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.
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. To find the constant a, we need to calculate the double integral of fX,Y(x,y) over the entire range of x and y, and set it equal to 1, since it represents the total probability of all possible outcomes.
∫∫fX,Y(x,y) dx dy = 1
∫∫(ax) dx dy = 1
To integrate with respect to x, we treat y as a constant since there is no dependence of x on y within the given range of x and y:
a∫[1 to 2] (x) dx ∫[0 to x] 1 dy = 1
a [x^2/2 ∣[1 to 2]] [y ∣[0 to x]] = 1
a [(2^2/2 - 1^2/2)] [x ∣[0 to 2]] = 1
a (3) (2) = 1
6a = 1
a = 1/6
Therefore, the constant a is 1/6.
2. To determine the marginal PDF fY(y), we integrate the joint PDF fX,Y(x,y) with respect to x, over the entire range of x, for each value of y.
For 0 ≤ y ≤ 1:
fY(y) = ∫[1 to y] (1/6)*(x) dx = (1/6) * (x^2/2 ∣[1 to y]) = (1/6) * (y^2/2 - 1/2)
fY(y) = (1/12) * (y^2 - 1)
For 1 < y ≤ 2:
fY(y) = ∫[1 to 2] (1/6)*(x) dx + ∫[2 to y] (1/6)*(x) dx
fY(y) = (1/6) * (x^2/2 ∣[1 to 2]) + (1/6) * (x^2/2 ∣[2 to y])
fY(y) = (1/6) * (2^2/2 - 1^2/2) + (1/6) * (y^2/2 - 2^2/2)
fY(y) = (1/12) * (5 + y^2 - 4)
fY(y) = (1/12) * (y^2 + 1)
Therefore, the marginal PDF fY(y) is:
- For 0 ≤ y ≤ 1: fY(y) = (1/12) * (y^2 - 1)
- For 1 < y ≤ 2: fY(y) = (1/12) * (y^2 + 1)
3. To determine the conditional expectation of 1/X given that Y = 3/2, denoted E[1/X | Y = 3/2], we need to calculate the conditional PDF fX | Y(x | y), and then calculate the conditional expectation using the formula:
E[1/X | Y = 3/2] = ∫(1/x) * fX | Y(x | 3/2) dx
To calculate the conditional PDF fX | Y(x | y), we use the formula:
fX | Y(x | y) = fX,Y(x, y) / fY(y)
= (1/6) * (x) / fY(y)
= (1/6) * (x) / [(1/12) * (y^2 - 1)] for 0 ≤ y ≤ 1
= (1/6) * (x) / [(1/12) * (y^2 + 1)] for 1 < y ≤ 2
Since Y = 3/2, we substitute y = 3/2 in the above formulas:
For 0 ≤ y ≤ 1:
fX | Y(x | 3/2) = (1/6) * (x) / [(1/12) * ((3/2)^2 - 1)]
fX | Y(x | 3/2) = 4/3x
For 1 < y ≤ 2:
fX | Y(x | 3/2) = (1/6) * (x) / [(1/12) * ((3/2)^2 + 1)]
fX | Y(x | 3/2) = 2/3x
Now, we can calculate the conditional expectation:
E[1/X | Y = 3/2] = ∫(1/x) * fX | Y(x | 3/2) dx
= ∫(1/x) * (4/3x) dx for 0 ≤ x ≤ 1
= 4/3 * ∫(1/x^2) dx for 0 ≤ x ≤ 1
= 4/3 * (-1/x) ∣[0 to 1]
= 4/3 * (-1/1 - (-1/0))
= 4/3 * (-1 + ∞)
As the integral diverges, E[1/X | Y = 3/2] is undefined.
To find the constant a, we need to solve the equation:
∫∫fX,Y(x,y) dx dy = 1
Since the joint PDF is only non-zero for 1 ≤ x ≤ 2 and 0 ≤ y ≤ x, we can rewrite the integral as follows:
∫∫ax dx dy = 1
Integrating with respect to x first, we have:
a∫[0 to x]x dx = 1
Simplifying the integral, we get:
a[x^2/2] [0 to x] = 1
a(x^2/2) = 1
Substituting the limits of integration, we get:
a(2^2/2) - a(1^2/2) = 1
2a - a/2 = 1
Multiplying through by 2 to eliminate the fraction:
4a - a = 2
3a = 2
a = 2/3
Therefore, the constant a is 2/3.
To find the marginal PDF fY(y), we need to integrate the joint PDF over all possible values of x:
fY(y) = ∫fX,Y(x, y) dx
We divide the calculation into two cases based on the value of y:
Case 1: For 0 ≤ y ≤ 1
In this case, the range of x that satisfies the condition 0 ≤ y ≤ x is y ≤ x ≤ 1. Therefore, the marginal PDF is:
fY(y) = ∫[y to 1] ax dx
= a/2[(1^2 - y^2)]
= a/2(1 - y^2)
Case 2: For 1 < y ≤ 2
In this case, the range of x that satisfies the condition 0 ≤ y ≤ x is y ≤ x ≤ 2. Therefore, the marginal PDF is:
fY(y) = ∫[y to 2] ax dx
= a/2[(2^2 - y^2)]
= a/2(4 - y^2)
So, the marginal PDF fY(y) is:
For 0 ≤ y ≤ 1: fY(y) = (2/3)(1 - y^2)
For 1 < y ≤ 2: fY(y) = (2/3)(4 - y^2)
Now, to determine the conditional expectation of 1/X given that Y = 3/2, we need to calculate the conditional PDF fX|Y(x|y) and use it to calculate the conditional expectation E[1/X|Y = 3/2]. However, since the joint PDF is not given for the specific value of Y = 3/2, we cannot directly calculate the conditional expectation in this case.
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