1. The random variables X and Y are distributed according to the joint PDF
fX,Y(x,y) = {ax2,0,if 1≤x≤2 and 0≤y≤x,otherwise.
Find the constant a.
2. Determine the marginal PDF fY(y) .
(Your answer can be either numerical or an algebraic function of y ).
a. If 0≤y≤1 :
fY(y)=
b. If 1<y≤2 :
fY(y)=
1. 4/15
2.
a. 28/45
b. 4/45*(8-y^3)
3. 0.2976
Enjoy!
1. 4/15
2.
a. 28/45
b. 4/45*(8-y^3)
3. 0.2976
good day mate
1. To find the constant a, we need to integrate the joint PDF over its entire support, which is the region where 1≤x≤2 and 0≤y≤x.
∫∫fX,Y(x,y) dy dx = 1
∫∫ax^2 dy dx
Since the joint PDF is only non-zero when 1≤x≤2 and 0≤y≤x, we can rewrite the integral as:
∫1^2 ∫0^x ax^2 dy dx
Integrating the inner integral:
∫1^2 (axy^2) | from 0 to x) dx
∫1^2 ax^3 dx
Integrating the outer integral:
[a(x^4)/4] | from 1 to 2
Substituting the limits:
[16a/4] - [a/4] = a
Since this integral is equal to 1, we have:
[16a/4] - [a/4] = 1
15a/4 = 1
Dividing both sides by 15/4:
a = 4/15
2. To find the marginal PDF fY(y), we need to integrate the joint PDF over the range of x for each value of y:
a. If 0≤y≤1:
∫∫fX,Y(x,y) dx dy
= ∫0^1 ∫y^2^2 ax^2 dx dy
= a∫0^1 [x^3/3] | from y^2 to 2 dy
= a[(2^3/3) - (y^6/3)] | from 0 to 1
= a[(8/3) - (1/3)]
= a(7/3)
= (4/15)(7/3)
= 28/135
Therefore, fY(y) = 28/135 for 0≤y≤1.
b. If 1<y≤2:
∫∫fX,Y(x,y) dx dy
= ∫1^2 ∫y^2^2 ax^2 dx dy
= a∫1^2 [x^3/3] | from y^2 to 2 dy
= a[(2^3/3) - (y^6/3)] | from y^2 to 2
= a[(8/3) - (2^6/3)]
= a[(8/3) - (64/3)]
= a(-56/3)
= -(4/15)(56/3)
= -224/45
Therefore, fY(y) = -224/45 for 1<y≤2.
To find the constant a in the joint PDF function, we need to integrate the PDF function over its entire domain and set it equal to 1.
1. Finding the constant a:
∫∫fX,Y(x,y)dxdy = 1
Since the joint PDF is non-zero only when 1≤x≤2 and 0≤y≤x, we can write the integral as follows:
∫∫ax^2 dxdy = 1
The limits for y are from 0 to x, and the limits for x are from 1 to 2. Rearranging the order of integration:
∫(∫ax^2 dy) dx = 1
∫[ayx^2]dy = 1
ay∫(x^2) dy = 1
ay[x^2y] with limits from 0 to x = 1
ay[x^3 - 0] = 1
ax^4 = 1
To satisfy the condition for all x values, we need the integral to be equal to 1 for every x. So, in order for ax^4 = 1 for all x, a must be the reciprocal of the integral of x^4 from 1 to 2.
Therefore, a = 1/∫(x^4)dx from 1 to 2.
Now, let's find the marginal PDF fY(y) for the given joint PDF.
2. Marginal PDF fY(y):
To find the marginal PDF of Y, we integrate the joint PDF fX,Y(x,y) over the domain of x while treating y as a constant.
a. If 0≤y≤1:
fY(y) = ∫fX,Y(x,y) dx
Since the joint PDF is non-zero only when 1≤x≤2 and 0≤y≤x, the integral becomes:
∫ax^2 dx, with limits from y to 2.
Integrating ax^2 with respect to x:
a(x^3/3) evaluated from y to 2
The marginal PDF fY(y) for 0≤y≤1 is:
fY(y) = a(8/3 - y^3/3)
b. If 1<y≤2:
Similarly, the integral becomes:
∫ax^2 dx, with limits from 1 to y.
Integrating ax^2 with respect to x:
a(x^3/3) evaluated from 1 to y
The marginal PDF fY(y) for 1<y≤2 is:
fY(y) = a(y^3/3 - 1/3)
Thus we have:
a. If 0≤y≤1:
fY(y) = a(8/3 - y^3/3)
b. If 1<y≤2:
fY(y) = a(y^3/3 - 1/3)