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Expert: - 4/29/2009


Question
QUESTION: Sir:

Good day!

thanks for your reply. you need to see more of the problem to solve it. here's the problem:

The joint density p of (x,z) is given by p(x,z)=exp(-x-z), x>0, z>0. With the transformation u=(x+z)/2, w=(x-z) the density q of (u,w) reads q(u,w)=exp(-2u), u> absolute value of w over 2. the conditional densities q(u)=4u exp(-2u), u>0.

How did it arrive to the density of q(u,w) and the conditional densities of q(u)?

Thanks...

jed

ANSWER: p(x,z) = exp(-x-z), x>0, z>0.

Let u=(x+z)/2, w=(x-z).

q(u,w)=exp(-2u), u>|w/2|.

Yes, I see how they got this far.

Conditional density q(u)=4u exp(-2u), u>0?

We read that u>|w/2|, 2u>|w|, so that we have -2u < w < 2u.

The conditional desnsity of q(u) would be
⌠2u
⌡-2u exp(-2u) dw,
= exp(-2u)w (from -2u up to 2u)
= exp(-2u)(2u+2u)
= 4u exp(-2u).

In case your curious, this is how to compute conditional densitities:
http://www-stat.stanford.edu/~susan/courses/s116/node76.html



---------- FOLLOW-UP ----------

QUESTION: good day sir!

i go straight to the point. What is the integral of (1+ square root of 2t)/2 times (exp(-square root of 2t)/square root of 2t) evaluated from c to positive infinity. hope you can give me a detailed solution. thanx!!!

Answer
Did I forget the attatchment to the answer?
Well, I can't attach it now, so here it is in a bad format.
The top row has the labels and below are the numbers.

t      2*t-1      e^-t      (2t-1)e^-t
1      1      0.367879441      0.367879441
2      3      0.135335283      0.40600585
4      7      0.018315639      0.128209472
8      15      0.000335463      0.005031939
16      31      1.12535E-07      3.48859E-06
32      63      1.26642E-14      7.97842E-13
64      127      1.60381E-28      2.03684E-26
128      255      2.57221E-56      6.55913E-54
256      511      6.6163E-112      3.3809E-109
512      1023      4.3775E-223      4.4782E-220
1024      2047      0          0

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