# Joint probability distribution

1. Nov 21, 2015

### toothpaste666

1. The problem statement, all variables and given/known data
1. The joint probability density function of X and Y is given by

f(x,y)=(6/7)(x^2+ xy/2) , 0<x<1, 0<y<2.

(a) Find the pdf of X.
(b) Find the cdf of X.
(c) FindP(X<.5).
(d) Determine the conditional pdf of Y given X = x.
3. The attempt at a solution
a) the pdf is what is stated in the probem, that P(x,y) = (6/7)(x^2+ xy/2) , 0<x<1, 0<y<2. and P(x,y) = 0 elsewhere
b)
$F(x,y) = \frac{6}{7} \int_0^Y \int_0^X(x^2 + \frac{xy}{2})dxdy$

$= \frac{6}{7} \int_0^Y (\frac{X^3}{3} + \frac{X^2y}{4})dy$

$= \frac{6}{7} (\frac{YX^3}{3} + \frac{X^2Y^2}{8})$

c)
$\frac{6}{7} \int_0^2 \int_0^.5(x^2 + \frac{xy}{2})dxdy$

$= \frac{6}{7} \int_0^2 (\frac{(.5)^3}{3} + \frac{(.5)^2y}{4})dy$

$= \frac{6}{7} (\frac{(2)(.5)^3}{3} + \frac{(.5)^2(2)^2}{8})$

d) fy(y|x) = f(x,y)/fx(x)

$fx(x) = \frac{6}{7}\int_0^2(x^2 + \frac{xy}{2})dy$

$fx(x) = \frac{6}{7}(2x^2 + x)$

fy(y|x) = f(x,y)/fx(x) =

$\frac{(6/7)(x^2+ xy/2)}{(6/7)(2x^2 + x)}$

$= \frac{x(x+ y/2)}{x(2x + 1)}$

$= \frac{(x+ y/2)}{(2x + 1)}$

Is this correct?

2. Nov 21, 2015

### Ray Vickson

No: (a) and (b) are wrong. You are given $f_{XY}(x,y)$, and have been asked to find $f_X(x)$ and $F_X(x)$, the marginal pdf and cdf of $X$ alone. Anyway, what you wrote in (b) makes no sense: you have numbers $x$ and $y$ on the left in $F(x,y)$ but have random variables $X$ and $Y$ on the right.

I have not checked the details of (c), but the basic calculations look OK. Your (d) looks OK as well.

3. Nov 21, 2015

### toothpaste666

a)
$fx(x,y) = \frac{6}{7} \int_0^2 (x^2 + \frac{xy}{2})dy$

$fx(x,y) = \frac{6}{7} ((2)x^2 + \frac{x(2)^2}{4})$

$fx(x,y) = \frac{6}{7} (2x^2 + x)$

b)
$Fx(x,y) = \frac{6}{7} \int_0^x(2u^2 + u)du$

$Fx(x,y) = \frac{6}{7} (\frac{2x^3}{3} + \frac{x^2}{2})$