Question 1192415: Given f(x, y) = e^(-x-y), for x>0 and y>0
(a) Find the marginal density of X
(b) Find the conditional density of Y given X = x
(c) Find the first joint moment of X and Y
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! **a) Find the marginal density of X**
* The marginal density of X, denoted by fX(x), is found by integrating the joint density function f(x, y) with respect to y:
fX(x) = ∫-∞∞ f(x, y) dy
Since f(x, y) = e-x-y for x > 0 and y > 0, and 0 otherwise, we have:
fX(x) = ∫0∞ e-x-y dy
= e-x ∫0∞ e-y dy
= e-x [-e-y]0∞
= e-x (0 - (-1))
= e-x for x > 0
and 0 otherwise.
**b) Find the conditional density of Y given X = x**
* The conditional density of Y given X = x, denoted by fY|X(y|x), is given by:
fY|X(y|x) = f(x, y) / fX(x)
Since we found fX(x) = e-x, we have:
fY|X(y|x) = (e-x-y) / (e-x)
= e-y for y > 0
and 0 otherwise.
**c) Find the first joint moment of X and Y**
* The first joint moment of X and Y, denoted by E[XY], is given by:
E[XY] = ∫-∞∞ ∫-∞∞ x*y*f(x, y) dx dy
= ∫0∞ ∫0∞ x*y*e-x-y dx dy
= ∫0∞ x*e-x [∫0∞ y*e-y dy] dx
* Note that ∫0∞ y*e-y dy is the expected value of an exponential random variable with parameter 1, which is equal to 1.
Therefore,
E[XY] = ∫0∞ x*e-x dx
* This integral also represents the expected value of an exponential random variable with parameter 1, which is equal to 1.
**So, the first joint moment of X and Y is E[XY] = 1.**
**In summary:**
* The marginal density of X is fX(x) = e-x for x > 0.
* The conditional density of Y given X = x is fY|X(y|x) = e-y for y > 0.
* The first joint moment of X and Y is E[XY] = 1.
|
|
|
