Question

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?

Answer 1

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.

Answer 2

Could you repeat that calculation and see if 2^4 is 8 or 16?

Answer 3

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.