemlins/smile.gif[/img] |
#6
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Re: who wants to help me?
[ QUOTE ]
Note: f(x,y) must first be normalized by adjusting the constant if necessary such that the double integral of f(x,y) for x from 0 to 1, and y from 0 to 1, is equal to 1. [/ QUOTE ] [ QUOTE ] a) double integrate f(x,y) for x from 0 to 1, and y from 1-x to 1. [/ QUOTE ] I got .02 from this. so far, so good I hope [ QUOTE ] b) integrate x*f(x,y) for x from 0 to 1. c) integrate y*f(x,y) for y from 0 to 1. [/ QUOTE ] when I do this, I am getting 3y/10 and 6/15x^2, respectively. can this be right? can I get rid of the variables from there? the problem is it's a multivariate equation and I'm only integrating for one variable each, so I'm left with one, right? [ QUOTE ] d) double integrate xy*f(x,y) for x from 0 to 1, and y from 0 to 1. [/ QUOTE ] I'm getting 1/10 for this [ QUOTE ] e) cov(x,y) = E(xy) - E(x)*E(Y) [/ QUOTE ] since my answers for E(x) and E(y) have the other variable in them, I'm not going to get a number here. I'm assuming this can't be correct. I'm going through my book now. you set me on the right path now. at least now I understand what I'm doing. before I was just mindlessly leafing through a book that I already read a couple of years ago |
#7
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Re: who wants to help me?
[ QUOTE ]
[ QUOTE ] I'm happy to help, but I think you have your paratheses wrong. I think this should be: f(x,y) = c*(x^2)*y where c = 6/5 to make the density integrate to 1. If not, can you tell me what the function c is and whether y is in the exponent or not? [/ QUOTE ] you're right. I'm an idiot for writing it the way I did I'm getting 6 and not 6/5 for some reason, if you're around, which I doubt. I'll check back before I hand it in in about 8 hours in case someone is so kind as to steer me in the right direction [img]/images/graemlins/smile.gif[/img] [/ QUOTE ] too late to edit but I see where I went wrong |
#8
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Re: who wants to help me?
[ QUOTE ]
[ QUOTE ] b) integrate x*f(x,y) for x from 0 to 1. c) integrate y*f(x,y) for y from 0 to 1. [/ QUOTE ] when I do this, I am getting 3y/10 and 6/15x^2, respectively. can this be right? can I get rid of the variables from there? the problem is it's a multivariate equation and I'm only integrating for one variable each, so I'm left with one, right? [/ QUOTE ] You should get 2x^2 and 3y/2 if f(x,y) = 6x^2*y. You then integrate out the other variable from 0 to 1. Or if you integrate f(x,y) for all y first, you get the marginal distribution of x or f(x) = 3x^2, and you then integrate x*f(x) to get E(x). Similarly for E(y), where f(y) = 2y. Use this method in general since you may not always get the same result by interchanging the order of the integrals. [ QUOTE ] [ QUOTE ] e) cov(x,y) = E(xy) - E(x)*E(Y) [/ QUOTE ] since my answers for E(x) and E(y) have the other variable in them, I'm not going to get a number here. I'm assuming this can't be correct. [/ QUOTE ] See above. The covariance should be 0 since x and y are independent. That is, you should find that f(x,y) = f(x)*f(y). |
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