Joint distribution doubts

[maahi]

1) given joint pmf
X/Y     0      1       2        3    pX(:)
0       1/16  1/16    0         0     2/16
1       1/16  3/16   2/16     0     6/16
2        0      2/16   3/16   1/16  6/16
3        0       0       1/16   1/16 2/16
pY (:) 2/16  6/16   6/16   2/16   1

how to find joint cdf?

2) The cumulative distribution function for the joint distribution of the contin-
uous random variables X and Y is FXY (x; y) = 0.2(3x^3y + 2x^2y^2); x belongs to [0,1]
y belongs [0,1]. Find fXY (1/2,1/2).

3) how do we solve double integrals ?

4 )Let X and Y be random variables with joint pdf
fXY (x; y) =1/4   , -1<= x , y <= 1
0 Otherwise
Determine
(a) P(X^2 + Y ^2 < 1);
(b) P(2X - Y > 0);
(c) P(mod X + mod Y  < 2):

[Sinistral]

1) you want cdf of marginal densities or joint densities??
  In the former case:
  F_x(x)=P_x(X<=x)
  F_x(x)= 0; x<0,
           = 2/16, 0<=x<1
           = 8/16 , 1<=x<2
           =14/16 , 2<=x<3
           = 1  , x>=3
  Similarly find for F_y(y)
for joint cdf :  Σ(x=0 to x)  Σ(y=0 to y) p(x,y) vary x and y accordingly. Please see we usually do not find joint cdfs for joint pmfs.

2) (∂ square F)/∂x∂y

3) read from any standard text book on calculus or just google search.

4) [Edited] plot the domain of f(x,y). For eg here it will be a rectangle with left hand vertical side at x=-1. right hand vertical side at x=  1. top horizontal side at y=1. bottom horizontal side at y = -1. [Edited]

now plot the required functions (whose probability needs to be found). see what portion of this graph comes inside the above rectangle. integrate f(x,y) for this overlapping region.
the good thing is f(x,y) is a constant so u do not need to actually integrate. just find the area of the overlapping region geometrically and multiply it by f(x,y) ie 1/4

eg P(X^2 +Y^2 <1). here X^2 +Y^2 <1 is the inside portion of a circle with centre at (0,0) and radius 1. see that this circle lies entirely within our (domain) rectangle. so basically we need to find the area of this circle and multiply it by 1/4. Area of circle =  π (since r=1). so  P(X^2 +Y^2 <1) = π/4   , π=pi

[maahi]

 thank u
4 ) can u tell   d ans for part  b ?

[Sinistral]

oh..i made a mistake in my above solution to ques no 4. I misinterpreted the domain of f(x,y).

please check the edited version above. correction is only wrt domain of f(x,y)

despite that goof up, the explanation of first part remains the same as the circle will still be perfectly inside our (corrected) domain.

as far as 2nd part (b) is concerned:
ans is 1/2

[maahi]

 right hand vertical side at x=  1 why ??? i thought d earlier one was right


 

[Sinistral]

Actually first I (mis)interpreted -1<= x , y <= 1  as x >=-1 & y <=1.
In reality however it meant -1<= x <=1  &  -1<= y <=1

when u asked the answer of part b then it struck to me that Integration of f(x,y) in its entire domain (incorrect domain) wasn't coming out to be 1. Hence I rectified it. :)

apologies for this goof up.  

[maahi]

sorry for bugging you but

 x > = -1 doesn't imply that x can take values from -1 to infinity ?//

[Sinistral]

but it is also bounded above.
-1<x<1
-1<y<1

[duck]

Hi Maahi.. :)

That pdf makes sense only when domain is -1≤x≤y≤1 , otherwise f(x,y) cannot serve as a pdf because it'll not integrate out to 1 (as pointed out by Sinistral).

So, work out the question taking the above domain.

[maahi]

how do we find this domain ??

[Sinistral]

the give pdf's domain is given to be -1<=x<=1 & -1<=y<=1. we dont need to find nething. it's given in the question.

I dont have any clue how Duck inferred that domain shud be -1≤x≤y≤1 because if we consider this domain the total integration of f(x,y) in its entire domain wont come out to be 1.

[maahi]

what part of d ques makes u conclude that ?

[Sinistral]

see ur quoted text below. the bold part clearly mentions the domain of f(x,y) where it is not equal to 0.

maahi wrote

1) given joint pmf
X/Y     0      1       2        3    pX(:)
0       1/16  1/16    0         0     2/16
1       1/16  3/16   2/16     0     6/16
2        0      2/16   3/16   1/16  6/16
3        0       0       1/16   1/16 2/16
pY (:) 2/16  6/16   6/16   2/16   1

how to find joint cdf?

2) The cumulative distribution function for the joint distribution of the contin-
uous random variables X and Y is FXY (x; y) = 0.2(3x^3y + 2x^2y^2); x belongs to [0,1]
y belongs [0,1]. Find fXY (1/2,1/2).

3) how do we solve double integrals ?

4 )Let X and Y be random variables with joint pdf
fXY (x; y) =1/4   , -1<= x , y <= 1 
0 Otherwise
Determine
(a) P(X^2 + Y ^2 < 1);
(b) P(2X - Y > 0);
(c) P(mod X + mod Y  < 2):