# multivariate conditional probability help

• Apr 10th 2008, 09:56 PM
lllll
multivariate conditional probability help
$\displaystyle f(x,y) = \left\{ \begin{array}{rcl} 6(1-y) & \mbox{for} & 0 \leq x \leq y \leq 1 \\ 0 & \mbox{for} & \mbox{other} \end{array}\right.$

so far I have:

for the marginal density functions:

$\displaystyle \begin{array}{l} f_1(x) = \int_{x}^{1} 6-6y \ dy \ 0 \leq x \leq 1 \\ \\ = 6y - 3y^2 \bigg{|}^{1}_{x} \\ \\ = 3 - (6x- 3x^2) \\ \\ = 3(x-1)^2 \\ \end{array}$

$\displaystyle f_1(x) = 3(x-1)^2 \ \ 0 \leq x \leq 1 \\$

$\displaystyle \begin{array}{l} f_2(y) = \int_{1}^{0} 6-6y \ dx \ x \leq y \leq 1 \\ \\ = 6x - 6xy \bigg{|}^{1}_{0} \\ \\ = 6(1-y) \\ \\ f_2(y) = 6(1-y) \ \ x \leq y \leq 1 \\ \end{array}$

so my marginals are:

$\displaystyle f_1(x) = \left\{ \begin{array}{rcl} 3(x-1)^2 & \mbox{for} & 0 \leq x \leq 1 \\ 0 & \mbox{for} & \mbox{other} \end{array}\right.$

$\displaystyle f_2(y) = \left\{ \begin{array}{rcl} 6(1-y) & \mbox{for} & x \leq y \leq 1 \\ 0 & \mbox{for} & \mbox{other} \end{array}\right.$

Find $\displaystyle P \left( Y \leq \frac{1}{2} \bigg{|} X \leq \frac{3}{4} \right)$

for $\displaystyle X \leq \frac{3}{4}$:

$\displaystyle \int^{3/4}_{0} 3(x-1)^2 \ dx = (x-1)^3 \bigg{|}^{3/4}_{0} = - \frac{1}{64}$ which is clearly wrong.

and $\displaystyle Y \leq \frac{1}{2}$:

$\displaystyle \int^{1/2}_{x} 6(1-y) \ dy = 6y-3y^2 \bigg{|}^{1/2}_{x} = \frac{9}{4} - 6x-3x^2$ which doesn't make any sense.

once you get the appropriate values do you just divide Y by X?

$\displaystyle P \left( Y \leq \frac{1}{2} \bigg{|} X \leq \frac{3}{4} \right) = \frac{32}{63}$ according to the solution in the back of the book.

I have no clue what I'm doing wrong and any help clarifying this would be greatly appreciated.
• Apr 10th 2008, 10:36 PM
mr fantastic
Quote:

Originally Posted by lllll
$\displaystyle f(x,y) = \left\{ \begin{array}{rcl} 6(1-y) & \mbox{for} & 0 \leq x \leq y \leq 1 \\ 0 & \mbox{for} & \mbox{other} \end{array}\right.$

so far I have:

for the marginal density functions:

$\displaystyle \begin{array}{l} f_1(x) = \int_{x}^{1} 6-6y \ dy \ 0 \leq x \leq 1 \\ \\ = 6y - 3y^2 \bigg{|}^{1}_{x} \\ \\ = 3 - (6x- 3x^2) \\ \\ = 3(x-1)^2 \\ \end{array}$

$\displaystyle f_1(x) = 3(x-1)^2 \ \ 0 \leq x \leq 1 \\$

$\displaystyle \begin{array}{l} f_2(y) = \int_{1}^{0} 6-6y \ dx \ x \leq y \leq 1 \\ \\ = 6x - 6xy \bigg{|}^{1}_{0} \\ \\ = 6(1-y) \\ \\ f_2(y) = 6(1-y) \ \ x \leq y \leq 1 \\ \end{array}$

so my marginals are:

$\displaystyle f_1(x) = \left\{ \begin{array}{rcl} 3(x-1)^2 & \mbox{for} & 0 \leq x \leq 1 \\ 0 & \mbox{for} & \mbox{other} \end{array}\right.$

$\displaystyle f_2(y) = \left\{ \begin{array}{rcl} 6(1-y) & \mbox{for} & x \leq y \leq 1 \\ 0 & \mbox{for} & \mbox{other} \end{array}\right.$

Find $\displaystyle P \left( Y \leq \frac{1}{2} \bigg{|} X \leq \frac{3}{4} \right)$

for $\displaystyle X \leq \frac{3}{4}$:

$\displaystyle \int^{3/4}_{0} 3(x-1)^2 \ dx = (x-1)^3 \bigg{|}^{3/4}_{0} = - \frac{1}{64}$ which is clearly wrong.

and $\displaystyle Y \leq \frac{1}{2}$:

$\displaystyle \int^{1/2}_{x} 6(1-y) \ dy = 6y-3y^2 \bigg{|}^{1/2}_{x} = \frac{9}{4} - 6x-3x^2$ which doesn't make any sense.

once you get the appropriate values do you just divide Y by X?

$\displaystyle P \left( Y \leq \frac{1}{2} \bigg{|} X \leq \frac{3}{4} \right) = \frac{32}{63}$ according to the solution in the back of the book.

I have no clue what I'm doing wrong and any help clarifying this would be greatly appreciated.

$\displaystyle \Pr \left( Y \leq \frac{1}{2} \bigg{|} X \leq \frac{3}{4} \right) = \frac{\Pr\left( Y \leq \frac{1}{2}, X \leq \frac{3}{4} \right)}{\Pr\left( X \leq \frac{3}{4} \right)} = \frac{\int_{y=0}^{y=1/2} \int_{x=0}^{x=y} 6(1 - y) \, dx \, dy}{\int_{0}^{3/4} 3 (x - 1)^2 dx}$

The integral limits for the double integral are best got by first sketching the region $\displaystyle 0 \leq x \leq y \leq 1$ and then shading the part corresponding to $\displaystyle 0 \leq y \leq \frac{1}{2}$ AND $\displaystyle 0 \leq x \leq \frac{3}{4}$.

By the way:

$\displaystyle f_2(y) = \int_{0}^{y} 6-6y \ dx, \, \, x \leq y \leq 1$ .......
• Apr 10th 2008, 10:45 PM
mr fantastic
By the way:

$\displaystyle \int^{3/4}_{0} 3(x-1)^2 \ dx = (x-1)^3 \bigg{|}^{3/4}_{0} = \left( \frac{3}{4} - 1 \right)^3 - (-1)^3 = \left( -1/4 \right)^3 + 1 = \frac{63}{64}$, not $\displaystyle - \frac{1}{64}$.
• Apr 11th 2008, 07:55 PM
lllll
one more question regarding this problem:

Find the conditional density function of $\displaystyle X$ given $\displaystyle Y=y$? I really have no clue on how to do this, so if any body can help that would be greatly appreciated.
• Apr 11th 2008, 08:04 PM
mr fantastic
Quote:

Originally Posted by lllll
one more question regarding this problem:

Find the conditional density function of $\displaystyle X$ given $\displaystyle Y=y$? I really have no clue on how to do this, so if any body can help that would be greatly appreciated.

By definition: $\displaystyle f(x | y) = \frac{f(x,y)}{f_2(y)}$.

And recall my earlier post: $\displaystyle f_2(y) = \int_{0}^{y} 6-6y \ dx, \, \, x \leq y \leq 1$ .... So $\displaystyle f_2(y) = 6y(1 - y), \, \, x \leq y \leq 1$ .....