Moment Generating Function - Integration Help


by ARLM
Tags: function, generating, integration, moment
ARLM
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#1
Dec6-12, 09:39 AM
P: 4
I am working on a probabilty theory problem:

Let (X,Y) be distributed with joint density
f(x,y)=(1/4)(1+xy(x^2-y^2)) if abs(x)≤1, abs(y)≤1; 0 otherwise

Find the MGF of (X,Y). Are X,Y independent? If not, find covariance.

I have set up the integral to find the mgf

∫∫e^(sx+ty)f(x,y)dx dy
with both integrals from -1 to 1.

I am having trouble integrating this though in order to move on with the problem. I began to try integration by parts and I do not think that is the best route but have no other ideas.
If anyone can help, I would greatly appreciate it!
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Ray Vickson
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#2
Dec6-12, 09:47 AM
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Quote Quote by ARLM View Post
I am working on a probabilty theory problem:

Let (X,Y) be distributed with joint density
f(x,y)=1/4(1+xy(x^2-y^2)) if abs(x)≤1, abs(y)≤1; 0 otherwise

Find the MGF of (X,Y). Are X,Y independent? If not, find covariance.

I have set up the integral to find the mgf

∫∫e^(sx+ty)f(x,y)dx dy
with both integrals from -1 to 1.

I am having trouble integrating this though in order to move on with the problem. I began to try integration by parts and I do not think that is the best route but have no other ideas.
If anyone can help, I would greatly appreciate it!
Please use parentheses. Do you mean
[tex] f(x,y) = \frac{1}{4}(1 + xy(x^2 - y^2)), [/tex]
or do you mean
[tex] f(x,y) = \frac{1}{4(1 + xy(x^2 - y^2))}?[/tex]
If you mean the first one, just write (1/4)(...) to make it clear; if you mean the second one, write 1/(4(...)).
ARLM
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#3
Dec6-12, 09:53 AM
P: 4
Yes, I mean the first one! Thank you.

Ray Vickson
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#4
Dec6-12, 09:58 AM
HW Helper
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P: 4,664

Moment Generating Function - Integration Help


Quote Quote by ARLM View Post
Yes, I mean the first one! Thank you.
OK. So given the conditions on |x| and |y|, what are the ranges of x and y themselves?

Since you have a 2-variable integration, you can integrate first over y (for any fixed x), then integrate the result over x; or you can do it in the other order. Is that what you have tried to do?

You really need to supply more details, since we have no way of helping if we do not know where you are stuck.
ARLM
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#5
Dec6-12, 10:38 AM
P: 4
I set up the integral
[itex]\frac{1}{4}[/itex][itex]\int[/itex][itex]\int[/itex]e[itex]^{sx+ty}[/itex](1+xy(x[itex]^{2}[/itex]+y[itex]^{2}[/itex]))dx dy

Both integrals are from -1 to 1 (I don't know how to show that)

I tried distributing xy and using integration by parts with u = 1+(x^3)y +x(y^3) and dv= e^(sx+ty) but not sure that this is the right idea because it would take numerous integrations before reducing to one function and then I would have to do it again for the second integral.

Other than that, I haven't tried anything.
pasmith
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#6
Dec6-12, 11:19 AM
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Quote Quote by ARLM View Post
I set up the integral
[itex]\frac{1}{4}[/itex][itex]\int[/itex][itex]\int[/itex]e[itex]^{sx+ty}[/itex](1+xy(x[itex]^{2}[/itex]+y[itex]^{2}[/itex]))dx dy

Both integrals are from -1 to 1 (I don't know how to show that)
The [itex]x^2 - y^2[/itex] in the original seems to have turned into [itex]x^2 + y^2[/itex].

I tried distributing xy and using integration by parts with u = 1+(x^3)y +x(y^3) and dv= e^(sx+ty) but not sure that this is the right idea because it would take numerous integrations before reducing to one function and then I would have to do it again for the second integral.

Other than that, I haven't tried anything.
Use the fact that [itex]e^{sx+ty} = e^{sx}e^{ty}[/itex], and that [itex]e^{sx}[/itex] does not depend on y to get
[tex]
\frac14 \int_{-1}^1 e^{sx}\left(\int_{-1}^1 e^{ty}(1 + xy(x^2 - y^2))\,\mathrm{d}y\right)\,\mathrm{d}x
[/tex]

Start by working out [itex]\int_{-1}^1 xe^{kx}dx[/itex], [itex]\int_{-1}^1 x^2e^{kx}dx[/itex], and [itex]\int_{-1}^1 x^3e^{kx}dx[/itex]. You're going to need these repeatedly, so you may as well do them once and then write in the values as needed.

You may also want to note that [itex]e^k - e^{-k} = 2\sinh k[/itex] and that [itex]e^k + e^{-k} = 2\cosh k[/itex]
ARLM
ARLM is offline
#7
Dec6-12, 01:40 PM
P: 4
Ok, thanks!


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