CDF, and setting the integral for this

  • #1
nacho-man
171
0
Please refer to the attached image.

I can't quite get the bounds for question a) right. it's so confusing. Would it be wise to split the double integral into two parts? I guess that's usually favourable with when dealing with absolute values. But the bounds are still confusing me.

would it be easier to integrate with respect to x first, or w.r.t y?For x, it would just be 0 to infinity.

for y it would be -infinity to x ? whilst splitting the integral with respect to x? is this correct.
Actually i'll have a crack at it now. some as i make these question threads i tend to find a solution/or at least get closer to it.

Still, help/correction is very much appreciated.
Thanks!

edit: first attempt i did was
from and . this was wrong, as it left me with a variable in my answer, so i will try again but int w.r.t y first.

edit 2: This also did not workedit 3: SOLVED PART A)
i used the bounds -x<y<x and 0<x<infinity
and obtained c= 1/2

edit 4: SOLVED PART B)
I got fy|x(x,y) =
and fx|y(x,y) =

edit 5: I'm having trouble with part d

E[Y|X=e] =
 

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  • #2
nacho said:
Please refer to the attached image.

I can't quite get the bounds for question a) right. it's so confusing. Would it be wise to split the double integral into two parts? I guess that's usually favourable with when dealing with absolute values. But the bounds are still confusing me.

would it be easier to integrate with respect to x first, or w.r.t y?For x, it would just be 0 to infinity.

for y it would be -infinity to x ? whilst splitting the integral with respect to x? is this correct.
Actually i'll have a crack at it now. some as i make these question threads i tend to find a solution/or at least get closer to it.

Still, help/correction is very much appreciated.
Thanks!

edit: first attempt i did was
from and . this was wrong, as it left me with a variable in my answer, so i will try again but int w.r.t y first.

edit 2: This also did not workedit 3: SOLVED PART A)
i used the bounds -x<y<x and 0<x<infinity
and obtained c= 1/2

edit 4: SOLVED PART B)
I got fy|x(x,y) =
and fx|y(x,y) =

What is not full clear is that the p.d.f. is defined as...



... for and is 0 in the rest of the x,y plane. In that case the condition for c is...



The remaining part can be attacked with the procedure we already saw...

Kind regards

 
  • #3
chisigma said:
What is not full clear is that the p.d.f. is defined as...



... for and is 0 in the rest of the x,y plane. In that case the condition for c is...



The remaining part can be attacked with the procedure we already saw...

Kind regards


thanks, i had just forgotten to multiply by 2 during a step, leading to me getting 1/2.

can you assist witg part d?i keep getting 0, which i think is wrong
 
  • #4
nacho said:
thanks, i had just forgotten to multiply by 2 during a step, leading to me getting 1/2.

can you assist witg part d?i keep getting 0, which i think is wrong

Answering to point d) is performed in three steps...

...first we conpute the marginal distribution of X...



... second the conditional distribution function of Y...



... and third...



Kind regards

 
  • #5
chisigma said:
Answering to point d) is performed in three steps...

...first we conpute the marginal distribution of X...



... second the conditional distribution function of Y...



... and third...



Kind regards


ah, i was right then, gues i should be more confident
 
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