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sthief09
09-14-2005, 09:06 AM
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 &gt; 1)
b) E[x]
c} E[y]
d) E[xy]
e) Cov(x,y)

any help/answers would be greatly appreciated

AaronBrown
09-14-2005, 06:49 PM
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?

BruceZ
09-14-2005, 07:46 PM
[ 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 &gt; 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)

sthief09
09-16-2005, 07:09 AM
hey, we got an extension until tomorrow and these responses have been really helpful. thanks

sthief09
09-16-2005, 07:14 AM
[ 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 /images/graemlins/smile.gif

sthief09
09-16-2005, 07:46 AM
[ 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

sthief09
09-16-2005, 08:04 AM
[ 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 /images/graemlins/smile.gif

[/ QUOTE ]

too late to edit but I see where I went wrong

BruceZ
09-16-2005, 04:14 PM
[ 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).