Integration of random variables

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SUMMARY

The discussion focuses on the calculation of the conditional expectation E[Y|X] for the joint probability density function f(x,y) = (4/5)(x+3y)exp(-x-2y) for x, y > 0. The user correctly identifies that E[Y|X] can be computed using the formula E[Y|X] = integral y * f_xy(x,y) / f_x(x) dy. The marginal density f_x(x) is derived as (2x+3)/(5exp(x)). The user clarifies that the outcome of the integral should indeed be a function of x, confirming the dependency of the conditional expectation on the variable X.

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Homework Statement



f(x,y)= (4/5)(x+3y)exp(-x-2y) for x,y, >0

Find E[Y|X]

Homework Equations



E[Y|X] =integral y *f_xy (x,y)/ f_x (x) dy

The Attempt at a Solution



f_x (x) = integral [o,∞] [4/5](x+3y)exp(-x-2y) dx = (2x+3)/(5exp(x))

When taking the integral of y[(4/5)(x+3y)exp(-x-2y)] / [(2x+3)/(5exp(x)) ] dy dx for [0,∞] for y (x+3)/(2x+3) is that correct?
 
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well, if it's conditional, and it's dependent on x, shouldn't your outcome be a function of x?
 
Oh my bad, I just got mixed up with definitions. Thanks! <3
 

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