My Math Forum

My Math Forum (http://mymathforum.com/math-forums.php)
-   Advanced Statistics (http://mymathforum.com/advanced-statistics/)
-   -   joint probability distribution. (http://mymathforum.com/advanced-statistics/345215-joint-probability-distribution.html)

zollen October 27th, 2018 02:37 PM

joint probability distribution.
 
Let X be a random variable with values 0, 1 and 3, and let Y be a random variable with values -1, 1, 2, with the following joint distribution:


Code:

                Y
          -1    1    2
    ----------------------
    0 | 0.12  0.08  0.10
X  1 | 0.20  0.04  0.25
    3 | 0.08  0.10  0.03



1) Compute the marginal distributions.
        Marg-Dist(X) = { 0.3, 0.49, 0.21 }
        Marg-Dist(Y) = { 0.4, 0.22, 0.38 }

2) Are X and Y independent? Justify your answer.
        No. X and Y are not independent.
        Recall If X and Y were independent, then P(X,Y) = P(X) * P(Y)
        P(X=1) = P(X=1,Y=-1)+P(X=1,Y=1)+P(X=1,Y=2) = 0.20+0.04+0.25 = 0.49
        P(Y=1) = P(Y=1,X=0)+P(Y=1,X=1)+P(Y=1,X=3) = 0.08+0.04+0.10 = 0.22
        P(X=1,Y=1) = 0.04
        P(X=1) * P(Y=1) = 0.49 * 0.22 = 0.1078
        P(X=1,Y=1) <> P(X=1) * P(Y=1)

3) Compute Exp(X) and Exp(Y)
        Exp(X) = Exp(X=0) + Exp(X=1) + Exp(X=3)
                  = (-1)(0.12) + (1)(0.08) + (2)(0.10) +
                            (-1)(0.20) + (1)(0.04) + (2)(0.25) +
                            (-1)(0.08) + (1)(0.10) + (2)(0.03)
                  = 0.58

        Exp(Y) = Exp(Y=-1) + Exp(Y=1) + Exp(Y=2)
                  = (0)(0.12) + (1)(0.20) + (3)(0.08) +
                            (0)(0.08) + (1)(0.04) + (3)(0.10) +
                            (0)(0.10) + (1)(0.25) + (3)(0.03)
                  = 1.12

4) Compute the distribution of X + Y.
        = { 0.12, 0.08, 0.10, 0.20, 0.04, 0.25, 0.08, 0.10, 0.03 }

5) Compute the joint distribution of X, Y.
        See Ans 4.

5) Compute P(X|Y=2) and P(Y|X=1).

        P(X|Y=2) = 0.1+0.25+0.03 = 0.38
        P(Y|X=1) = 0.2+0.04+0.25 = 0.49

Would anyone check my math please?? I am not sure I did it correctly...

romsek October 27th, 2018 07:19 PM

You seem to have $E[X]$ and $E[Y]$ flipped.

let $X=(0,1,3)$

and the marginal of $X$ was computed correctly as $p_X=(0.3, 0.49, 0.21 )$

$E[X] = X.p_X = 1.12$

similarly $E[Y]=0.58$

romsek October 27th, 2018 07:29 PM

For (4) you need to include the values.

5) you've done incorrectly. You should end up with 3 element distributions.

zollen October 28th, 2018 03:08 AM

Quote:

Originally Posted by romsek (Post 601676)
For (4) you need to include the values.

5) you've done incorrectly. You should end up with 3 element distributions.

Would you provide the correct answers so I know what I did wrong?

Thanks!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

zollen October 28th, 2018 07:03 AM

Code:

4) Compute the distribution of X + Y.
Answer:
Dist(X + Y) = { P(X=0,Y=-1), P(X=0,Y=1), P(X=0,Y=2),
                    P(X=1,Y=-1), P(X=1,Y=1), P(X=1,Y=2),
                    P(X=3,Y=-1), P(X=3,Y=-1), P(X=3,Y=2) }
             

5) Compute P(X|Y=2) and P(Y|X=1).
Answer:

P(X|Y=2) = { 0.1, 0.25, 0.03 }
P(Y|X=1) = { 0.2, 0.04, 0.25 }

Better?

romsek October 28th, 2018 09:04 AM

Quote:

Originally Posted by zollen (Post 601690)
Code:

4) Compute the distribution of X + Y.
Answer:
Dist(X + Y) = { P(X=0,Y=-1), P(X=0,Y=1), P(X=0,Y=2),
                    P(X=1,Y=-1), P(X=1,Y=1), P(X=1,Y=2),
                    P(X=3,Y=-1), P(X=3,Y=-1), P(X=3,Y=2) }
             

5) Compute P(X|Y=2) and P(Y|X=1).
Answer:

P(X|Y=2) = { 0.1, 0.25, 0.03 }
P(Y|X=1) = { 0.2, 0.04, 0.25 }

Better?

does either of those "distributions" sum to 1?

zollen October 28th, 2018 09:54 AM

If you were talking about question 5. No.
Let me think about it...

romsek October 28th, 2018 10:26 AM

Quote:

Originally Posted by zollen (Post 601695)
If you were talking about question 5. No.
Let me think about it...

hint - if you have a group of numbers you'd like to cast as a probability distribution how do you usually normalize them?

zollen October 28th, 2018 11:21 AM

0.1+0.25+0.03=0.38
p(x|y=2) = { 0.1/0.38, 0.25/0.38, 0.03/0.38 }

0.2+0.04+0.25=0.49
p(y|x=1) = { 0.2/0.49, 0.04/0.49, 0.25/0.49 }

romsek October 28th, 2018 11:43 AM

https://vignette.wikia.nocookie.net/...20100604214309


All times are GMT -8. The time now is 06:46 AM.

Copyright © 2019 My Math Forum. All rights reserved.