The random variables X and Y are distributed according to the joint PDF:
It has the value of a*x^2 if 1 <= x <= 2 and 0 <= y <= x, and 0 otherwise. What's the value of the constant a?
To find the value of the constant a, we need to use the condition that the joint probability density function integrates to one over the entire domain of X and Y.
The domain of X is from 1 to 2, and the domain of Y is from 0 to X. Therefore, we can write:
∫∫ f(x,y) dx dy = 1, where the integral is taken over the region D in the xy-plane defined by D={(x,y): 1<=x<=2, 0<=y<=x}
Substituting the given PDF, we get:
∫∫ a*x^2 dx dy = 1
Integrating with respect to y first, we get:
∫1^x a*x^2 dy = a*x^3/3
Substituting the limits of y, we get:
∫∫ a*x^2 dx dy = ∫1^2 (a*x^3/3) dx
Integrating with respect to x, we get:
∫1^2 (a*x^3/3) dx = a/3 * [(2^4)/4 - (1^4)/4] = a/3 * (8/4 - 1/4) = a/3 * 7/4
Therefore, we have:
a/3 * 7/4 = 1
Multiplying both sides by 12/7, we get:
a = 36/7
Hence, the value of the constant a is 36/7.
Could you repeat that calculation and see if 2^4 is 8 or 16?
You are correct - 2^4 is 16, not 8. Therefore, the correct calculation is:
∫1^2 (a*x^3/3) dx = a/3 * [(2^4)/4 - (1^4)/4] = a/3 * (16/4 - 1/4) = a/3 * 15/4
Therefore, we have:
a/3 * 15/4 = 1
Multiplying both sides by 12/15, we get:
a = 4/5
Hence, the value of the constant a is 4/5.
To determine the value of the constant a, we need to integrate the joint PDF over its entire domain and set the resulting value equal to 1, since the total probability over the domain of a PDF should be equal to 1.
First, let's find the bounds for the integration:
- For x, the bounds are from 1 to 2 (as given: 1 <= x <= 2).
- For y, the bounds are from 0 to x (as given: 0 <= y <= x).
The joint PDF is given by f(x,y) = a*x^2 if 1 <= x <= 2 and 0 <= y <= x, and 0 otherwise.
Let's set up the integral:
∫∫[R^2] f(x,y) dA = 1,
where [R^2] represents the region of integration.
Now, integrating the joint PDF over the given domain:
∫[1,2] ∫[0,x] a*x^2 dy dx = 1.
Now, let's evaluate the double integral:
∫[1,2] [ay*x^2]dy dx
= a∫[1,2] [xy^2]dy
= a[∫[1,2] [(1/3)*(x*y^3)] from y=0 to y=x
= a[∫[1,2] [(1/3)*(x*x^3)] - [(1/3)*(x*0^3)] dx
= a[∫[1,2] [(1/3)*(x^4)] dx
= a[(1/3)*[∫[1,2] x^4 dx
= a[(1/3)*[(x^5)/5] from x=1 to x=2
= a[(1/3)*[((2^5)/5) - ((1^5)/5)]
= a[(1/3)*[(32/5) - (1/5)]
= a[(1/3)*[(31/5)]
= a * (31/15).
To find the value of a, we set the result equal to 1:
a * (31/15) = 1.
Now, solving for a:
a = (15/31).
Therefore, the value of the constant a is (15/31).