The random variables X and Y are distributed according to the joint PDF:
The joint PDF has the value of a*x^2 if 1 <= x <= 2 and 0 <= y <= x, and 0 otherwise. What's the conditional expectation of 1/(X^2*Y) given that Y = 5/4
First, we need to find the marginal PDF of Y by integrating the joint PDF with respect to x over the range 1 <= x <= 2 and y <= x:
f_Y(y) = ∫ f_XY(x,y) dx = ∫ a*x^2 dx over the range y <= x <= 2
= a/3 * (2^3 - y^3) for 0 <= y <= 2
= 0 otherwise
Next, we can use the definition of conditional probability to find the conditional PDF of X given Y = 5/4:
f_X|Y(x|y) = f_XY(x,y) / f_Y(y) for y = 5/4
= (a*x^2) / (a/3 * (2^3 - (5/4)^3))
= 12/(31*(8-y^3)) * x^2 for 1 <= x <= 2
Finally, we can use the definition of conditional expectation to find E[1/(X^2*Y) | Y = 5/4]:
E[1/(X^2*Y) | Y = 5/4]
= ∫ (1/(x^2*y)) * f_X|Y(x|y) dx over the range 1 <= x <= 2
= ∫ (1/(x^2*(5/4))) * (12/(31*(8-(5/4)^3))) * x^2 dx over the range 1 <= x <= 2
= (12/(31*(8-(5/4)^3))) * ∫ (1/(x^(3/2))) dx over the range 1 <= x <= 2
= 8/(31*(8-(5/4)^3)) * (2^(1/2) - 1)
Therefore, the conditional expectation of 1/(X^2*Y) given that Y = 5/4 is 8/(31*(8-(5/4)^3)) * (2^(1/2) - 1).
To find the conditional expectation of 1/(X^2 * Y) given that Y = 5/4, we need to calculate the conditional expectation using the conditional probability density function (PDF).
First, let's find the conditional PDF f(x|y) by dividing the joint PDF by the marginal PDF of Y.
The marginal PDF of Y can be found by integrating the joint PDF over the range of X:
f_Y(y) = ∫[1 to 2] (a * x^2) dx = a * [x^3 / 3] | [1 to 2] = a * (2^3 / 3 - 1^3 / 3) = a * (8/3 - 1/3) = a * (7/3)
Next, let's calculate the conditional PDF f(x|y):
f(x|y) = f(x, y) / f_Y(y) = (a * x^2) / (a * (7/3)) = (3/7) * x^2
Given that Y = 5/4, we can substitute this value into f(x|y):
f(x|Y=5/4) = (3/7) * x^2
Now, we can find the conditional expectation E[1/(X^2 * Y)|Y = 5/4] by integrating the conditional PDF multiplied by the function 1/(x^2 * y) with respect to x over the range from 1 to 2:
E[1/(X^2 * Y)|Y = 5/4] = ∫[1 to 2] (3/7) * x^2 * (1 / (x^2 * (5/4))) dx
Simplifying, we get:
E[1/(X^2 * Y)|Y = 5/4] = (3/7) * (1 / (5/4)) * ∫[1 to 2] dx
E[1/(X^2 * Y)|Y = 5/4] = (3/7) * (4/5) * (2 - 1)
E[1/(X^2 * Y)|Y = 5/4] = (12/35)
So, the conditional expectation of 1/(X^2 * Y) given that Y = 5/4 is 12/35.