Stats - Conditional Expectation

[Ted123]

Homework Statement

[PLAIN]http://img222.imageshack.us/img222/2781/statsqk.jpg

Homework Equations

$f_{X} (x) = \int^{\infty}_{-\infty} f_{X,Y} (x,y)\;dy$

$f_{Y} (y) = \int^{\infty}_{-\infty} f_{X,Y} (x,y)\;dx$

$f_{X|Y} (x|y) = \frac{f_{X,Y} (x,y)}{f_Y (y)}$

$f_{Y|X} (y|x) = \frac{f_{X,Y} (x,y)}{f_X (x)}$

$\mathbb{E}\;[X|Y] = \int^{\infty}_{-\infty} x f_{X|Y} (x|y)\;dx$

$\mathbb{E}\;[Y|X] = \int^{\infty}_{-\infty} y f_{Y|X} (y|x)\;dy$

The Attempt at a Solution

Can anyone check that I've done this correctly (specifically check the limits)?

$\displaystyle f_X (x) = \int^{\infty}_x x(y-x)e^{-y}\;dy = xe^{-x} \;\; \text{for} \; x\in [0,y]$

$\displaystyle f_Y (y) = \int^y_0 x(y-x)e^{-y}\;dx = \frac{y^3}{6} e^{-y} \;\; \text{for} \; y\in [x,\infty )$

$\displaystyle f_{X|Y} (x|y) = \frac{x(y-x)e^{-y}}{\frac{y^3}{6} e^{-y}} = \frac{6x(y-x)}{y^3}\;\; \text{for} \; x\in [0,y]$

$\displaystyle f_{Y|X} (y|x) = \frac{x(y-x)e^{-y}}{xe^{-x}} = e^{x-y} (y-x)\;\; \text{for} \; y\in [x,\infty )$

$\displaystyle\mathbb{E}\;[X|Y]= \int^{\infty}_x ye^{x-y} (y-x)\;dy = x+2$

$\displaystyle\mathbb{E}\;[Y|X]=\int^{y}_0 \frac{6x^2 (y-x)}{y^3} \;dx = \frac{y}{2}$

 

[statdad]

Homework Statement

[PLAIN]http://img222.imageshack.us/img222/2781/statsqk.jpg

Homework Equations

$f_{X} (x) = \int^{\infty}_{-\infty} f_{X,Y} (x,y)\;dy$

$f_{Y} (y) = \int^{\infty}_{-\infty} f_{X,Y} (x,y)\;dx$

$f_{X|Y} (x|y) = \frac{f_{X,Y} (x,y)}{f_Y (y)}$

$f_{Y|X} (y|x) = \frac{f_{X,Y} (x,y)}{f_X (x)}$

$\mathbb{E}\;[X|Y] = \int^{\infty}_{-\infty} x f_{X|Y} (x|y)\;dx$

$\mathbb{E}\;[Y|X] = \int^{\infty}_{-\infty} y f_{Y|X} (y|x)\;dy$

The Attempt at a Solution

Can anyone check that I've done this correctly (specifically check the limits)?

$\displaystyle f_X (x) = \int^{\infty}_x x(y-x)e^{-y}\;dy = xe^{-x} \;\; \text{for} \; x\in [0,y]$

$\displaystyle f_Y (y) = \int^y_0 x(y-x)e^{-y}\;dx = \frac{y^3}{6} e^{-y} \;\; \text{for} \; y\in [x,\infty )$

$\displaystyle f_{X|Y} (x|y) = \frac{x(y-x)e^{-y}}{\frac{y^3}{6} e^{-y}} = \frac{6x(y-x)}{y^3}\;\; \text{for} \; x\in [0,y]$

$\displaystyle f_{Y|X} (y|x) = \frac{x(y-x)e^{-y}}{xe^{-x}} = e^{x-y} (y-x)\;\; \text{for} \; y\in [x,\infty )$

$\displaystyle\mathbb{E}\;[X|Y]= \int^{\infty}_x ye^{x-y} (y-x)\;dy = x+2$

$\displaystyle\mathbb{E}\;[Y|X]=\int^{y}_0 \frac{6x^2 (y-x)}{y^3} \;dx = \frac{y}{2}$

The domains for the marginal distributions aren't correct: think about $$f_x(x)$$ to have a specific topic. You obtain this distribution by integrating out $$y$$
from the joint distribution, so $$y$$ shouldn't be in the domain of definition. you can see this the following way: for your answer,

$$\int_0^y f_X(x) \, dx \ne 1$$

You seem to have the other quantities correct.

 

[Ted123]

The domains for the marginal distributions aren't correct: think about $$f_x(x)$$ to have a specific topic. You obtain this distribution by integrating out $$y$$
from the joint distribution, so $$y$$ shouldn't be in the domain of definition. you can see this the following way: for your answer,

$$\int_0^y f_X(x) \, dx \ne 1$$

You seem to have the other quantities correct.

So should it be $f_X (x)\;\text{for}\;x\geq 0$ (or $x\in [0,\infty )$ )?

and $f_Y (y)\;\text{for}\;y\geq 0$ (or $y\in [0,\infty )$ )?

Are the conditional density domains correct?

 

[statdad]

Integrate them out - if the domains are correct the integrals should equal 1.

 

[Ted123]

Integrate them out - if the domains are correct the integrals should equal 1.

which they do