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March 21st, 2015, 12:47 PM   #1
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Math Focus: Calculus
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
Attached Images ex.jpg (8.3 KB, 21 views)

Last edited by szz; March 21st, 2015 at 01:00 PM. March 21st, 2015, 02:00 PM #2 Global Moderator   Joined: May 2007 Posts: 6,804 Thanks: 715 From looking at the picture I got $\displaystyle P(Y\le y|X\le 1) = \frac{y}{2}$. March 21st, 2015, 02:15 PM #3 Senior Member   Joined: Oct 2014 From: EU Posts: 224 Thanks: 26 Math Focus: Calculus 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. March 21st, 2015, 03:13 PM #4 Global Moderator   Joined: May 2007 Posts: 6,804 Thanks: 715 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. Thanks from szz March 21st, 2015, 04:11 PM #5 Senior Member   Joined: Oct 2014 From: EU Posts: 224 Thanks: 26 Math Focus: Calculus 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. March 21st, 2015, 04:35 PM #6 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 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. Thanks from szz March 22nd, 2015, 06:53 AM #7 Senior Member   Joined: Oct 2014 From: EU Posts: 224 Thanks: 26 Math Focus: Calculus 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. March 22nd, 2015, 12:55 PM   #8
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Quote:
 Originally Posted by szz 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. $\displaystyle P(Y \le y|X \le 1)= P(Y \le y)$ March 26th, 2015, 12:27 PM #9 Senior Member   Joined: Oct 2014 From: EU Posts: 224 Thanks: 26 Math Focus: Calculus I finally solved also following your sugggestions. Thank you guy and sorry if I have been stressed. : ( March 27th, 2015, 07:40 AM   #10
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Quote:
 Originally Posted by szz 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". Tags divergence, probability Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post jamesmith134 Calculus 1 January 29th, 2015 11:51 AM black8ghost Calculus 2 January 4th, 2015 10:31 PM SeanRobinson Calculus 0 December 7th, 2014 09:32 AM vysero Calculus 0 December 5th, 2014 09:21 PM FreaKariDunk Real Analysis 28 April 30th, 2012 08:22 PM

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