document.write( "Question 1192049: Given f(x, y) = e^(-x-y), for x>0 and y>0
\n" ); document.write( "(a) Find the marginal density of X
\n" ); document.write( "(b) Find the conditional density of Y given X = x
\n" ); document.write( "(c) Find the first joint moment of X and Y\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "
\n" ); document.write( "

Algebra.Com's Answer #849040 by CPhill(1959)\"\" \"About 
You can put this solution on YOUR website!
Certainly, let's break down the problem step-by-step.\r
\n" ); document.write( "\n" ); document.write( "**Given:**\r
\n" ); document.write( "\n" ); document.write( "* Joint probability density function: f(x, y) = e^(-x-y) for x > 0 and y > 0 \r
\n" ); document.write( "\n" ); document.write( "**a) Find the marginal density of X:**\r
\n" ); document.write( "\n" ); document.write( "* To find the marginal density of X, we integrate the joint density function with respect to y over its entire range:\r
\n" ); document.write( "\n" ); document.write( " f_X(x) = ∫[from 0 to ∞] f(x, y) dy
\n" ); document.write( " = ∫[from 0 to ∞] e^(-x-y) dy
\n" ); document.write( " = e^(-x) ∫[from 0 to ∞] e^(-y) dy
\n" ); document.write( " = e^(-x) * [-e^(-y)] [from 0 to ∞]
\n" ); document.write( " = e^(-x) * (0 - (-1))
\n" ); document.write( " = e^(-x) \r
\n" ); document.write( "\n" ); document.write( " Therefore, the marginal density of X is:
\n" ); document.write( " f_X(x) = e^(-x) for x > 0 \r
\n" ); document.write( "\n" ); document.write( "**b) Find the conditional density of Y given X = x:**\r
\n" ); document.write( "\n" ); document.write( "* The conditional density of Y given X = x is defined as:\r
\n" ); document.write( "\n" ); document.write( " f_Y|X(y|x) = f(x, y) / f_X(x)\r
\n" ); document.write( "\n" ); document.write( "* Substituting the given values:\r
\n" ); document.write( "\n" ); document.write( " f_Y|X(y|x) = e^(-x-y) / e^(-x)
\n" ); document.write( " = e^(-y) \r
\n" ); document.write( "\n" ); document.write( " Therefore, the conditional density of Y given X = x is:
\n" ); document.write( " f_Y|X(y|x) = e^(-y) for y > 0\r
\n" ); document.write( "\n" ); document.write( "**c) Find the first joint moment of X and Y:**\r
\n" ); document.write( "\n" ); document.write( "* The first joint moment of X and Y, E[XY], is given by:\r
\n" ); document.write( "\n" ); document.write( " E[XY] = ∫∫[over the support of (X, Y)] x * y * f(x, y) dx dy\r
\n" ); document.write( "\n" ); document.write( "* In this case, the support of (X, Y) is x > 0 and y > 0:\r
\n" ); document.write( "\n" ); document.write( " E[XY] = ∫[from 0 to ∞] ∫[from 0 to ∞] x * y * e^(-x-y) dy dx\r
\n" ); document.write( "\n" ); document.write( "* We can solve this double integral using integration by parts. However, a quicker approach is to recognize that the joint density function is the product of the marginal densities of X and Y:\r
\n" ); document.write( "\n" ); document.write( " f(x, y) = e^(-x-y) = e^(-x) * e^(-y) = f_X(x) * f_Y(y)\r
\n" ); document.write( "\n" ); document.write( "* This indicates that X and Y are independent random variables. \r
\n" ); document.write( "\n" ); document.write( "* For independent random variables, E[XY] = E[X] * E[Y]\r
\n" ); document.write( "\n" ); document.write( "* We know that X and Y both follow an exponential distribution with parameter 1. The expected value of an exponential distribution with parameter λ is 1/λ.\r
\n" ); document.write( "\n" ); document.write( "* Therefore,
\n" ); document.write( " E[X] = 1/1 = 1
\n" ); document.write( " E[Y] = 1/1 = 1\r
\n" ); document.write( "\n" ); document.write( "* Hence, the first joint moment of X and Y is:
\n" ); document.write( " E[XY] = E[X] * E[Y] = 1 * 1 = 1\r
\n" ); document.write( "\n" ); document.write( "**In summary:**\r
\n" ); document.write( "\n" ); document.write( "* Marginal density of X: f_X(x) = e^(-x) for x > 0
\n" ); document.write( "* Conditional density of Y given X = x: f_Y|X(y|x) = e^(-y) for y > 0
\n" ); document.write( "* First joint moment of X and Y: E[XY] = 1\r
\n" ); document.write( "\n" ); document.write( "I hope this comprehensive explanation is helpful!
\n" ); document.write( "
\n" );