a Determine the marginal pdfs of X and Y b Are X and Y indep

a. Determine the marginal p.d.f.\'s of X and Y.

b. Are X and Y independent??

Solution

a) Given that the joint p.d.f. of X and Y is as follows,

f(x,y) = 2xe^-y   for 0 x 1 and 0 < y < infinity.

= 0 otherwise

We have to prove that whether X and Y are independent.

This we can prove by using independent definition of events.

The events X and Y are independent iff,

P(X,Y) = P(X)*P(Y)

Here we have given joint p.d.f. of X and Y i.e. f(x,y) so the definition is ,

The marginal p.d.f. x and y are independent iff,

f(x,y) = f(x)*f(y)

First we find marginal p.d.f. of X and Y.

Marginal p.d.f. of X :

f(x) = f(x,y) dy (y is from 0 to infinity)

=   2xe^-y dy ( y is from 0 to infinity)

= 2x    e^-y dy ( y is from 0 to infinity)

= 2x [e^-y /-1] ( y is from 0 to infinity)

= 2x [ (e^-infinity / -1)-(e^-0 / -1)]

= 2x [ (0 / -1) - (1 / -1) ]

= 2x(1)

f(x) = 2x    0 x 1

Marginal p.d.f. of Y :

f(y) = f(x,y) dx (x is from 0 to 1)

=   2xe^-y dx ( x is from 0 to 1)

= 2e^-y  x dx ( x is from 0 to 1)

= 2e^-y  [x² / 2] ( y is from 0 to 1)

= 2e^-y  [ (1²/2) - (0²/2) ]

= 2e^-y  [ (1/2) - 0]

= 2e^-y /2

f(y) = e^-y   for 0<y<infinity   

Consider, f(x,y) = 2xe^-y  

f(x)*f(y) = 2x * e^-y This is nothing but the joint p.d.f. of X and Y.

So the condition for independency is satisfied.

X and Y are independent.

Now we can prove second part of example.

The joint p.d.f. of X and Y is as follows,

f(x,y) = (3/2)*y² 0 x 2 and 0 y 1

Now first her we have to find marginal of X and Y.

Marginal of X:

f(x) =   f(x,y) dy   0 y 1

=    (3/2)*y² dy 0 y 1

= (3/2) y² dy 0 y 1

= (3/2) [y³/3] 0 y 1

=(3/2) [(1³/3) - (0³/3) ]

f(x) = 1/2   0 x 2

Marginal p.d.f. of Y :

f(y) = f(x,y) dx 0 y 2

= (3/2)*y²   1 dx 0 x 2

= (3/2)*y² [x] 0 x 2

= (3/2)*y² [ 2 - 0 ]    0 x 2

= 3/2 *y²*2

f(y) = 3y²    0 x 2

For checking independencywe have to check whether these marginals are independent or not.

Consider,

f(x,y) = (3/2)*y²    0 x 2

f(x)*f(y) = (1/2)*3y² This is also nothing but the joint p.d.f. of X and Y.

Next we have to find the probabilities:

P(X<1) = f(x)dx    0 x 1 because we have given that X is less than 1.

= (1/2) dx    0 x 1   

= (1/2) [x]    0 x 1   

= (1/2) [1-0]

P(X<1) = 1/2

P(Y 1/2) =   f(y) dy   1/2 x 1

= 3y² dy   1/2 x 1

= 3 y² dy   1/2 x 1

= 3 [ y³/3]   1/2 x 1

= 3 [((1/2)³ / 3) - (1³/3) ]

P(Y 1/2) = 7/8