# Thread: multivariate conditional probability help

1. ## multivariate conditional probability help

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

so far I have:

for the marginal density functions:

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

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

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

so my marginals are:

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

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

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

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

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

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

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

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

$P \left( Y \leq \frac{1}{2} \bigg{|} X \leq \frac{3}{4} \right) = \frac{32}{63}$ 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.

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

so far I have:

for the marginal density functions:

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

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

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

so my marginals are:

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

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

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

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

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

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

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

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

$P \left( Y \leq \frac{1}{2} \bigg{|} X \leq \frac{3}{4} \right) = \frac{32}{63}$ 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.
$\Pr \left( Y \leq \frac{1}{2} \bigg{|} X \leq \frac{3}{4} \right) = \frac{\Pr\left( Y \leq \frac{1}{2}, X \leq \frac{3}{4} \right)}{\Pr\left( X \leq \frac{3}{4} \right)} = \frac{\int_{y=0}^{y=1/2} \int_{x=0}^{x=y} 6(1 - y) \, dx \, dy}{\int_{0}^{3/4} 3 (x - 1)^2 dx}$

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

By the way:

$f_2(y) = \int_{0}^{y} 6-6y \ dx, \, \, x \leq y \leq 1$ .......

3. By the way:

$\int^{3/4}_{0} 3(x-1)^2 \ dx = (x-1)^3 \bigg{|}^{3/4}_{0} = \left( \frac{3}{4} - 1 \right)^3 - (-1)^3 = \left( -1/4 \right)^3 + 1 = \frac{63}{64}$, not $- \frac{1}{64}$.

4. one more question regarding this problem:

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

5. Originally Posted by lllll
one more question regarding this problem:

Find the conditional density function of $X$ given $Y=y$? I really have no clue on how to do this, so if any body can help that would be greatly appreciated.
By definition: $f(x | y) = \frac{f(x,y)}{f_2(y)}$.

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