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April 22nd, 2012, 04:19 AM   #1
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Joint Distribution help please

Can't get my head around this joint distribution problem.

Suppose X and Y have joint distribution given by:



Find the distribution of X+Y

Having difficulties finding the boundaries for my integration. I know that I need do declare a dummy variable t = X + Y
Been stuck on this for a few hours now, help is much appreciated! Thanks.
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April 22nd, 2012, 01:50 PM   #2
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Re: Joint Distribution help please

Quote:
Originally Posted by batman350z
Can't get my head around this joint distribution problem.

Suppose X and Y have joint distribution given by:



Find the distribution of X+Y

Having difficulties finding the boundaries for my integration. I know that I need do declare a dummy variable t = X + Y
Been stuck on this for a few hours now, help is much appreciated! Thanks.
f(x,y) cannot be either a density function or a distribution function without further limits on x and y. y can be negative giving negative values for f for y < -x. x can be arbitrarily large - distribution would eventually be > 1.
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April 22nd, 2012, 02:02 PM   #3
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Re: Joint Distribution help please

Hmm, I might be misunderstanding the boundaries. The question actually has them written as,

, so I am guessing that would be interpreted as

If the above is the case, how would i go about it?
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April 23rd, 2012, 03:50 PM   #4
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Re: Joint Distribution help please

First f(x,y) is a density function, not a distribution function.

Let T = X + Y. What you want to calculate is P(T < t) = P(X+Y < t) = P(X < t and Y < t - X).

To do this simply integrate the density function over the domain specified, 0 < x < t, 0 < y < t - x. The main thing you have to be careful of is the fact that f(x,y) = 0 outside the unit square (in x,y), so that you have to break up the integration into the parts where f(x,y) = x + y and f(x,y) = 0. For 0 < t < 1, there is no problem, but for 1 < t < 2 you need to do the breakup. For t > 2, the probability = 1.

Post script: for t > 1, it is easier to work with P(T > t) and the use P(T < t) = 1 - P(T > t).
In this case P(T > t) = P(X > t-1 and Y > t - X) so the integrals of f(x,y) is over t-1 < x < 1 and t-x < y < 1.

For t < 1 the integral to get P(T < t) directly has 0 < x < t and 0 < y < t-x.
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