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).
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
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