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March 21st, 2015, 12:47 PM   #1
szz
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Exercise of joint probabilty

Hi all,

I am solving a problem and I would like to be sure that my solution is correct.
So I appreciate any suggestion and correction in case it should be needed.

Given a random vector $\displaystyle (X,Y)$ with a uniform density area shown as follows:



the exercise asks to calculate $\displaystyle P(Y \leq 1.5 \,|\, Y \leq 1 )$.

I defined the area and calculated $\displaystyle f_{XY}(x,y)$

$\displaystyle \begin{aligned}
&a = {xy\over 2} = 2\\
&f_{X,Y}(x,y) = \left\{
\begin{matrix}
{1 \over a} = & \text{if} & x,y \in T\\
0 & & \text{otherwise}
\end{matrix}
\right.
\end{aligned}$


Then I calculated the marginal density functions over the interested area (the left side of the triangle $\displaystyle T$):

$\displaystyle \begin{aligned}
f_{X}(x) & = \int_{-\infty}^{+\infty} f_{x,y}(x,y)\,\text dy\\
& = \int_0^{2x} {1 \over 2}\,\text dy \\
& = \left.\left({y \over 2} \right )\right|_0^{2x}\\
& = x \\
f_Y(y) &= \int_{-\infty}^{+\infty} f_{x,y}(x,y)\,\text dx\\
& = \int_y^{1}{1 \over 2}\,\text dx \\
& = \left.\left({x \over 2} \right )\right|_y^{1}\\
& = {1 - y \over 2}
\end{aligned}$


Then, I calculate the conditional density function. So, if


$\displaystyle \begin{aligned}
& a = {xy \over 2} = 2\\
& f_{XY}(x,y) = {1 \over a} = {1 \over 2} = {1 \over {xy\over 2}} = {2 \over xy}
\end{aligned}$


$\displaystyle \begin{aligned}
& f(y|x) = {f_{XY}(x,y)\over f_X(x)}
& = {2 \over y}
\end{aligned}$


But the integral:

$\displaystyle P(Y \leq 1.5 | X \leq 1) = \int_0^{1.5}{2 \over y}\,\text dy$

does not converges and I think this is impossible in my situation..
So I do as follows:

$\displaystyle
\begin{aligned}
P(Y \leq 1.5 | X \leq 1) & = \int_{0}^{1.5}{(1/2)\over x}\,\text dy \\
& = \int_0^{1.5}{x \over 2}\,\text dy \\
& = \left.\left({xy\over2} \right )\right|_0^{1.5} = 0.75x
\end{aligned}$


Can a probability be a function of another variable ?

Thank you in advance.
szz
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Last edited by szz; March 21st, 2015 at 01:00 PM.
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March 21st, 2015, 02:00 PM   #2
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From looking at the picture I got $\displaystyle P(Y\le y|X\le 1) = \frac{y}{2}$.
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March 21st, 2015, 02:15 PM   #3
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I think I understand your result since I am working on a half of a probability area.

Could you show me where I am wrong ?
Thank you.
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March 21st, 2015, 03:13 PM   #4
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I blundered. $\displaystyle P(Y \le y|X \le 1) = y-\frac{y^2}{4}$

Your expression $\displaystyle f_{XY}(x,y) = \frac{1}{a} = \frac {2}{xy} $ is wrong. To get a, you used the upper limits of x and y, but now you are using them as variables.
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March 21st, 2015, 04:11 PM   #5
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I don't understand nothing.

I am fighting against this topic and it's really frustrating for me. I have a few formulas on my professor notes and a few words explaining (or trying to do do so) the concepts with a very few exercises and if I ask him about my doubts the answers are: "yes, no, maybe". Maybe whar ?!



I did not believe there exist professors as such but now I do.

I am not the kind of person who ask for a solution and wants the exercises done, but this topics are really being painfull.

I am stressed and I feel I want to cry because I spent all the day trying to understand and I realize I don't.
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March 21st, 2015, 04:35 PM   #6
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I wouldn't use integration at all- this is a simple geometry problem.

Since x= 1 is the midpoint of that triangle, the area corresponding to "$\displaystyle x\le 1$ is the left half of that triangle so a right triangle with height 2 and base 1 so area 1. "$\displaystyle y\le 1.5$ given that [mqth]x\le 1[/math]" (At the beginning of your post you have "$\displaystyle y\le 1.5$ given that $\displaystyle y\le 1$". That threw me off for a moment!) is the area of the trapezoid formed by that part of that right triangle below y= 1.5. Since in that right triangle y goes from 0 to 2 while the width of the triangle goes from down from 1 to 0. y= 1.5 is 3/4 of the way from 0 to 2 so the width is 3/4 of the way from 1 to 0: 1/4. That is, we have a trapezoid with bases of length 1 and 1/4 and height 1.5. That has area 1.5(1+ 1/4)/2= 15/16.
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March 22nd, 2015, 06:53 AM   #7
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Country Boy I understand your explanation.

But I would like to solve it using probabilty concepts of marginal density as I think it should be the way I should solve it.
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March 22nd, 2015, 12:55 PM   #8
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Quote:
Originally Posted by szz View Post
Country Boy I understand your explanation.

But I would like to solve it using probabilty concepts of marginal density as I think it should be the way I should solve it.
You need to use the joint density function (= 1/2 in the triangle and 0 otherwise). To get what you want is basic integration - just set up the limits properly. You might want to use the fact that the X distribution is symmetric around X = 1. $\displaystyle P(Y \le y|X \le 1)= P(Y \le y)$
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March 26th, 2015, 12:27 PM   #9
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I finally solved also following your sugggestions.

Thank you guy and sorry if I have been stressed. : (
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March 27th, 2015, 07:40 AM   #10
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Quote:
Originally Posted by szz View Post
Country Boy I understand your explanation.

But I would like to solve it using probabilty concepts of marginal density as I think it should be the way I should solve it.
?? What I suggested was "using probability concepts of marginal density".
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