Random variables X and Y are distributed according to the joint PDF

fX,Y(x,y) = {ax,0,if 1≤x≤2 and 0≤y≤x,otherwise.

1 Find the constant a.
a=

2 Determine the marginal PDF fY(y). (Your answer can be either numerical or algebraic functions of y).
For 0≤y≤1,

fY(y)=

For 1<y≤2,

fY(y)=

3 Determine the conditional expectation of 1/X given that Y=3/2.

E[1/X∣Y=3/2]=

1. a= 0.4286

2.
For 0≤y≤1, fY(y)= 0.6429

For 1<y≤2, fY(y)= 0.2142*(4-y^2)

3. E[1/X∣Y=3/2]= 0.5714

1 To find the constant a, we can integrate the joint PDF over its entire support:

∫∫fX,Y(x,y) dxdy = 1

Let's integrate over the given limits:

∫∫ax dxdy = 1

∫[1,2]∫[0,x] ax dydx = 1

Using the limits of integration, the equation becomes:

a∫[1,2] (x∫[0,x] dy) dx = 1

a∫[1,2] (xy|0,x) dx = 1

Now, let's evaluate the inner integral:

a∫[1,2] (x(0-x)) dx = 1

a∫[1,2] (-x^2) dx = 1

Using the limits of integration, the equation becomes:

a (-∫[1,2] x^2 dx) = 1

a ([-(1/3)x^3] |[1,2]) = 1

a ((-8/3) - (-1/3)) = 1

a (-8/3 + 1/3) = 1

a (-7/3) = 1

a = -3/7

Therefore, a = -3/7.

2 To determine the marginal PDF fY(y), we need to integrate the joint PDF with respect to x:

fY(y) = ∫fX,Y(x,y) dx

Let's calculate the marginal PDF for each range of y:

For 0≤y≤1:
fY(y) = ∫[1,2] ax dx
= [-3/7 * (1/2)x^2] |[1,2]
= (-3/7 * (1/2) * 2^2) - (-3/7 * (1/2) * 1^2)
= (-3/7 * 2) - (-3/7 * 1)
= (-6/7) - (-3/7)
= -6/7 + 3/7
= -3/7

For 1<y≤2:
fY(y) = ∫[y,2] ax dx
= [-3/7 * (1/2)x^2] |[y,2]
= (-3/7 * (1/2) * 2^2) - (-3/7 * (1/2) * y^2)
= (-3/7 * 2) - (-3/7 * (1/2) * y^2)
= (-6/7) - (-3/7 * (1/2) * y^2)
= (-6/7) - (-3/14 * y^2)
= (-12/14) + (3/14 * y^2)
= (-6/7) + (3/14 * y^2)

Therefore, for 0≤y≤1, fY(y) = -3/7, and for 1<y≤2, fY(y) = (-6/7) + (3/14 * y^2).

3 To determine the conditional expectation of 1/X given that Y = 3/2, we need to find E[1/X|Y=3/2]. The conditional expectation is given by:

E[1/X|Y=3/2] = ∫(1/X) * fX|Y(x|3/2) dx

To find fX|Y(x|3/2), we need to normalize the joint PDF given the condition Y = 3/2:

fX|Y(x|3/2) = fX,Y(x,3/2) / fY(3/2)

First, let's find fY(3/2) using the marginal PDF we obtained earlier:

fY(3/2) = (-6/7) + (3/14 * (3/2)^2)
= (-6/7) + (3/14 * (9/4))
= (-6/7) + (27/56)
= (-48/56) + (27/56)
= -21/56

Now, we can find fX|Y(x|3/2):

fX|Y(x|3/2) = fX,Y(x,3/2) / fY(3/2)
= (-(3/7)x) / (-21/56)
= (56/21) * (3/7)x
= (8/3)x

Finally, let's calculate the conditional expectation by integrating (1/X) * fX|Y(x|3/2):

E[1/X|Y=3/2] = ∫(1/X) * (8/3)x dx
= (8/3)∫(1/X)x dx
= (8/3) * log(x) |[1,2]
= (8/3) * (log(2) - log(1))
= (8/3) * log(2)

Therefore, E[1/X|Y=3/2] = (8/3) * log(2)

1. To find the constant a, we need to calculate the integral of the joint PDF over its entire support. In this case, the support is defined by 1≤x≤2 and 0≤y≤x.

∫∫fX,Y(x, y) dx dy = ∫∫ax dx dy

Integration limits:
∫(1 to 2) ∫(0 to x) a x dy dx

Performing the integrations:
∫(1 to 2) [a x y] (0 to x) dx
= ∫(1 to 2) a x^2 dx
= [a/3 x^3] (1 to 2)
= (8a/3 - a/3)
= 7a/3

Since the integral of the joint PDF over its entire support should be equal to 1, we have:
7a/3 = 1

Solving for a:
7a = 3
a = 3/7

Therefore, the constant a is 3/7.

2. To determine the marginal PDF fY(y), we need to integrate the joint PDF over x while considering the different ranges of y.

For 0≤y≤1:
∫(y to 2) fX,Y(x, y) dx
= ∫(y to 2) (3/7)x dx
= (3/7) [x^2/2] (y to 2)
= (3/7) [(2^2/2) - (y^2/2)]
= (3/7) [2 - (y^2/2)]
= 6/7 - (3/14)y^2

Therefore, for 0≤y≤1, the marginal PDF fY(y) is 6/7 - (3/14)y^2.

For 1<y≤2:
∫(1 to 2) fX,Y(x, y) dx
= ∫(1 to 2) (3/7)x dx
= (3/7) [x^2/2] (1 to 2)
= (3/7) [(2^2/2) - (1^2/2)]
= (3/7) [2 - 1/2]
= 3/7 - 3/14
= (9/14)

Therefore, for 1<y≤2, the marginal PDF fY(y) is 9/14.

3. The conditional expectation of 1/X given that Y=3/2 can be calculated using the conditional PDF, which is obtained by dividing the joint PDF by the marginal PDF of Y.

The conditional PDF fX|Y(x|3/2) is given by:
fX|Y(x|3/2) = fX,Y(x, 3/2) / fY(3/2)

Using the provided joint PDF and marginal PDF values, we have:
fX|Y(x|3/2) = (3/7)x / (9/14)

Simplifying, we get:
fX|Y(x|3/2) = (2/7)x

The conditional expectation E[1/X∣Y=3/2] is calculated as:
E[1/X∣Y=3/2] = ∫(1 to 2) (1/x) fX|Y(x|3/2) dx

E[1/X∣Y=3/2] = ∫(1 to 2) (1/x)(2/7)x dx

E[1/X∣Y=3/2] = (2/7) ∫(1 to 2) dx

E[1/X∣Y=3/2] = (2/7)[x] (1 to 2)

E[1/X∣Y=3/2] = (2/7)[2 - 1]

E[1/X∣Y=3/2] = 2/7

Therefore, the conditional expectation of 1/X given that Y=3/2 is 2/7.